FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH ...
FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH ...
FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH ...
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<strong>FUNCTIONAL</strong> <strong>ANALYSIS</strong> <strong>LECTURE</strong> <strong>NOTES</strong><br />
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES<br />
CHRISTOPHER HEIL<br />
1. Elementary Properties and Examples<br />
Notation 1.1. Throughout, F will denote either the real line R or the complex plane C.<br />
All vector spaces are assumed to be over the field F.<br />
Definition 1.2. Let X be a vector space over the field F. Then a semi-norm on X is a<br />
function · : X → R such that<br />
(a) x ≥ 0 for all x ∈ X,<br />
(b) αx = |α| x for all x ∈ X and α ∈ F,<br />
(c) Triangle Inequality: x + y ≤ x + y for all x, y ∈ X.<br />
A norm on X is a semi-norm which also satisfies:<br />
(d) x = 0 =⇒ x = 0.<br />
A vector space X together with a norm · is called a normed linear space, a normed vector<br />
space, or simply a normed space.<br />
Definition 1.<strong>3.</strong> Let I be a finite or countable index set (for example, I = {1, . . . , N} if<br />
finite, or I = N or Z if infinite). Let w : I → [0, ∞). Given a sequence of scalars x = (xi)i∈I,<br />
set<br />
xp,w =<br />
⎧<br />
<br />
⎪⎨ |xi|<br />
i∈I<br />
⎪⎩<br />
p w(i) p<br />
1/p ,<br />
sup |xi| w(i),<br />
i∈I<br />
0 < p < ∞,<br />
p = ∞,<br />
where these quantities could be infinite. Then we set<br />
<br />
<br />
x = (xi)i∈I : xp < ∞ .<br />
ℓ p w (I) =<br />
We call ℓ p w(I) a weighted ℓ p space, and often denote it just by ℓ p w (especially if I = N). If<br />
w(i) = 1 for all i, then we simply call this space ℓ p (I) or ℓ p and write · p instead of · p,w.<br />
Date: April 11, 2006.<br />
These notes closely follow and expand on the text by John B. Conway, “A Course in Functional Analysis,”<br />
Second Edition, Springer, 1990.<br />
1
2 CHRISTOPHER HEIL<br />
Exercise 1.4. Show that if 1 ≤ p ≤ ∞ then · p,w defines a semi-norm on ℓp w, and it is a<br />
norm if w(i) > 0 for all i.<br />
In particular, if I = {1, . . . , n} then ℓp w = Fn , and each choice of p and w gives a semi-norm<br />
or norm on Fn .<br />
Hints: The Triangle Inequality on ℓp is often called Minkowski’s Inequality. It is easy to<br />
prove if p = 1 or p = ∞. There are several ways to prove it for other p. One ways is to begin<br />
with<br />
<br />
= |xi + yi| p = <br />
|xi + yi| p−1 |xi + yi|<br />
x + y p p<br />
i∈I<br />
i∈I<br />
≤ <br />
|xi + yi| p−1 |xi| + <br />
|xi + yi| p−1 |yi|.<br />
i∈I<br />
Then apply Hölder’s Inequality to each sum using the exponent p ′ on the first factor and p<br />
for the second (recall that p ′ = p/(p − 1)). Then divide both sides by x + y p−1<br />
p .<br />
Definition 1.5. Let (X, Ω, µ) be a measure space. Given a measurable f : X → [−∞, ∞]<br />
(if F = R) or f : X → C (if F = C), set<br />
⎧<br />
⎪⎨ |f(x)|<br />
fp = X<br />
⎪⎩<br />
p 1/p dµ(x) , 0 < p < ∞,<br />
ess sup |f(x)|,<br />
x∈X<br />
p = ∞,<br />
where these quantities could be infinite. Define<br />
L p <br />
<br />
(X) = f : X → [−∞, ∞] or C : fp < ∞ .<br />
Other notations for L p (X) are L p (µ), L p (X, µ), L p (dµ), L p (X, dµ), etc.<br />
When we write L p (R n ), it will be assumed that µ is Lebesgue measure on R n , unless<br />
specifically stated otherwise. In this case we will write dx instead of dµ(x).<br />
The space ℓ p w (I) is a special case of Lp (X), where X = I and µ is a weighted counting<br />
measure on I.<br />
Exercise 1.6. Show that if 1 ≤ p ≤ ∞ then · p is a semi-norm on L p (X), and it is a norm<br />
if we identify functions that are equal almost everywhere.<br />
The Triangle Inequality on L p is often called Minkowski’s Inequality, and its proof is similar<br />
to the proof of Minkowski’s Inequality for ℓ p .<br />
Exercise 1.7. Show that every subspace of a normed space is itself a normed space (using<br />
the same norm).<br />
Definition 1.8 (Distance). Let · be a norm on X. Then the distance from x to y in X<br />
is d(x, y) = x − y.<br />
i∈I
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 3<br />
Exercise 1.9. Show that d(·, ·) defines a metric on X (see Appendix A).<br />
Since X has a metric and hence has an associated topology, all the standard topological<br />
notions (open/closed sets, convergence, etc.) apply to X. For convenience, we give some<br />
explicit definitions and facts relating to these topics for the setting of normed spaces.<br />
Definition 1.10 (Convergence). Let X be a normed linear space (such as an inner product<br />
space), and let {fn}n∈N be a sequence of elements of X.<br />
(a) We say that {fn}n∈N converges to f ∈ X, and write fn → f, if<br />
i.e.,<br />
lim<br />
n→∞ f − fn = 0,<br />
∀ ε > 0, ∃ N > 0 such that n > N =⇒ f − fn < ε.<br />
(b) We say that {fn}n∈N is Cauchy if<br />
∀ ε > 0, ∃ N > 0 such that m, n > N =⇒ fm − fn < ε.<br />
Exercise 1.11. Let X be a normed linear space. Prove the following.<br />
(a) Reverse Triangle Inequality: f − g ≤ f − g.<br />
(b) Continuity of the norm: fn → f =⇒ fn → f.<br />
(c) Continuity of vector addition: fn → f and gn → g =⇒ fn + gn → f + g.<br />
(d) Continuity of scalar multiplication: fn → f and αn → α =⇒ αnfn → αf.<br />
(e) All convergent sequences are bounded, and the limit of a convergent sequence is unique.<br />
(f) Every Cauchy sequence is bounded.<br />
(g) Every convergent sequence is Cauchy.<br />
Exercise 1.12. Let {fn}n∈N be a sequence of vectors in a normed space X. Show that if<br />
fn − fn+1 < 2 −n for every n, then {fn}n∈N is Cauchy.<br />
Exercise 1.1<strong>3.</strong> Let ℓ p w (I) be the weighted ℓp space defined in Exercise 1.3, where we assume<br />
w(i) > 0 for all i ∈ I. Let {xn}n∈N be a sequence of vectors in ℓ p w(I), and let x be a vector in<br />
ℓ p w(I). Write the components of xn and x as xn = (xn(1), xn(2), . . . ) and x = (x(1), x(2), . . . ).<br />
(a) Prove that if xn → x (i.e., x − xnp,w → 0), then xn converges componentwise to x,<br />
i.e., for each fixed k we have limn→∞ xn(k) = x(k).<br />
(b) Prove that if I is finite then the converse is also true, i.e., componentwise convergence<br />
implies convergence with respect to the norm · p,w.<br />
(c) Prove that if I is infinite then componentwise convergence does not imply convergence<br />
in the norm of ℓ w p (I).
4 CHRISTOPHER HEIL<br />
It is not true in an arbitrary normed space that every Cauchy sequence must converge.<br />
Normed spaces which do have the property that all Cauchy sequences converge are given a<br />
special name.<br />
Definition 1.14 (Banach Space). A normed space X is called a Banach space if it is<br />
complete, i.e., if every Cauchy sequence is convergent. That is,<br />
{fn}n∈N is Cauchy in X =⇒ ∃ f ∈ X such that fn → f.<br />
Exercise 1.15. Show that the weighted ℓp space ℓp w (I) defined in Exercise 1.3 is a Banach<br />
space if w(i) > 0 for all i ∈ I.<br />
Hints: Consider the case I = N. Suppose that {xn}n∈N is a Cauchy sequence in ℓp w<br />
xn is a sequence of scalars. Write out the components of xn as<br />
Prove that for a fixed component k we have<br />
xn = (xn(1), xn(2), . . . ).<br />
|xm(k) − xn(k)| ≤ C xm − xnp,w,<br />
. Each<br />
where C is a fixed constant (determined by the weight and by k but independent of m and<br />
n). Conclude that with k fixed, {xn(k)}n∈N is a Cauchy sequence of scalars, and hence converges.<br />
Define x(k) = limn→∞ xn(k) and set x = (x(1), x(2), . . . ). Then we have constructed<br />
a candidate limit for the sequence {xn}n∈N. However, so far we only have that each individual<br />
component of xn converges to the corresponding component of x, i.e., xn converges<br />
componentwise to x. This is not enough: to complete the proof you must show that xn → x<br />
in the norm of ℓ p w . Use the fact that {xn}n∈N is Cauchy together with the componentwise<br />
convergence to show that x − xn ℓ p w → 0 as n → ∞.<br />
Compare this proof to the proof of Theorem 2.18 given below.<br />
Exercise 1.16. Show that the space Lp (X) defined in Example 1.5 is a Banach space if<br />
we identify functions that are equal almost everywhere. This is called the Riesz–Fisher<br />
Theorem.<br />
Hint: The argument is similar in spirit but more subtle than the one used to prove that<br />
ℓp w (I) is a Banach space. First find a candidate limit and then show that the sequence<br />
converges in norm to this limit.<br />
The next two exercises will be useful to us later.<br />
Exercise 1.17. Let X be a normed linear space. If {fn}n∈N is a Cauchy sequence in X and<br />
there exists a subsequence {fnk }k∈N that converges to f ∈ X, then fn → f.<br />
Solution<br />
Choose any ε > 0. Since {fn}n∈N is Cauchy, there is an N such that fm − fn < ε for m,<br />
2<br />
ε<br />
n > N. Also, there is a k such that nk > N and f − fnk < . Hence for n > N we have<br />
2<br />
f − fn ≤ f − fnk + fnk − fn < ε ε<br />
+ = ε.<br />
2 2
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 5<br />
Thus fn → f. <br />
Exercise 1.18. Let {fn}n∈N be a Cauchy sequence in a normed space X. Show that there<br />
exists a subsequence {fnk }k∈N such that<br />
∀ k ∈ N, fnk+1 − fnk < 2−k .<br />
Solution<br />
Since {fn}n∈N is Cauchy, we can find an n1 such that<br />
Then we can find an n2 > n1 such that<br />
m, n ≥ n1 =⇒ fm − fn < 2 −1 .<br />
m, n ≥ n2 =⇒ fm − fn < 2 −2 .<br />
Continuing in this way, we inductively construct n1 < n2 < · · · such that for each k,<br />
m, n ≥ nk =⇒ fm − fn < 2 −k .<br />
In particular, since nk+1 > nk, we have fnk+1 − fnk < 2−k . <br />
Definition 1.19 (Convergent Series). Let {fn}n∈N be a sequence of elements of a normed<br />
linear space X. Then the series ∞ n=1 fn converges and equals f ∈ X if the partial sums<br />
sN = N n=1 fn converge to f, i.e., if<br />
f − sN =<br />
N<br />
<br />
f −<br />
<br />
<br />
→ 0 as N → ∞.<br />
n=1<br />
fn<br />
Definition 1.20 (Absolutely Convergent Series). Let X be a normed space and let {fn}n∈N<br />
be a sequence of elements of X. If<br />
∞<br />
fn < ∞,<br />
then we say that the series <br />
n fn is absolutely convergent in X.<br />
n=1<br />
Note that the definition of absolute convergence does not by itself tell us that the series<br />
<br />
n fn actually converges. That is always true if X is a Banach space, but need not be true<br />
if X is not complete.<br />
Exercise 1.21. Let X be a Banach space and let {fn}n∈N be a sequence of elements of X.<br />
Prove that if <br />
n fn < ∞ then the series <br />
n fn does converge in X.<br />
Hint: You must show that the sequence of partial sums {sN}N∈N converges. Since X is a<br />
Banach space, you just have to show that this sequence is Cauchy.<br />
The converse of this exercise is also true, and is often a useful method for proving that a<br />
given normed space is a Banach space.
6 CHRISTOPHER HEIL<br />
Proposition 1.22. Let X be a normed space. Prove that X is a Banach space if and only<br />
if every absolutely convergent series in X converges in X.<br />
Proof. ⇒. This is Exercise 1.21.<br />
⇐. Suppose that every absolutely convergent series is convergent. Let {fn}n∈N be a<br />
Cauchy sequence in X. By Exercise 1.18, there exists a subsequence {fnk }k∈N such that<br />
fnk+1 − fnk < 2−k for every k. The series <br />
(fnk+1 k − fnk ) is absolutely convergent,<br />
because<br />
∞<br />
− fnk ≤<br />
∞<br />
2 −k = 1 < ∞.<br />
fnk+1<br />
k=1<br />
k=1<br />
By hypothesis, we conclude that <br />
(fnk+1 k − fnk ) converges in X, say to f. In terms of the<br />
partial sums, this says that<br />
fnk+1 − fn1 =<br />
k<br />
j=1<br />
(fnj+1<br />
− fnj ) → f as k → ∞,<br />
or fnk → g = f +fn1 as k → ∞. Thus {fn}n∈N is a Cauchy sequence which has a subsequence<br />
that converges to g. It therefore follows from Exercise 1.17 that fn → g. Therefore X is<br />
complete. <br />
Example 1.23 (The Harmonic Series). To illustrate the difference between convergence and<br />
absolute convergence, consider the one-dimensional case, i.e., X = F, the field of scalars. Let<br />
xn = 1 <br />
. Then n n xn = 1<br />
n n is the harmonic series, and the partial sums sN = N n=1<br />
this series are unbounded. Thus the harmonic series does not converge. Since sN → ∞, we<br />
usually say that <br />
<br />
1<br />
1<br />
n diverges to infinity, and write n n = ∞. n<br />
On the other hand, consider the alternating series 1<br />
n (−1)n . Since the terms alternate<br />
n<br />
signs and since 1 → 0, it follows that this series does converge to a finite scalar (in fact, it<br />
n<br />
converges to − ln 2). However, it does not converge absolutely.<br />
Definition 1.24 (Unconditionally Convergent Series). Let X be a Banach space and let<br />
{fn}n∈N be a sequence of elements of X. The series f = ∞ n=1 fn is said to converge unconditionally<br />
if every rearrangement of the series converges. That is, f = ∞ n=1 fn converges<br />
unconditionally if for each bijection σ : N → N the series<br />
∞<br />
converges.<br />
n=1<br />
fσ(n)<br />
Remark 1.25. It is not obvious, but it can be shown that if ∞ n=1 fσ(n) is unconditionally<br />
convergent, then every rearrangement of the series must converge to the same sum, i.e., there<br />
is a single f such that f = ∞ n=1 fσ(n) for every permutation σ.<br />
1<br />
n of
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 7<br />
Exercise 1.26. Let X be a Banach space. Prove that if a series f = ∞<br />
n=1 fn converges<br />
absolutely, then it converges unconditionally.<br />
Remark 1.27. In finite dimensions the converse to Exercise 1.26 is true, i.e., if X is finitedimensional<br />
and a series f = ∞ n=1 fn converges unconditionally, then it converges absolutely.<br />
However, this fails in infinite dimensions.<br />
Example 1.28 (The Harmonic Series Revisited). To illustrate the importance of unconditional<br />
convergence, again consider X = F and the alternating series 1<br />
n (−1)n . We know<br />
n<br />
that this series converges, but does not converge absolutely.<br />
Now consider what happens if we change the order of summation. Let pn = 1<br />
2n and<br />
qn = 1<br />
2n+1 , i.e., the pn are the positive terms from the alternative series and the qn are the<br />
absolute values of the negative terms. Each series <br />
n qn diverges. Hence there<br />
must exist an m1 > 0 such that<br />
p1 + · · · + pm1 > 1.<br />
Then, there must exist an m2 > m1 such that<br />
Continuing in this way, we see that<br />
n pn and <br />
p1 + · · · + pm1 − q1 + pm1+1 + · · · + pm2 > 2.<br />
p1 + · · · + pm1 − q1 + pm1+1 + · · · + pm2 − q2 + · · ·<br />
is a rearrangement of 1<br />
n (−1)n which diverges to +∞.<br />
n<br />
In the same way, we can construct a rearrangement which diverges to −∞, which converges<br />
to any given real number r, or which simply oscillates without ever converging. Moreover,<br />
the same can be done for any series of real scalars which converges conditionally.<br />
The following is an equivalent formulation of unconditional convergence.<br />
Proposition 1.29. Let X be a Banach space and let {fn}n∈N be a sequence of elements<br />
of X. Then the series f = ∞ n=1 fn converges unconditionally if and only if it converges with<br />
respect to the net of finite subsets of N, i.e., if<br />
<br />
<br />
∀ ε > 0, ∃ finite F0 ⊆ N such that ∀ finite F ⊇ F0, f − <br />
<br />
< ε.<br />
Definition 1.30 (Topology). Let X be a normed linear space.<br />
(a) The open ball in X centered at x ∈ X with radius r > 0 is<br />
(b) A subset U ⊆ X is open if<br />
Br(x) = B(x, r) = {y ∈ X : x − y < r}.<br />
∀ x ∈ U, ∃ r > 0 such that Br(x) ⊆ U.<br />
(c) A subset F ⊆ X is closed if X \ F is open.<br />
n∈F<br />
fn
8 CHRISTOPHER HEIL<br />
Definition 1.31 (Limit Points, Closure, Density). Let X be a normed linear space and let<br />
A ⊆ X.<br />
(a) A point f ∈ A is called a limit point of A if there exist fn ∈ A with fn = f such that<br />
fn → f.<br />
(b) The closure of A is the smallest closed set Ā such that A ⊆ Ā. Specifically,<br />
Ā = {F ⊆ X : F is closed and F ⊇ A}.<br />
(c) We say that A is dense in X if Ā = X.<br />
Exercise 1.32. (a) The closure of A equals the union of A and all limit points of A:<br />
Ā = A ∪ {x ∈ X : x is a limit point of A} = {z ∈ X : ∃ yn ∈ A such that yn → z}.<br />
(b) If X is a normed linear space, then the closure of an open ball Br(x) is the closed ball<br />
Br(x) = {y ∈ X : x − y ≤ r}.<br />
(c) Prove that A is dense if and only if<br />
∀ x ∈ X, ∀ ε > 0, ∃ y ∈ A such that x − y < ε.<br />
Example 1.3<strong>3.</strong> The set of rationals Q is dense in the real line R.<br />
Exercise 1.34. Let X be a normed linear space and let F ⊆ X. Then<br />
F is closed ⇐⇒ F contains all its limit points.<br />
Solution<br />
⇒. Suppose that F is closed but that there exists a limit point f that does not belong to<br />
F . By definition, there must exist fn ∈ F such that fn → f. However, f ∈ X \ F , which is<br />
open, so there exists some r > 0 such that B(f, r) ⊆ X \ F . Yet there must exist some fn<br />
with f − fn < r, so this fn will belong to X \ F , which is a contradiction.<br />
⇐. Exercise. <br />
Remark 1.35. Some authors make the restriction that, when dealing with normed spaces,<br />
the terminology “subspace” is used only for closed subspaces. Other authors use the terminology<br />
“linear manifold” to denote a subspace that need not be closed. To avoid ambiguity,<br />
we will use the following terminology.<br />
Definition 1.36. (a) A subset Y of a vector space X is a subspace of X if it is closed under<br />
both vector addition and scalar multiplication, i.e., if for all u, v ∈ Y and α, β ∈ F we have<br />
αu + βv ∈ Y .<br />
(b) A subset Y of a normed linear space X is a closed subspace of X if it is a subspace<br />
and it is closed with respect to the norm of X.
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 9<br />
Exercise 1.37. In finite dimensions, all subspaces are closed sets (this will be proved in<br />
Proposition <strong>3.</strong>5). This exercise demonstrates that, in infinite dimensions, subspaces need<br />
not be closed sets.<br />
(a) Fix 1 ≤ p ≤ ∞. Prove that<br />
c00 = {x = (x1, . . . , xN, 0, 0, . . . ) : N > 0, x1, . . . , xN ∈ F}<br />
is a subspace of ℓ p (N) that is not closed (with respect to the ℓ p -norm). Prove that c00 is<br />
dense in ℓ p (N) if p < ∞, but that it is not dense in ℓ ∞ (N).<br />
(b) Define<br />
c0 = {x = (xk) ∞ k=1 : lim<br />
k→∞ xk = 0}.<br />
Prove c0 is a closed subspace of ℓ ∞ (N) (in ℓ ∞ -norm). Prove that c0 is the closure of c00<br />
(under the ℓ ∞ -norm).<br />
(c) Fix 1 ≤ p ≤ ∞. Prove that Cc(R n ), the space of continuous, compactly supported<br />
functions on R n , is a subspace of L p (R n ) that is not closed. Prove that Cc(R n ) is dense in<br />
L p (R n ) if p < ∞, but not if p = ∞.<br />
Hints: For a continuous function we have f∞ = sup |f(x)|. Hence, if {fn}n∈N is a<br />
sequence of continuous functions in L ∞ (R n ) that converges in L ∞ -norm, then it converges<br />
uniformly. From undergraduate real analysis, we know that the limit of a uniformly convergent<br />
sequence of continuous functions is continuous.<br />
(d) Let C0(R n ) be the set of all continuous functions f : R n → F such that<br />
lim f(x) = 0. (1.1)<br />
|x|→∞<br />
More precisely, (1.1) means that for every ε > 0 there exists a compact set K such that<br />
|f(x)| < ε for all x /∈ K. Prove that C0(R n ) is a closed subspace of L ∞ (R n ) (closed in<br />
L ∞ -norm). Prove that C0(R n ) is the closure of Cc(R n ) (under the L ∞ -norm).<br />
(e) Let Cb(R n ) be the set of all bounded, continuous functions f : R n → F. Prove that<br />
Cb(R n ) is a closed subspace of L ∞ (R n ).<br />
(f) Fix 1 ≤ p ≤ ∞, and let E be a (Lebesgue) measurable subset of R n . Let M = {f ∈<br />
L p (R n ) : supp(f) ⊆ E}. Prove that M is a closed subspace of L p (R n ).<br />
Remark 1.38. (a) We took the domain in the preceding exercise to be R n just for convenience;<br />
the definitions of the spaces Cc, C0, Cb can be extended to domains that are more<br />
general topological spaces.<br />
(b) Beware that some authors use the notation C0 for the space that we are calling Cc!<br />
Exercise 1.39. Let X be a Banach space and let M be a subspace of X. Then M is itself<br />
a Banach space (using the norm from X) if and only if M is closed.
10 CHRISTOPHER HEIL<br />
Solution<br />
⇒. Suppose that M is a Banach space. Let f be any limit point of M, i.e., suppose<br />
fn ∈ M and fn → f. Then {fn} is a convergent sequence in X, and hence is Cauchy in<br />
X. Since each fn belongs to M, it is also Cauchy in M. Since M is a Banach space, {fn}<br />
must therefore converge in M, i.e., fn → g for some g ∈ M. However, limits are unique, so<br />
f = g ∈ M. Therefore M contains all its limit points and hence is closed.<br />
⇐. Exercise. <br />
Notation 1.40 (Notation for Closed Subspaces). Since we will often deal with closed subspaces<br />
of a Banach space, we declare that the notation<br />
M ≤ X<br />
means that M is a closed subspace of the Banach space X.<br />
Exercise 1.41. Find examples of normed spaces that are not Banach spaces.<br />
Hint: Look for examples of normed spaces Y which are subspaces of a larger Banach<br />
space X.<br />
Remark 1.42. Is every normed space Y a subspace of a larger Banach space? The answer<br />
is yes, given a normed space Y it is always possible to construct a Banach space X ⊇ Y such<br />
that the norm on X, when restricted to Y , is the same as the norm on Y , and Y is dense in<br />
X with respect to that norm. This space is called the completion of Y .<br />
Definition 1.4<strong>3.</strong> Suppose that X is a normed linear space with respect to a norm · a<br />
and also with respect to another norm · b. Then we say that these norms are equivalent<br />
if there exist constants C1, C2 > 0 such that<br />
∀ f ∈ X, C1 fa ≤ fb ≤ C2 fa. (1.2)<br />
Observe that equation (1.2) can be rearranged to read 1<br />
C2 fb ≤ fa ≤ 1<br />
C1 fb.<br />
Exercise 1.44. Let X be a vector space. If · a and · b are two norms on X, define<br />
· a ∼ · b if · a and · b are equivalent. Prove that ∼ is an equivalence relation on<br />
the class of norms on X.<br />
The next result will show that equivalent norms define the same topology and the same<br />
convergence criterion.<br />
Proposition 1.45. Let ·a and ·b be two norms on a vector space X. Then the following<br />
statements are equivalent.<br />
(a) · a and · b are equivalent norms.<br />
(b) · a and · b define the same topologies on X. That is, if U ⊆ X, then U is open<br />
with respect to · a if and only if it is open with respect to · b.
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 11<br />
(c) ·a and ·b define the same convergence criterion. That is, if {fn}n∈N is a sequence<br />
in X and f ∈ X, then<br />
lim<br />
n→∞ f − fna = 0 ⇐⇒ lim f − fnb = 0.<br />
n→∞<br />
Proof. (b) ⇒ (a). Assume that statement (b) holds. Let Ba r (f) and Bb r(f) denote the open<br />
balls of radius r centered at f ∈ X with respect to · a and · b. Since Ba 1 (0) is open<br />
with respect to · a, the hypothesis that statement (b) holds implies that Ba 1 (0) is open<br />
with respect to · b. Therefore, since 0 ∈ Ba 1 (0), there must exist some r > 0 such that<br />
Bb r(0) ⊂ Ba 1(0).<br />
Now choose any f ∈ X and any ε > 0. Then<br />
(r − ε) f<br />
∈ B b r (0) ⊆ Ba 1 (0),<br />
fb<br />
so<br />
<br />
(r<br />
− ε) f<br />
<br />
<br />
< 1.<br />
fb a<br />
Rearranging, this implies (r − ε) fa < fb. Since this is true for every ε, we conclude<br />
that r fa ≤ fb.<br />
A symmetric argument, interchanging the roles of the two norms, shows that there exists<br />
an s > such that fb ≤ s fa for every f ∈ X. Hence the two norms are equivalent<br />
The remaining implications are exercises.<br />
Hints on (c) ⇒ (a): Suppose that statement (c) holds. Show that this implies that the<br />
function ν : X → R given by ν(f) = fb is continuous with respect to the norm ·a. Since<br />
(−1, 1) is an open subset of R, it follows that ν −1 (−1, 1) is an open subset of X (with respect<br />
to · a). Since 0 ∈ ν −1 (−1, 1), there must exist an r > 0 such that B b r(0) ⊂ ν −1 (−1, 1).<br />
But ν −1 (−1, 1) = B a 1(0), so the remainder of the proof proceeds exactly like the proof of (b)<br />
⇒ (a). <br />
2. Linear Operators on Normed Spaces<br />
Definition 2.1 (Notation for Operators). Let X, Y be vector spaces. Let T : X → Y be a<br />
function (= operator = transformation) mapping X into Y . We write either T (f) or T f to<br />
denote the image of an element f ∈ X.<br />
(a) T is linear if T (αf + βg) = αT (f) + βT (g) for every f, g ∈ X and α, β ∈ F.<br />
(b) T is injective if T (f) = T (g) implies f = g.<br />
(c) The kernel or nullspace of T is ker(T ) = {f ∈ X : T (f) = 0}.<br />
(c) The range of T is range(T ) = {T (f) : f ∈ X}.<br />
(d) The rank of T is the dimension of its range, i.e., rank(T ) = dim(range(T )). In<br />
particular, T is finite-rank if range(T ) is finite-dimensional.<br />
(d) T is surjective if range(T ) = Y .
12 CHRISTOPHER HEIL<br />
(e) T is a bijection if it is both injective and surjective.<br />
(f) We use the notation I or IX to denote the identity map of a space X onto itself.<br />
Definition 2.2 (Continuous and Bounded Operators). Let X, Y be normed linear spaces,<br />
and let L: X → Y be a linear operator.<br />
(a) L is continuous at a point f ∈ X if fn → f in X implies Lfn → Lf in Y .<br />
(b) L is continuous if it is continuous at every point, i.e., if fn → f in X implies Lfn → Lf<br />
in Y for every f.<br />
(c) L is bounded if there exists a finite K ≥ 0 such that<br />
∀ f ∈ X, Lf ≤ K f.<br />
Note that Lf is the norm of Lf in Y , while f is the norm of f in X.<br />
(d) The operator norm of L is<br />
L = sup Lf.<br />
f=1<br />
(e) We let B(X, Y ) denote the set of all bounded linear operators mapping X into Y ,<br />
i.e.,<br />
B(X, Y ) = {L: X → Y : L is bounded and linear}.<br />
If X = Y then we write B(X) = B(X, X).<br />
(f) If Y = F then we say that L is a functional. The set of all bounded linear functionals<br />
on X is the dual space of X, and is denoted<br />
X ′ = B(X, F) = {L: X → F : L is bounded and linear}.<br />
Another common notation for the dual space is X ∗ .<br />
Note that since the norm on F is just absolutely value, the operator norm of a linear<br />
|Lf|.<br />
functional L ∈ X ′ = B(X, F) is L = sup<br />
f=1<br />
Exercise 2.<strong>3.</strong> Show that if T : X → Y is linear and continuous, then ker(T ) is a closed<br />
subspace of X and that range(T ) is a subspace of Y . Must range(T ) be a closed subspace?<br />
Exercise 2.4. Let X, Y be normed linear spaces. Let L: X → Y be a linear operator.<br />
(a) L is injective if and only if ker L = {0}.<br />
(b) If L is a bijection then L −1 : Y → X is also a linear bijection.<br />
(c) L is bounded if and only if L < ∞.<br />
(d) If L is bounded then Lf ≤ L f for every f ∈ X (note that three different<br />
meanings of the symbol · appear in this statement!).
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 13<br />
(e) If L is bounded then L is the smallest value of K such that Lf ≤ Kf holds<br />
for all f ∈ X.<br />
Lf<br />
(f) L = sup Lf = sup<br />
f≤1<br />
f=0 f .<br />
Exercise 2.5. Show that B(X, Y ) is a subspace of the vector space V consisting of ALL<br />
functions A: X → Y . Moreover, show that the operator norm is a norm on the space<br />
B(X, Y ), i.e.,<br />
(a) 0 ≤ L < ∞ for all L ∈ B(X, Y ),<br />
(b) L = 0 if and only if L = 0 (the zero operator that sends every element of X to the<br />
zero vector in Y ),<br />
(c) αL = |α| L for every L ∈ B(X, Y ) and every α ∈ F,<br />
(d) L + K ≤ L + K for every L, K ∈ B(X, Y ).<br />
The preceding exercise shows that B(X, Y ) is a normed linear space, and we will show<br />
in Theorem 2.18 that it is a Banach space if Y is a Banach space. However, in addition to<br />
vector addition and scalar multiplication operations, there is a third operation that we can<br />
perform with functions: composition.<br />
Exercise 2.6. Prove that the operator norm is submultiplicative, i.e., prove if A ∈ B(X, Y )<br />
and B ∈ B(Y, Z), then BA ∈ B(X, Z) and<br />
BA ≤ B A. (2.1)<br />
In particular, when X = Y = Z, we see that B(X) is closed under compositions. The<br />
space B(X) is an example of an algebra.<br />
Exercise 2.7. Let F n be n-dimensional Euclidean space over F, under the Euclidean norm,<br />
and let Y be any normed linear space. Prove that if L: F n → Y is linear, then L is bounded.<br />
Hint: If x = (x1, . . . , xn) ∈ F n then x = x1e1+· · ·+xnen where {e1, . . . , en} is the standard<br />
basis for F n . Use the Triangle and the Cauchy-Schwarz Inequalities.<br />
Solution<br />
Given x ∈ Fn , we have<br />
<br />
<br />
n<br />
<br />
<br />
Lx = <br />
L(xkek) <br />
≤<br />
k=1<br />
n<br />
|xk| Lek ≤<br />
k=1<br />
n <br />
k=1<br />
|xk| 2<br />
1/2 <br />
n<br />
k=1<br />
Lek 2<br />
1/2 = C x,<br />
where C = n<br />
k=1 Lek 2 1/2 . Hence L is bounded. <br />
Remark 2.8. We will prove later that if X is any finite-dimensional vector space and · is<br />
any norm on X, then any linear function L: X → Y is bounded. To do this we will use the<br />
fact (that we will prove later) that all norms on a finite-dimensional space are equivalent.
14 CHRISTOPHER HEIL<br />
The next lemma is a standard fact about continuous functions. L −1 (U) denotes the inverse<br />
image of U ⊆ Y , i.e., L −1 (U) = {f ∈ X : Lf ∈ U}.<br />
Exercise 2.9. Let X, Y be normed linear spaces. Let L: X → Y be linear. Then<br />
<br />
L is continuous ⇐⇒ U open in Y =⇒ L −1 <br />
(U) open in X .<br />
Solution<br />
⇒. Suppose that L is continuous and that U is an open subset of Y . We will show that<br />
X \ L −1 (U) is closed by showing that it contains all its limit points.<br />
Suppose that f is a limit point of X \L −1 (U). Then there exist fn ∈ X \L −1 (U) such that<br />
fn → f. Since L is continuous, this implies Lfn → Lf. However, fn /∈ L −1 (U), so Lfn /∈ U,<br />
i.e., Lfn ∈ Y \ U, which is a closed set. Therefore Lf ∈ Y \ U, and hence f ∈ X \ L −1 (U).<br />
Thus X \ L −1 (U) is closed, so L −1 (U) is open.<br />
⇐. Exercise. <br />
Theorem 2.10 (Equivalence of Bounded and Continuous Linear Operators). Let X, Y be<br />
normed linear spaces, and let L: X → Y be a linear mapping. Then the following statements<br />
are equivalent.<br />
(a) L is continuous at some f ∈ X.<br />
(b) L is continuous at f = 0.<br />
(c) L is continuous.<br />
(d) L is bounded.<br />
Proof. (c) ⇒ (d). Suppose that L is continuous but unbounded. Then L = ∞, so<br />
there must exist fn ∈ X with fn = 1 such that Lfn ≥ n. Set gn = fn/n. Then<br />
gn − 0 = gn = fn/n → 0, so gn → 0. Since L is continuous and linear, this implies<br />
Lgn → L0 = 0, and therefore Lgn → 0 = 0. But<br />
Lgn = 1<br />
n Lfn ≥ 1<br />
· n = 1<br />
n<br />
for all n, which is a contradiction. Hence L must be bounded.<br />
(d) ⇒ (c). Suppose that L is bounded, so L < ∞. Suppose that f ∈ X and that<br />
fn → f. Then fn − f → 0, so<br />
i.e., Lfn → Lf. Thus L is continuous.<br />
Lfn − Lf = L(fn − f) ≤ L fn − f → 0,<br />
The remaining implications are exercises. <br />
Definition 2.11 (Isometries and Isometric Isomorphisms). Let X, Y be normed linear<br />
spaces and let L: X → Y be linear.
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 15<br />
(a) If Lf = f for all f ∈ X then L is called an isometry or is said to be normpreserving.<br />
(b) An isometry L: X → Y that is a bijection is called an isometric isomorphism. In<br />
this case we say that X and Y are isometrically isomorphic.<br />
Exercise 2.12. (a) Suppose that L: X → Y is an isometry. Prove that L is injective and<br />
find L.<br />
(b) Find an example of an isometry that is not surjective. Contrast this with the fact that<br />
if A: C n → C n is linear, then A is injective if and only if it is surjective.<br />
(c) Prove that if L: X → Y is an isometric isomorphism, then L −1 : Y → X is also an<br />
isometric isomorphism.<br />
Exercise 2.13 (Unilateral Shift Operators). Fix 1 ≤ p ≤ ∞.<br />
(a) Define L: ℓ p (N) → ℓ p (N) by L(x) = (x2, x3, . . . ) for x = (x1, x2, . . . ) ∈ ℓ p (N). Prove<br />
that this left-shift operator is bounded, linear, surjective, not injective, and is not an isometry.<br />
Find L.<br />
(b) Define R: ℓ p (N) → ℓ p (N) by R(x) = (0, x1, x2, x3, . . . ) for x = (x1, x2, . . . ) ∈ ℓ p (N).<br />
Prove that this right-shift operator is bounded, linear, injective, not surjective, and is an<br />
isometry. Find R.<br />
(c) Compute LR and RL. Contrast this computation with the fact that in finite dimensions,<br />
if A, B : C n → C n are linear maps (hence correspond to multiplication by n × n<br />
matrices), then AB = I implies BA = I and conversely.<br />
Definition 2.14 (Topological Isomorphisms). Let X, Y be normed linear spaces. If L: X →<br />
Y is a linear bijection such that both L and L −1 are bounded, then L is called a topological<br />
isomorphism. In this case we say that X and Y are topologically isomorphic.<br />
Remark 2.15. We will see later that if X and Y are Banach spaces and L: X → Y<br />
is a bounded bijection, then L −1 is automatically bounded and hence L is a topological<br />
isomorphism. Thus, when X and Y are Banach spaces, every continuous invertible map is a<br />
topological isomorphism. Sometimes the abbreviation isomorphism or invertible map is used<br />
to mean a topological isomorphism, but it should be noted that these terms are ambiguous.<br />
Exercise 2.16. Prove that if L: X → Y is a topological isomorphism, then<br />
1<br />
∀ f ∈ X,<br />
L−1 f ≤ Lf ≤ L f.<br />
<br />
Exercise 2.17. Let X be a Banach space and Y a normed linear space. Suppose that<br />
L: X → Y is bounded and linear. Prove that if there exists c > 0 such that Lf ≥ cf<br />
for all f ∈ X, then L is injective and range(L) is closed.
16 CHRISTOPHER HEIL<br />
The next theorem shows that B(X, Y ) is a Banach space whenever Y is a Banach space.<br />
Theorem 2.18. If X is a normed space and Y is a Banach space, then B(X, Y ) is a Banach<br />
space.<br />
Proof. Assume that {An}n∈N is a sequence of operators An ∈ B(X, Y ) that is Cauchy in<br />
operator norm. For any given f ∈ X, we have<br />
Amf − Anf ≤ Am − An f,<br />
so we conclude that {Anf}n∈N is a Cauchy sequence of vectors in Y . Since Y is complete,<br />
this sequence must converge, say Afn → g ∈ Y . Define Af = g. This gives us a candidate<br />
limit operator A.<br />
Exercise: Show that A defined in this way is a linear operator.<br />
To show that A is bounded, first recall that all Cauchy sequences are bounded. Hence we<br />
must have C = sup An < ∞. If f ∈ X, then since Anf → Af we have<br />
Af = lim Anf ≤ sup<br />
n→∞<br />
n∈N<br />
Anf ≤ sup An f = C f.<br />
n∈N<br />
Hence A is bounded, and A ≤ C.<br />
Finally, we must show that An → A in operator norm. Fix any ε > 0. Then there exists<br />
an N such that<br />
m, n > N =⇒ Am − An < ε<br />
2 .<br />
Choose any f ∈ X with f = 1. Then since Amf → Af, there exists an m > N such that<br />
Af − Amf < ε<br />
2 .<br />
Hence for any n > N we have<br />
Af −Anf ≤ Af −Amf+Amf −Anf ≤ Af −Amf+Am−An f < ε ε<br />
+ = ε.<br />
Taking the supremum over all unit vectors, we conclude that A − An ≤ ε for all n > N.<br />
Thus An → A. <br />
Corollary 2.19. If X is a normed space, then its dual space X ′ = B(X, F) is a Banach<br />
space.<br />
The next exercise deals with the problem of extending an operator defined only a dense<br />
subspace to the entire space.<br />
Exercise 2.20 (Extension of Bounded Operators). Let Y be a dense subspace of a normed<br />
space X, and let Z be a Banach space.<br />
(a) Suppose that L: Y → Z is a bounded linear operator. Show that there exists a<br />
unique bounded linear operator ˜ L: X → Z whose restriction to Y is L. Prove that<br />
˜ L = L.<br />
(b) Prove that if L: Y → range(L) is a topological isomorphism, then ˜ L: X → range(L)<br />
is also.<br />
2<br />
2
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 17<br />
(c) Prove that if L: Y → range(L) is an isometry, ˜ L: X → range(L) is also.<br />
Solution<br />
(a) Fix any f ∈ X. Since Y is dense in X, there exist gn ∈ Y such that gn → f. Since<br />
L is bounded, we have Lgm − Lgn ≤ L gm − gn. But {gn}n∈N is Cauchy in X, so this<br />
implies that {Lgn}n∈N is Cauchy in Z. Since Z is a Banach space, we conclude that there<br />
exists an h ∈ Z such that Lgn → h. Define ˜ Lf = h.<br />
To see that ˜ L is well-defined, suppose that we also had g ′ n → f for some g ′ n ∈ Y . Then<br />
Lg ′ n−Lgn ≤ L g ′ n−gn → 0. Since Lgn → h, it follows that Lg ′ n = Lgn+(Lg ′ n−Lgn) →<br />
h + 0 = h. Thus ˜ L is well-defined.<br />
To see that ˜ L is linear, suppose that f, g ∈ X are given and c ∈ F. Then there exist fn,<br />
gn ∈ Y such that fn → f and gn → g. Since cfn + gn → cf + g and<br />
L(cfn + gn) = cLfn + Lgn → c ˜ Lf + ˜ Lg,<br />
by definition we have that ˜ L(cf + g) = c ˜ Lf + ˜ Lg.<br />
To see that ˜ L is an extension of L, suppose that g ∈ Y is fixed. If we set gn = g, then<br />
gn → g and Lgn → Lg, so by definition we have ˜ Lg = Lg. Hence the restriction of ˜ L to Y<br />
is L. Consequently,<br />
˜ L = sup <br />
f∈X, f=1<br />
˜ Lf ≥ sup <br />
f∈Y, f=1<br />
˜ Lf = sup Lf = L.<br />
f∈Y, f=1<br />
Finally, suppose that f ∈ X. Then there exist gn ∈ Y such that gn → f and Lgn → ˜ Lf,<br />
so<br />
˜ Lf = lim<br />
n→∞ Lgn ≤ lim<br />
n→∞ L gn = L f.<br />
Hence ˜ L ≤ L. Combining this with the opposite inequality derived above, we conclude<br />
that ˜ L = L.<br />
(b) Suppose that L: Y → range(Y ) is a topological isomorphism. We already know that<br />
˜L: X → range( ˜ L) is bounded. We need to show that ˜ L is injective, that ˜ L −1 : range( ˜ L) → X<br />
is bounded, and that range( ˜ L) = range(L).<br />
Fix any f ∈ X. Then there exist gn ∈ Y such that gn → f and Lgn → ˜ Lf. Since L is a<br />
topological isomorphism, we have by Exercise 2.16 that gn ≤ L −1 Lgn. Hence<br />
˜ Lf = lim<br />
n→∞ Lgn ≥ lim<br />
n→∞<br />
gn<br />
L −1 <br />
Consequently, ˜ L is injective and for any h ∈ range( ˜ L) we have<br />
= f<br />
L −1 .<br />
˜ L −1 h ≤ L −1 ˜ L( ˜ L −1 h) = L −1 h.<br />
Therefore ˜ L −1 : range( ˜ L) → X is bounded.<br />
It remains only to show that the range of ˜ L is the closure of the range of L. If f ∈ X, then<br />
by definition there exist gn ∈ Y such that gn → f and Lgn → ˜ Lf. Hence ˜ Lf ∈ range(L), so<br />
range( ˜ L) ⊂ range(L).
18 CHRISTOPHER HEIL<br />
On the other hand, suppose that h ∈ range(L) Then there exist gn ∈ Y such that Lgn → h.<br />
Since ˜ L −1 is bounded and ˜ L extends L, we conclude that gn = ˜ L −1 (Lgn) → ˜ L −1 (h). Hence<br />
f = ˜ L −1 (h), so f ∈ range( ˜ L). <br />
<strong>3.</strong> Finite-Dimensional Normed Spaces<br />
In this section we will prove some basic facts about finite-dimensional spaces.<br />
First, recall that a finite-dimensional vector space has a finite basis, which gives us a<br />
natural notion of coordinates of a vector, which in turn yields a linear bijection of X onto<br />
F n for some n.<br />
Example <strong>3.</strong>1 (Coordinates). Let X be a finite-dimensional vector space over F. Then X<br />
has a finite basis, say B = {e1, . . . , en}. Every element of X can be written uniquely in this<br />
basis, say,<br />
x = c1(x) e1 + · · · + cn(x) en, x ∈ X.<br />
Define the coordinates of x with respect to the basis B to be<br />
[x]B =<br />
⎡ ⎤<br />
c1(x)<br />
⎣<br />
.<br />
⎦ .<br />
cn(x)<br />
Then the mapping T : X → F n given by x ↦→ [x]B is, by definition of basis, a linear bijection<br />
of X onto F n .<br />
Since we already know how to construct many norms on F n , by transferring these to X<br />
we obtain a multitude of norms for X.<br />
Exercise <strong>3.</strong>2 (ℓ p w Norms on X). Let X be a finite-dimensional vector space over F and let<br />
B = {e1, . . . , en} be any basis. Fix any 1 ≤ p ≤ ∞ and any weight w : {1, . . . , n} → (0, ∞).<br />
Using the notation of Example <strong>3.</strong>1, given x ∈ X define<br />
xp,w = ⎧<br />
⎪⎨<br />
n<br />
<br />
|ck(x)|<br />
[x]B<br />
= p,w k=1<br />
⎪⎩<br />
p w(k) p<br />
1/p , 1 ≤ p < ∞,<br />
max |ck(x)| w(k), p = ∞.<br />
k<br />
Note that while we use the same symbol · p,w to denote a function on X and on F n , by<br />
context it has different meanings depending on whether it is being applied to an element of<br />
X or to an element of F n .<br />
Prove the following.<br />
(a) · p,w is a norm on X.<br />
(b) x ↦→ [x]B is a isometric isomorphism of X onto F n (using the norm · p,w on X and<br />
the norm · p,w on F n ).
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 19<br />
(c) Let {xn}n∈N be a sequence of vectors in X and let x ∈ X. Prove that xn → x with<br />
respect to the norm · p,w on X if and only if the coordinate vectors [xn]B converge<br />
componentwise to the coordinate vector [x]B.<br />
(d) X is complete in the norm · p,w.<br />
Now we can show that all norms on a finite-dimensional space are equivalent.<br />
Theorem <strong>3.</strong><strong>3.</strong> If X is a finite-dimensional vector space over F, then any two norms on X<br />
are equivalent.<br />
Proof. Let B = {e1, . . . , en} be any basis for X, and let · ∞ be the norm on X defined in<br />
Exercise <strong>3.</strong>2. Since equivalence of norms is an equivalence relation, it suffices to show that<br />
an arbitrary norm · on X is equivalent to · ∞.<br />
Using the notation of Exercise <strong>3.</strong>1, given x ∈ X we can write x uniquely as x = c1(x) e1 +<br />
· · · + cn(x) en. Therefore,<br />
n<br />
x ≤ |ck(x)| ek ≤<br />
k=1<br />
n <br />
k=1<br />
<br />
max<br />
ek |ck(x)|<br />
k<br />
= C2 x∞,<br />
where C2 = n k=1 ek is a nonzero constant independent of x.<br />
It remains to show that there is a constant C1 > 0 such that C1 x∞ ≤ x for every x.<br />
First, let<br />
D = {x ∈ X : x∞ = 1}<br />
be the ℓ∞-unit circle in X.<br />
Exercise: Show that D is compact (with respect to the norm · ∞). Hints: Suppose<br />
that {xn}n∈N is a sequence of vectors in D. Then for each n, we have |ck(xn)| = 1 for some<br />
k ∈ {1, . . . , n}. Hence there must be some k such that |ck(xnj )| = 1 for infinitely many j.<br />
Since {c1(xnj )}j∈N is an infinite sequence of scalars in the compact set {c ∈ F : |c| ≤ 1}, we<br />
can select a subsequence whose first coordinates converge. Repeat for each coordinate, and<br />
remember that the kth coordinate is always 1. Hence we can construct a subsequence that<br />
converges to x ∈ D with respect to the ℓ ∞ -norm.<br />
Our next goal is to show that D is also compact with respect to the norm · . Let<br />
{xn}n∈N be any sequence of vectors in D. Since D is compact with respect to · ∞,<br />
there exists a subsequence {xnk }k∈N and an x ∈ D such that x − xnk ∞ → 0. But then<br />
x − xnk ≤ C2 x − xnk ∞ → 0, so {xnk }k∈N is a subsequence that converges to x ∈ X with<br />
respect to · . Hence D is compact with respect to · .<br />
Now, · is a continuous function with respect to the convergence criteria defined by · <br />
(this is part (b) of Exercise 1.11). The set D is compact with respect to the topology defined<br />
by · . A real-valued continuous function on a compact set must achieve a maximum and<br />
minimum on that set. Hence, there must exist constants m and M such that m ≤ x ≤ M<br />
for all x ∈ D. Since x ∈ D if and only if x∞ = 1, this implies that<br />
∀ x ∈ X, m x∞ ≤ x ≤ M x∞.<br />
If we had m = 0, then this would imply that there is an x ∈ D such that x = 0. But then<br />
x = 0, which implies x∞ = 0, contradicting the fact that x ∈ D. Hence m > 0, so we can<br />
take C1 = m.
20 CHRISTOPHER HEIL<br />
Consequently, from now on we need not specify the norm on a finite-dimensional vector<br />
space X—we can take any norm that we like whenever we need it.<br />
Exercise <strong>3.</strong>4. Let F m×n be the space of all m × n matrices with entries in F. F m×n is<br />
naturally isomorphic to F mn .<br />
Prove that if · a is any norm on F m×n , · b is any norm on F n×k , and · c is any norm<br />
on F m×k , then there exists a constant C > 0 such that<br />
∀ A ∈ F m×n , ∀ B ∈ F n×k , ABc ≤ C Aa Bb.<br />
Proposition <strong>3.</strong>5. If M is a subspace of a finite-dimensional vector space X, then M is<br />
closed.<br />
Proof. Let · be any norm on X. Suppose that xn ∈ M and that xn → y ∈ X. Set<br />
M1 = span{M, y} = {m + cy : m ∈ M, c ∈ F}.<br />
Then M1 is finite-dimensional subspace of X. Moreover, every element z ∈ M1 can be<br />
written uniquely as z = m(z) + c(z)y where m(z) ∈ M and c(z) is a scalar. For z ∈ M1<br />
define<br />
zM1 = m(z) + |c(z)|.<br />
Exercise: Show that · M1 is a norm on M1.<br />
Since · is also a norm on M1 and all norms on a finite-dimensional space are equivalent,<br />
we conclude that there is a constant C > 0 such that zM1 ≤ C z for all z ∈ M1. Since<br />
c(xn) = 0 for every n, we therefore have<br />
|c(y)| = |c(y) − c(xn)| = |c(y − xn)|<br />
≤ m(y − xn) + |c(y − xn)|<br />
= y − xnM1<br />
≤ C y − xn → 0.<br />
Therefore c(y) = 0, so y ∈ M. <br />
If X is any normed vector space and M is a finite-dimensional subspace of X, then a proof<br />
identical to the one used in the preceding proposition, except using the given norm · on<br />
X, shows that M is closed.<br />
Exercise <strong>3.</strong>6. Let X be a normed linear space. If M is a finite-dimensional subspace of X,<br />
then M is closed.<br />
Exercise <strong>3.</strong>7. Let X be a finite-dimensional normed space, and let Y be a normed linear<br />
space. Prove that if L: X → Y is linear, then L is bounded.<br />
The following lemma will be needed for Exercise <strong>3.</strong>9.
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 21<br />
Lemma <strong>3.</strong>8 (F. Riesz’s Lemma). Let M be a proper, closed subspace of a normed space X.<br />
Then for each ε > 0, there exists g ∈ X with g = 1 such that<br />
dist(g, M) = inf g − f > 1 − ε.<br />
f∈M<br />
Proof. Choose any u ∈ X \ M. Since M is closed, we have<br />
Fix δ > 0 small enough that a<br />
a+δ<br />
that a ≤ u − v < a + δ. Set<br />
a = dist(u, M) = inf u − f > 0.<br />
f∈M<br />
> 1 − ε. By definition of infimum, there exists v ∈ M such<br />
g =<br />
u − v<br />
u − v ,<br />
and note that g = 1. Given f ∈ M we have h = v + u − v f ∈ M, so<br />
g − f =<br />
<br />
<br />
<br />
<br />
u − v − u − v f <br />
<br />
u − h<br />
u − v = ><br />
u − v<br />
a<br />
> 1 − ε.<br />
a + δ<br />
<br />
Exercise <strong>3.</strong>9. Let X be a normed linear space. Let B = {x ∈ X : x ≤ 1} be the closed<br />
unit ball in X. Prove that if B is compact, then X is finite-dimensional.<br />
Hints: Suppose that X is infinite-dimensional. Given any nonzero e1 with e1 ≤ 1, by<br />
Lemma <strong>3.</strong>8 there exists e2 ∈ X \span{e1} with e2 ≤ 1 such that e2 −e1 > 1.<br />
Continue in<br />
2<br />
this way to construct vectors ek such that {e1, . . . , en} are independent for any n. Conclude<br />
that X is infinite-dimensional.<br />
Definition <strong>3.</strong>10. We say that a normed linear space X is locally compact if for each f ∈ X<br />
there exists a compact K ⊂ X with nonempty interior K ◦ such that f ∈ K ◦ .<br />
In other words, X is locally compact if for every f ∈ X there is a neighborhood of f that<br />
is contained in a compact subset of X. For example, F n is locally compact.<br />
With this terminology, we can reword Exercise <strong>3.</strong>9 as follows.<br />
Exercise <strong>3.</strong>11. Let X be a normed linear space.<br />
(a) Prove that if X is locally compact, then X is finite-dimensional.<br />
(b) Prove that if X is infinite-dimensional, then no nonempty open subset of X has<br />
compact closure.
22 CHRISTOPHER HEIL<br />
4. Quotients and Products of Normed Spaces<br />
Any vector space is an abelian group under the operation of vector addition. So, if you are<br />
familiar with the basic notions of abstract algebra, the concept of a coset will be familiar to<br />
you. However, even if you have not studied abstract algebra, the idea of a coset in a vector<br />
space is very natural.<br />
Example 4.1 (Cosets in R 2 ). Consider the vector space X = R 2 . Let M be any onedimensional<br />
subspace of R 2 , i.e., M is a line in R 2 through the origin. A coset of M is<br />
simply a rigid translate of M by a vector in R 2 . For concreteness, let us specifically consider<br />
the case where M is the x1-axis in R 2 , i.e., M = {(x1, 0) : x1 ∈ R}. Then given a vector<br />
y = (y1, y2) ∈ R 2 , the coset y + M is the set<br />
y + M = {y + m : m ∈ M} = {(y1 + x1, y2 + 0) : x1 ∈ R} = {(x1, y2) : x1 ∈ R},<br />
which is the horizontal line at height y2. This is not a subspace of R 2 , but it is a rigid<br />
translate of the x1-axis. Note that there are infinitely many different choices of y that give<br />
the same coset. Furthermore, we have the following facts for this particular setting.<br />
(a) Two cosets are either identical or entirely disjoint.<br />
(b) The union of all the cosets is all of R 2 .<br />
(c) The set of distinct cosets is a partition of R 2 .<br />
The preceding example is entirely typical.<br />
Definition 4.2 (Cosets). Let M be a subspace of a vector space X. Then the cosets of M<br />
are the sets<br />
f + M = {f + m : m ∈ M}, f ∈ M.<br />
Exercise 4.<strong>3.</strong> Let X be a vector space, and let M be a subspace of X. Given f, g ∈ M,<br />
define f ∼ g if f − g ∈ M. Prove the following.<br />
(a) ∼ is an equivalence relation on X.<br />
(b) The equivalence class of f under the relation ∼ is [f] = f + M.<br />
(c) If f, g ∈ M then either f + M = g + M or (f + M) ∩ (g + M) = ∅.<br />
(d) f + M = g + M if and only if f − g ∈ M.<br />
(e) f + M = M if and only if f ∈ M.<br />
(f) If f ∈ X and m ∈ M then f + M = f + m + M.<br />
(g) The set of distinct cosets of M is a partition of X.<br />
Definition 4.4 (Quotient Space). If M is a subspace of a vector space X, then the quotient<br />
space X/M is<br />
X/M = {f + M : f ∈ X}.
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 23<br />
Since two cosets of M are either identical or disjoint, the quotient space X/M is simply<br />
the set of all the distinct cosets of M.<br />
Example 4.5. Again let M = {(x1, 0) : x1 ∈ R} be the x1-axis in R 2 . Then, by Example 4.1,<br />
we have that<br />
R 2 /M = {y + M : y ∈ R 2 } = {(x1, 0) + M : x1 ∈ R},<br />
i.e., R 2 /M is the set of all horizontal lines in R 2 . Note that R 2 /M is in 1-1 correspondence<br />
with the set of distinct heights, i.e., there is a natural bijection of R 2 /M onto R. This is a<br />
special case of a more general fact that we will explore.<br />
Next we define two natural operations on the set of cosets: addition of cosets and multiplication<br />
of a coset by a scalar. These are defined formally as follows.<br />
Definition 4.6. Let M be a subspace of a vector space X. Given f, g ∈ X, define addition<br />
of cosets by<br />
(f + M) + (g + M) = (f + g) + M.<br />
Given f ∈ X and c ∈ F, define scalar multiplication by<br />
c(f + M) = cf + M.<br />
Remark 4.7. Before proceeding, we must show that these operations are actually welldefined.<br />
After all, there need not be just one f that determines the coset f + M—how do<br />
we know that if we choose different vectors that determine the same cosets, we will get the<br />
same result when we compute (f + g) + M? We must show that f1 + M = f2 + M and<br />
g1 + M = g2 + M then (f1 + g1) + M = (f2 + g2) + M in order to know that Definition 4.6<br />
makes sense.<br />
Proposition 4.8. If M is a subspace of a vector space X, then the addition of cosets of M<br />
given in Definition 4.6 is well-defined.<br />
Proof. Suppose that f1 + M = f2 + M and g1 + M = g2 + M. Then by Exercise 4.3(d)<br />
we know that f1 − f2 = k ∈ M and g1 − g2 = l ∈ M. If h ∈ (f1 + g1) + M then we have<br />
h = f1 + g1 + m for some m ∈ M. Hence<br />
h = (f2 + k) + (g2 + l) + m = (f2 + g2) + (k + l + m) ∈ (f2 + g2) + M.<br />
Thus (f1 + g1) + M ⊂ (f2 + g2) + M, and the converse inclusion is symmetric. <br />
Exercise 4.9. Show that scalar multiplication is likewise well-defined.<br />
Now we can show that the quotient space is actually a vector space under the operations<br />
just defined.<br />
Proposition 4.10. If M is a subspace of a vector space X, then X/M is a vector space<br />
with respect to the operations given in Definition 4.6.
24 CHRISTOPHER HEIL<br />
Proof. Addition of cosets is commutative because<br />
(f + M) + (g + M) = (f + g) + M = (g + f) + M = (g + M) + (f + M).<br />
The zero vector in X/M is the coset 0+M = M, because (f +M)+(0+M) = (f +0)+M =<br />
f + M.<br />
Exercise: Show that the remaining axioms of a vector space are satisfied. <br />
Definition 4.11 (Codimension). If M is a subspace of a vector space X, then the codimension<br />
of M is the dimension of X/M, i.e.,<br />
codim(M) = dim(X/M).<br />
Example 4.12. Let C(R) be space of continuous functions on R, and let P be the subspace<br />
containing the polynomials. Given f ∈ C(R), the coset f + P is f + P = {f + p :<br />
p is a polynomial}. Further, f + P = g + P if and only if f − g is a polynomial. Thus, f + P<br />
can be thought of as “f modulo the polynomials,” i.e., it is the equivalence class obtained<br />
by identifying functions which differ by a polynomial.<br />
In the same way, a coset f + M can be thought of as the equivalence class obtained by<br />
identifying vectors which differ by an element of M. We can imagine the mapping that takes<br />
f to f + M as “collapsing information modulo M.” 1<br />
Definition 4.1<strong>3.</strong> If M is a subspace of a vector space X, then the canonical projection or<br />
the canonical mapping of X onto X/M is π : X → X/M defined by<br />
π(f) = f + M, f ∈ X.<br />
Exercise 4.14. Let M be a subspace of a vector space X.<br />
(a) Prove that the canonical projection π is linear.<br />
(b) Prove that π is surjective and ker(π) = M.<br />
(c) Prove that if E ⊂ X, then the inverse image of π(E) is<br />
π −1 π(E) = E + M = {u + m : u ∈ E, m ∈ M}.<br />
Solution<br />
(c) Suppose that u ∈ E and m ∈ M are given. Then<br />
π(u + m) = u + m + M = u + M = π(u) ∈ π(E).<br />
Hence u + m ∈ π −1 π(E) .<br />
Now suppose that v ∈ π −1 π(E) . Then, by definition, π(v) ∈ π(E) = {u + M : u ∈ E}.<br />
Hence v + M = π(v) = u + M for some u ∈ E. But then m = v − u ∈ M, so v = u + m<br />
with u ∈ E and m ∈ M. <br />
1 Conway calls this map Q, but I prefer to call it π for “projection.”
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 25<br />
We will mostly be interested in the case where X is a normed space. The following result<br />
shows that X/M is a semi-normed space in general, and is a normed space if M is closed.<br />
Proposition 4.15. Let M be a subspace of a normed linear space X. Given f ∈ X, define<br />
f + M = dist(f, M) = inf f − m.<br />
m∈M<br />
Then the following statements hold.<br />
(a) · is well-defined.<br />
(b) · is a semi-norm on X/M.<br />
(c) If M is closed, then · is a norm on X/M.<br />
Proof. (a) Exercise.<br />
Hint: Show that if f1 + M = f2 + M then {f1 − m : m ∈ M} = {f2 − m : m ∈ M}.<br />
(b) Exercise.<br />
(c) Suppose that M is closed, and that f + M = 0. Then inf m∈M f − m = 0. Hence<br />
there exist vectors gn ∈ M such that f − gn → 0 as n → ∞. But M is closed, so this<br />
implies f ∈ M. By Exercise 4.3(e), we therefore have f + M = M = 0 + M, which is the<br />
zero vector in X/M. <br />
Now we derive some basic properties of the canonical projection π of X onto X/M.<br />
Proposition 4.16. Let M be a closed subspace of a normed linear space X. Then the<br />
following statements hold.<br />
(a) π(f) = f + M ≤ f for each f ∈ X.<br />
(b) Let B X r<br />
(f) denote the open ball of radius r in X centered at f, and let BX/M r (f + M)<br />
denote the open ball of radius r in X/M centered at f + M. Then for any f ∈ X<br />
and r > 0 we have<br />
π B X r (f) = B X/M<br />
r (f + M).<br />
(c) W ⊂ X/M is open in X/M if and only if π −1 (W ) = {f ∈ X : f + M ∈ W } is open<br />
in X.<br />
(d) π is an open mapping, i.e., if U is open in X then π(U) is open in X/M.<br />
Proof. (a) Choose any f ∈ X. Since 0 is one of the elements of M, we have<br />
π(f) = f + M = inf f − m ≤ f − 0 = f.<br />
m∈M<br />
(b) First consider the case f = 0 and r > 0. Suppose that g + M ∈ π BX r (0) . Then<br />
(0), i.e., h < r. Hence g + M = h + M ≤ h < r,<br />
(0 + M).<br />
(0 + M). Then infm∈M g − m = g + M < r. Hence<br />
g + M = h + M for some h ∈ BX r<br />
so g + M ∈ B X/M<br />
r<br />
Now suppose that g + M ∈ B X/M<br />
r<br />
there exists m ∈ M such that g − m < r. Thus g − m ∈ BX r (0), so<br />
g + M = g − m + M = π(g − m) ∈ π B X r (0) .
26 CHRISTOPHER HEIL<br />
Exercise: Show that statement (b) holds for an arbitrary f ∈ X.<br />
(c) ⇒. Part (a) implies that π is continuous. Hence π −1 (W ) must be open in X if W is<br />
open in X/M.<br />
⇐. Suppose that W is a subset of X/M such that π −1 (W ) is open in X. We must show<br />
that W is open in X/M. Choose any point f + M ∈ W . Then f ∈ π −1 (W ), which is open<br />
in X. Hence, there exists an r > 0 such that B X r (f) ⊂ π−1 (W ). By part (b) we therefore<br />
have<br />
B X/M<br />
r (f + M) = π B X r (f) ⊆ π π −1 (W ) = W.<br />
Therefore W is open.<br />
(d) Suppose that U is an open subset of X. Then by Exercise 4.14(c), we have<br />
π −1 π(U) = U + M = {u + m : u ∈ U, m ∈ M} = <br />
(U + m).<br />
But each set U + m, being the translate of the open set U, is itself open. Hence π −1 π(U) <br />
is open, since it is a union of open sets. Part (c) therefore implies that π(U) is open in<br />
X/M. <br />
Exercise 4.17. Let M be a closed subspace of a normed space X, and let π be the canonical<br />
projection π of X onto X/M. Prove that π = 1.<br />
Hint: Lemma <strong>3.</strong>8.<br />
Now we can prove that if X is a Banach space, then X/M inherits a Banach space structure<br />
from X.<br />
Theorem 4.18. If M is a closed subspace of a Banach space X, then X/M is a Banach<br />
space.<br />
Proof. We have already shown that X/M is a normed space, so we must show that it is<br />
complete in that norm.<br />
Suppose that {fn + M}n∈N is a Cauchy sequence in X/M. It would be convenient if this<br />
implies that {fn}n∈N is a Cauchy sequence in X, but this need not be the case. For, the<br />
vectors fn are not unique in general: if we replace fn by any vector fn + m with m ∈ M,<br />
then we obtain the same coset. We will show that by choosing an appropriate subsequence<br />
{fnk }k∈N and replacing the fnk by appropriate vectors that determine the same cosets fnk +M,<br />
we can create a sequence in X that is Cauchy and hence converges, and then use this to<br />
show that the original sequence of cosets {fn + M}n∈N converges in X/M.<br />
We begin by applying Exercise 1.18: there exists a subsequence {fnk + M}k∈N such that<br />
m∈M<br />
∀ k ∈ N, (fnk+1 − fnk ) + M = (fnk+1 + M) − (fnk + M) < 2−k .<br />
Now we seek to create vectors gk ∈ M so that {fnk − gk}k∈N will converge in X. Note that<br />
the cosets determined by fnk and by fnk − gk are identical.<br />
Set g1 = 0. Then<br />
inf<br />
g∈M (fn1 − g1) − (fn2 − g) = inf<br />
g∈M (fn1 − fn2) + g = (fn1 − fn2) + M < 1<br />
2 .
Therefore, there exists a g2 ∈ M such that<br />
Then, since g2 ∈ M,<br />
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 27<br />
(fn1 − g1) − (fn2 − g2) < 2 · 1<br />
2 .<br />
inf<br />
g∈M (fn2 − g2) − (fn3 − g) = inf<br />
g∈M (fn2 − fn3) + g = (fn2 − fn3) + M < 1<br />
.<br />
22 Therefore, there exists a g3 ∈ M such that<br />
(fn2 − g2) − (fn3 − g3) < 2 · 1<br />
.<br />
22 Continuing in this way, by induction we construct hk = fnk − gk such that<br />
hk − hk+1 < 1<br />
.<br />
2k−1 Exercise 1.12 therefore implies that {hk}k∈N is a Cauchy sequence in X. Since X is complete,<br />
this sequence converges, say hk → h.<br />
Since<br />
(fnk + M) − (h + M) = fnk − gk − h + M (since gk ∈ M)<br />
= hk − h + M<br />
≤ hk − h → 0,<br />
we see that {fnk + M}k∈N is a convergent subsequence of {fn + M}n∈N. Since we know that<br />
{fn + M}n∈N is Cauchy, it follows from Exercise 1.17 that fnk + M → h + M. Thus X/M<br />
is complete. <br />
The following exercise shows that the converse of the preceding theorem is true as well.<br />
Exercise 4.19. Let M be a closed subspace of a normed space X. Prove that if M and<br />
X/M are both complete, then X must be complete.<br />
Hints: Suppose that {fn}n∈N is a Cauchy sequence in X. Show that {fn + M}n∈N is a<br />
Cauchy sequence in X/M, hence converges to some coset f + M. Thus f − fn + M → 0.<br />
Does this imply that there exists a g ∈ M such that f − fn + g → 0? Or perhaps vectors<br />
gn ∈ M such that f − fn + gn → 0?<br />
The quotient space and canonical map will be useful tools for proving many later results.<br />
The following proof illustrates their utility.<br />
Proposition 4.20. Let X be a normed linear space. If M is a closed subspace of X and N<br />
is finite-dimensional, then M + N is a closed subspace of X.<br />
Proof. Let π be the canonical projection of X onto X/M. Since N is finite-dimensional, it<br />
has a finite basis, say {e1, . . . , en}. Then since π is linear,<br />
π(N) = π span{e1, . . . , en} = span{π(e1), . . . , π(en)} = span{e1 + M, . . . , en + M}.
28 CHRISTOPHER HEIL<br />
Thus π(N) is a finite-dimensional subspace of X/M, and therefore is closed by Proposition<br />
<strong>3.</strong>6. Since π is continuous, it follows that π −1 (π(N)) is closed in X. However, by<br />
Exercise 4.14(c), we have π −1 (π(N)) = M + N. <br />
Exercise 4.21. Let M be a closed subspace of a normed linear space X.<br />
(a) Prove that if X is separable, then X/M is separable.<br />
(b) Prove that if X/M and M are both separable, then X is separable.<br />
(c) Give an example of X, M such that X/M is separable, but X is not separable.<br />
Solution<br />
(b) Let {fn + M}n∈N be a countable dense subset of X/M, and let {gn}n∈N be a countable<br />
dense subset of M. Then S = {fm + gn}m,n∈N is a countable subset of X, and we claim that<br />
it is dense.<br />
To see this, fix any f ∈ X. Then there exists an m such that<br />
inf<br />
h∈M f − fm + h = f − fm + M = (f + M) − (fm + M) < ε<br />
2 .<br />
Hence, there exists some h ∈ M such that f − fm + h < ε<br />
. Since h ∈ M, there exists an<br />
2<br />
n such that h − gn < ε.<br />
Therefore<br />
2<br />
f − (fm + gn) ≤ f − fm − h + h − gn < ε ε<br />
+ = ε. <br />
2 2<br />
Exercise 4.22. Recall from Exercise 1.37 that c0 is a closed subspace of the Banach space ℓ ∞ .<br />
(a) Prove that ℓ ∞ is not separable.<br />
Hint: Consider<br />
S = {(x1, x2, . . . ) : xk = 0 or 1 for every k}.<br />
What is the distance between two distinct elements of S?<br />
(b) Let x, y be vectors in S Prove that if xk = yk for at most finitely many k, then<br />
x + c0 = y + c0. Prove that if xk = yk for infinitely many k, then x − y + c0 = 1.<br />
(c) Use part (b) to prove directly that ℓ ∞ /c0 is not separable.<br />
(e) Prove that the standard basis {en}n∈N is a Schauder basis for c0 (with respect to the<br />
ℓ ∞ -norm).<br />
(f) Use part (e) to show that c0 is separable.<br />
Hint: Use one of the exercises about Schauder bases from the Chapter 1 lecture notes.<br />
(g) Use part (e) and Exercise 5.4 to show that ℓ ∞ /c0 is not separable.<br />
In the Hilbert space case, there is a very close relationship between π and the orthogonal<br />
projection of H onto M ⊥ .
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 29<br />
Exercise 4.2<strong>3.</strong> Let M be a closed subspace of a Hilbert space H, and let π be the canonical<br />
projection of H onto H/M. Prove that the restriction of π to M ⊥ is an isometric isomorphism<br />
of M ⊥ onto H/M.<br />
Remark 4.24. There is no analog of the preceding result for arbitrary Banach spaces. If X<br />
is a Banach space and M is a closed subspace then we say that M is complemented in X if<br />
there exists another closed subspace N such that M ∩ N = {0} and M + N = X.<br />
It is not true that every closed subspace of every Banach space is complemented. In<br />
particular, c0 is not complemented in ℓ ∞ . Also, if 1 < p ≤ ∞ and p = 2, then ℓ p has<br />
uncomplemented subspaces.<br />
Next we will define the product or direct sum of normed spaces. An analogous definition<br />
holds for the case of a finite collection of spaces.<br />
Definition 4.25. Let {Xi}i∈N be a countable family of Banach spaces, and let · i denote<br />
the norm on Xi. Define<br />
For 1 ≤ p < ∞, define<br />
For p = ∞, define<br />
<br />
p Xi =<br />
∞<br />
i=1<br />
<br />
∞ Xi =<br />
Xi = <br />
f = (f1, f2, . . . ) : fi ∈ Xi .<br />
<br />
f ∈<br />
∞<br />
Xk : fp =<br />
i=1<br />
<br />
f ∈<br />
∞<br />
i=1<br />
∞ <br />
i=1<br />
fi p<br />
1/p <br />
i < ∞ .<br />
Xk : f∞ = sup fii < ∞<br />
i<br />
<br />
.<br />
Exercise 4.26. Let {Xi}i∈N be a countable family of Banach spaces and fix 1 ≤ p ≤ ∞.<br />
Let X = <br />
p Xi. Prove the following.<br />
(a) X is a normed space.<br />
(b) For each i, the projection Pi : X → Xi given by Pi(f1, f2, . . . ) = fi is continuous, and<br />
Pi = 1.<br />
(c) X is a Banach space if and only if each Xi is a Banach space.<br />
Exercise 4.27. Let X1, . . . , Xn be finitely many normed spaces. Prove that the spaces ⊕pXi<br />
are equal for 1 ≤ p ≤ ∞, and that all the norms · p are equivalent. For this reason, we<br />
often denote this space by X1 × · · · × Xn.
30 CHRISTOPHER HEIL<br />
Exercise 4.28. Let X and Y be normed vector spaces. Define T : B(X, Y ) × X → Y by<br />
T (A, f) = Af.<br />
(a) Prove that T is continuous if and only if An → A and fn → f implies Anfn → Af.<br />
(b) Prove that T is continuous. Conclude that T ∈ B(B(X, Y ) × X, Y ).<br />
5. Linear Functionals<br />
Definition 5.1 (Hyperplane). Let M be a subspace of a vector space X. Then we say that<br />
M is a hyperplane if it has codimension 1, i.e., if codim(M) = dim(X/M) = 1.<br />
Example 5.2. Let H be a Hilbert space, and let L be a bounded linear functional on H.<br />
That is, L: H → F is a bounded linear operator, so L ∗ : F → H is also a bounded linear<br />
operator. Now, F is one-dimensional, and a linear operator is entirely determined by what<br />
it does to the basis elements. In particular, {1} is a basis for F, so L ∗ (c) = L ∗ (c1) = c L ∗ (1).<br />
Hence range(L ∗ ) = span{L ∗ (1)}. If L ∗ (1) = 0 then L ∗ is the zero operator, and hence L is<br />
the zero operator as well (why?). Otherwise, range(L ∗ ) must be one-dimensional, and hence<br />
is a closed subspace of H.<br />
Therefore,<br />
H = range(L ∗ ) ⊕ range(L ∗ ) ⊥ = range(L ∗ ) ⊕ ker(L).<br />
Hence, every vector f ∈ H can be written uniquely as<br />
f = c L ∗ (1) + k<br />
for some scalar c ∈ range(L ∗ ) and k ∈ ker(L). Hence, if π is the canonical projection of H<br />
onto H/ ker(L), then since k ∈ ker(L) we have<br />
Therefore<br />
f + ker(L) = π(f) = cL ∗ (1) + k + ker(L) = cL ∗ (1) + ker(L).<br />
H/ ker(L) = {f + ker(L) : f ∈ H} = {cL ∗ (1) + k + ker(L) : c ∈ F, k ∈ ker(L)}<br />
= {cL ∗ (1) + ker(L) : c ∈ F}<br />
= span{L ∗ (1) + ker(L)},<br />
which is one-dimensional.<br />
Thus, if L is a bounded linear functional on a Hilbert space H, then we conclude that<br />
ker(L) is a hyperplane in H.<br />
We will show that a similar result holds for arbitrary normed spaces. We will need the<br />
following tool. In abstract algebra, the group version of the next result is called the First<br />
Isomorphism Theorem or the Homomorphism Theorem. In the group setting, an isomorphism<br />
is a bijective homomorphism. In the vector space setting, an isomorphism is a linear<br />
bijection.
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 31<br />
Exercise 5.3 (Isomorphism Theorem). Let X and Y be vector spaces, and let ϕ: X → Y<br />
be a linear surjection. Let M = ker(ϕ). Prove that ψ : X/M → Y given by ψ(f +M) = ϕ(f)<br />
is a well-defined linear bijection, and that ϕ = ψ ◦ π.<br />
Exercise 5.4. Fix 1 ≤ p ≤ ∞, and set<br />
M = {x ∈ ℓ p : x(2k) = 0 for every k}.<br />
Prove that M is a closed subspace of ℓ p , and that ℓ p /M is isometrically isomorphic to ℓ p .<br />
Now we can show that every hyperplane is the kernel of some (not necessarily continuous)<br />
linear functional.<br />
Proposition 5.5. Let M be a subspace of a normed space X. Then the following statements<br />
are equivalent.<br />
(a) M is a hyperplane.<br />
(b) M = ker(µ) for some nonzero linear functional µ: X → F.<br />
Proof. (b) ⇒ (a). Suppose that M = ker(µ) where µ is a nonzero linear functional. Then<br />
by the Isomorphism Theorem, there is a linear bijection ψ : X/ ker(µ) → F. Since F is<br />
one-dimensional, we conclude that X/ ker(µ) is as well.<br />
(a) ⇒ (b). Assume that M is a hyperplane. Then X/M is one-dimensional, so there<br />
exists a linear bijection ψ : X/M → F. Set ϕ = ψ ◦ π, then ψ : X → F. If f ∈ M, then<br />
π(f) = f + M = 0 + M, so ψ(f) = ψ(π(f)) = ψ(0 + M) = 0, so f ∈ ker(ϕ). Conversely,<br />
if f ∈ ker(ϕ) then we have ψ(π(f)) = ϕ(0 + M) = 0. But ψ is a bijection, so this implies<br />
f + M = π(f) = 0 + M. Hence f ∈ M, so ker(ϕ) ⊂ M. <br />
Proposition 5.6. Let X be a normed linear space. If µ, ν : X → F are nonzero linear<br />
functionals, then<br />
Proof. ⇐. Trivial.<br />
ker(µ) = ker(ν) ⇐⇒ µ = cν for some nonzero scalarc.<br />
⇒. Suppose that ker(µ) = ker(ν). Since µ = 0, there exists some f ∈ X such that<br />
µ(f) = 0, and by rescaling, we can assume that µ(f) = 1. Since f /∈ ker(µ) = ker(ν), we<br />
have ν(f) = 0. Given any g ∈ X, we have<br />
so g − µ(g)f ∈ ker(µ) = ker(ν). Therefore<br />
µ g − µ(g)f = µ(g) − µ(g) · µ(f) = 0,<br />
ν(g) − µ(g) · ν(f) = ν g − µ(g)f = 0,<br />
so after rearranging we see that ν = ν(f)µ. <br />
Proposition 5.7. If M is a hyperplane in a normed linear space X, then M is either closed<br />
or is dense in X.
32 CHRISTOPHER HEIL<br />
Proof. We are given that X/M is one-dimensional. Let π be the canonical projection of X<br />
onto X/M. The closure M of M is a subspace of X, so since π is linear we know that π(M)<br />
is a subspace of X/M. But X/M is one-dimensional, so there only two possibilities.<br />
First, we could have π(M) = {0+M}. In this case, M ⊂ ker(π) = M, so we have M = M<br />
and M is closed.<br />
Second, we could have π(M) = X/M. In this case, we have by Exercise 4.14(c) that<br />
X = π −1 (X/M) = π −1 (π(M)) = M + M = M,<br />
and thus M is dense. <br />
Proposition 5.8. Let µ: X → F be a linear functional on a normed space X. Then:<br />
Proof. ⇒. Exercise.<br />
µ is continuous ⇐⇒ ker(µ) is closed.<br />
⇐. Suppose that ker(µ) is closed. We know by Proposition 5.5 that ker(µ) is a hyperplane,<br />
so X/ ker(µ) is one-dimensional. Let π be the canonical projection of X onto X/ ker(µ).<br />
Because M is closed, we know that π is continuous. By the Isomorphism Theorem, there<br />
exists a linear bijection ψ : X/ ker(µ) → F. Since X/ ker(µ) is a one-dimensional normed<br />
linear space, and linear map from X/ ker(µ) into another normed space is continuous by<br />
Exercise <strong>3.</strong>7. Therefore ψ is continuous, and hence µ = ψ ◦ π is continuous. <br />
Recall now the definition of the dual space of a normed space X:<br />
X ∗ = X ′ = B(X, F) = {L: X → F : L is bounded and linear}.<br />
Since the norm on F is just absolute value, the operator norm of a linear functional L ∈<br />
X ′ = B(X, F) is<br />
L = sup |Lf|.<br />
f=1<br />
Since F is a Banach space, the dual space X ′ is a Banach space even if X is not.<br />
In order to give some examples of dual spaces, we recall Hölder’s Inequality.<br />
Theorem 5.9 (Hölder’s Inequality). Let (X, Ω, µ) be a measure space. If 1 ≤ p ≤ ∞ and<br />
1<br />
p<br />
+ 1<br />
p ′ = 1, then<br />
∀ f ∈ L p (X), ∀ g ∈ L p′<br />
(X), fg1 ≤ fp gp ′.<br />
In particular, if f ∈ Lp (X) and g ∈ Lp′ (X), then fg ∈ L1 (X).
<strong>CHAPTER</strong> <strong>3.</strong> <strong>BANACH</strong> SPACES 33<br />
Appendix A. Appendix: Topological and Metric Spaces<br />
Definition A.1 (Topological Space). A topological space (X, T ) is a nonempty set X together<br />
with a family T of subsets of X such that the following statements hold.<br />
(a) ∅, X ∈ T .<br />
(b) Closure under finite intersections: If U, V ∈ T , then U ∩ V ∈ T .<br />
(c) Closure under arbitrary unions: If I is any index set and Ui ∈ T for i ∈ I, then<br />
∪i Ui ∈ T .<br />
We call T a topology on X. The elements of T are the open subsets of X.<br />
Definition A.2 (Metric Space). Let X be a nonempty set. A metric on X is a function<br />
d(·, ·) → R such that<br />
(a) d(f, g) ≥ 0 for all f, g ∈ X,<br />
(b) d(f, g) = 0 if and only if f = g,<br />
(c) Triangle Inequality: d(f, h) ≤ d(f, g) + d(g, h) for all f, g, h ∈ X.<br />
A space X together with a metric d(·, ·) is called a metric space.<br />
Definition A.3 (Topology on a Metric Space). Let X be a metric space.<br />
(a) The open ball in X centered at x ∈ X with radius r > 0 is<br />
(b) A subset U ⊆ X is open if<br />
Br(x) = B(x, r) = {y ∈ X : d(x, y) < r}.<br />
∀ x ∈ U, ∃ r > 0 such that Br(x) ⊆ U.<br />
(c) The topology on X is T = {U ⊆ X : U is open}.<br />
Exercise A.4. Prove that if X is a metric space, then (X, T ) is a topological space using<br />
the preceding definition.