Solutions to Final Exam Practice Questions

Math 361: Real and Abstract Analysis
Fall 2012, Professor Levandosky
Solutions to Final Exam Practice Questions
1. For each sequence of functions fn, determine the pointwise limit function f and whether ir not fn
converges uniformly to f on [0, 1].
(a) fn(x) = nx
nx 2 + 1
Solution. For x ̸= 0 we have lim
n→∞
pointwise limit function is
nx
nx 2 + 1
f(x) =
x 1
= =
x2 x , and for x = 0, fn(0) = 0 → 0. Thus the
{
1
x
x ̸= 0
0 x = 0
Since this is discontinuous at x = 0 and each fn is continuous on [0, 1], the convergence cannot
be uniform.
(b) fn(x) = nx3
nx 2 + 1
nx
Solution. For x ̸= 0 we have lim
n→∞
3
nx2 + 1
the pointwise limit function is f(x) = x.
To see if the convergence is uniform, consider
Then, on [0, 1],
= x3
x 2 = x and for x = 0 we have fn(0) = 0 → 0, so
fn(x) − f(x) = nx3
nx2 −x
− x =
+ 1 nx2 + 1 .
g(x) ≡ |fn(x) − f(x)| =
satisfies g(0) = 0 and g(1) = 1
n2 +1 . Since g′ (x) = 1−nx2
√
1/n, it follows that
1
2
∥fn − f∥∞ = max
and thus fn converges to f uniformly on [0, 1].
x
nx 2 + 1
(nx 2 +1) 2 = 0 when x = √ 1/n and g( √ 1/n) =
{
1
n2 √ }
1 1
, =
+ 1 2 n
1
√
1
→ 0
2 n
(c) fn(x) = nx n (1 − x)
Solution. For 0 ≤ x < 1, nxn → 0 so fn(x) → 0. For x = 1, fn(1) = 0 → 0 so the limit
function is f(x) = 0. To compute ∥fn − f∥∞ = ∥fn∥∞, first rewrite fn(x) = n(xn − xn+1 ). Then
f ′ n(x) = n(nxn−1 − (n + 1)xn ) = nxn−1 (n − (n + 1)x). So f ′ n(x) = 0 when x = n
n+1 . Since
fn(0) = fn(1) = 0 it then follows that
( ) ( ) n ( )
n
n 1
∥fn∥∞ = fn = n
n + 1 n + 1 n + 1
Since
( ) n
n
lim
= lim
n + 1
1
( 1 + 1
n
) n = 1
e
it follows that lim ∥fn∥∞ = 1
e ̸= 0, so fn does not converge to f uniformly on [0, 1].
2. Use the M-test to show that the series
∞∑
n=1
x
n 2 x + 1
converges uniformly on [0, ∞).

Solution. Let gn(x) = x
n2x+1 . Then g′ n(x) = 1
lim
x→∞ gn(x) = 1
n2 it follows that Mn = ∥gn∥∞ = 1
n2 . Since the series
∞∑
gn(x) converges uniformly on [0, ∞).
n=1
3. Consider the series
∞∑
k=1
x
k 2 x 2 + 1 .
(n 2 x+1) 2 > 0 for all x ∈ [0∞). Since gn(0) = 0 and
(a) Show that for each x ∈ R, the series converges.
Solution. For x = 0, the series clearly converges (to 0). For x ̸= 0, we have
x
k
2x2
|x|
+ 1
≤
k2 1
=
x2 |x|k2 and since the series
converges, the series
∞∑
k=1
x
k 2 x 2 + 1
∞∑
k=1
1 1
=
|x|k2 |x|
∞∑
k=1
1
k 2
∞∑
n=1
converges (absolutely) by the comparison test.
1
converges, the series
n2 (b) Let f(x) be the sum of the series. Prove that the series does not converge uniformly to f on R.
Hint: Show that f(1/n) ≥ 1
2 for all n.
Solution. For each n, we have
f(1/n) =
∞∑
k=1
Now when 1 ≤ k ≤ n, 1 ≤ k 2 /n 2 + 1 ≤ 2, so
n∑
k=1
1/n
k 2 /n 2 + 1 ≥
1/n
k 2 /n 2 +1
1/n
k 2 /n 2 + 1 ≥
n∑
k=1
n∑
k=1
1/n
k 2 /n 2 + 1
1/n
≥ 2 . Therefore
1 1
=
2n 2 .
Thus we have the sequence xn = 1
n → 0 with f(xn) ≥ 1
2 for all n, so that f(xn) ̸→ 0 = f(0), and
therefore f cannot be continuous at x = 0. Since each partial sum of the series
continuous, the convergence cannot be uniform.
4. Let A = {f ∈ C([0, 2]) : |f(x)| ≤ 4 for all x ∈ [0, 2]}.
∞∑
k=1
x
k 2 x 2 + 1 is
(a) Show that A is closed and bounded.
Solution. A is the closed ball of radius 4 centered at the zero function, and is therefore closed.
It is bounded since ∥f∥∞ ≤ 4 for each f ∈ A.
(b) Is A compact?
Solution. No, A is not compact. The sequence fn(x) = (x/2) n in A converges pointwise to the
discontinuous function f defined by f(x) = 0 for 0 ≤ x < 2 and f(2) = 1. Any subsequence (fnk )
of (fn) also converges pointwise to f and since each fnk is continuous, the convergence cannot be
uniform. Hence fnk does not converge uniformly to any element of A. Hence A is not compact.
(c) Is A complete?
Solution. Yes, A is complete because it is a closed subset of the complete metric space C([0, 2]).

5. Let Aϵ = {f ∈ C([−ϵ, ϵ]) : |f(x)| ≤ 4} and define
T (f)(x) = 1 +
(a) Find ϵ > 0 such that T : Aϵ → Aϵ.
Solution. Given f ∈ Aϵ, let g = T (f). Then
g(x) = 1 +
∫ x
0
∫ x
0
(f(t)) 2 cos(t) dt.
(f(t)) 2 cos(t) dt.
By the Fundamental Theorem of Calculus, since f is continuous, g is differentiable, and therefore
continuous on [−ϵ, ϵ]. We need to show that |g(x)| ≤ 4 for all x ∈ [−ϵ, ϵ] for some ϵ > 0. To that
end, for 0 ≤ x ≤ ϵ we have
∫
x
|g(x)| ≤ 1 +
(f(t)) 2 ∫
x
cos(t) dt
≤ 1 + (f(t)) 2 ∫ x
| cos(t)| dt ≤ 1 + 16 dt = 1 + 16x ≤ 1 + 16ϵ
0
0
Likewise |g(x)| ≤ 1 + 16ϵ when −ϵ ≤ x ≤ 0. Thus |g(x)| ≤ 4 for all x ∈ [−ϵ, ϵ] provided ϵ ≤ 3
16 .
(b) Find ϵ > 0 such that T is a contraction mapping on Aϵ.
Solution. We need to show ∥T (f)−T (g)∥∞ ≤ α∥f −g∥∞ for all f, g ∈ Aϵ for some α < 1. When
0 ≤ x ≤ ϵ,
∫
x
|T (f)(x) − T (g)(x)| =
[f(t) 2 − g(t) 2
] cos(t) dt
≤
0
∫ x
0
≤ ∥f − g∥∞
|f(t) + g(t)||f(t) − g(t)| dt
∫ x
0
= 8x∥f − g∥∞
≤ 8ϵ∥f − g∥∞.
The same inequality holds for −ϵ ≤ x ≤ 0, so ∥T (f) − T (g)∥∞ ≤ 8ϵ∥f − g∥∞ and therefore T is
a contraction mapping provided ϵ < 1
8 .
(c) Show that T has a unique fixed point in Aϵ for the ϵ from part (b).
Solution. The space Aϵ is complete since it is a closed subset of the complete space C([−ϵ, ϵ]).
Therefore, since T is a contraction mapping on Aϵ, the contraction mapping principle implies that
T has a unique fixed point in Aϵ.
(d) Find the fixed point.
Solution. The fixed point f must satisfy T (f) = f and thus
1 +
∫ x
0
8 dt
(f(t)) 2 cos(t) dt = f(x).
Taking the derivative of both sides gives f ′ (x) = f(x) 2 cos(x). This separable differential equation
has solution f(x) =
−1
sin(x)+C
implies C = −1, so f(x) = −1
sin(x)−1
6. According to Maple,
∫ π
1
π −π
. The initial condition f(0) = 1 (from the integral equation above)
is the unique fixed point of T .
(e x + e −x ) cos(kx) dx = 2(eπ − e−π )(−1) k
π(1 + k2 .
)
0

(a) Let f(x) = e x + e −x . Use the equation above to write the Fourier series for f.
Solution. The integral above gives a formula for ak. Since f is an even function bk = 0 for all k.
Thus the Fourier series for f is
eπ − e−π ∞∑ 2(e
+
π
π − e−π )(−1) k
π(1 + k2 cos(kx)
)
k=1
(b) Does the Fourier series for f converge uniformly to f?
Solution. Yes, f is differentiable on [−π, π] and since f is even, its 2π-periodic extension is
continuous, so the Fourier series converges uniformly to f on [−π, π].
(c) Evaluate the series at x = 0 and at x = π and use this to determine the sum of the series
∞∑
k=1
(−1) k
k 2 + 1 and
∞∑
k=1
1
k 2 + 1 .
Solution. At x = 0 we have
2 = f(0) = eπ − e−π ∞∑ 2(e
+
π
π − e−π )(−1) k
π(1 + k2 )
so
∞∑
k=1
At x = π, since cos(kπ) = (−1) k , we have
so
(−1) k
=
1 + k2 k=1
e π + e −π = f(π) = eπ − e −π
∞∑
k=1
π
eπ 1
−
− e−π 2 .
π
+
∞∑
k=1
1
1 + k2 = π(eπ + e−π )
2(eπ − e−π 1
−
) 2 .
(d) Use Parseval’s equation to find the sum of the series
Solution. The square of the L 2 -norm of f is
∥f∥ 2 2 =
∞∑
k=1
2(e π − e −π )
π(1 + k 2 )
1
(k2 .
+ 1) 2
∫ π
(e
−π
x + e −x ) 2 dx = e 2π + 4π − e −2π .
By Parseval’s equation, this is equal to
(
a
π
2 0
2 +
∞∑
a 2 )
k = π · 4(eπ − e−π ) 2
π2 k=1
Solving for the summation gives
∞∑
k=1
(
1
2 +
∞∑
k=1
1
(1 + k2 ) 2 = π(e2π + 4π − e−2π )
4(eπ − e−π ) 2 − 1
2 .
1
(1 + k2 ) 2
)
.
7. Determine whether or not the Fourier series for each function converges uniformly. You don’t need to
compute any Fourier coefficients to answer this question.
(a) f(x) = x cos(x)
Solution. The 2π-periodic extension of f is discontinuous, so the Fourier series cannot converge
uniformly.
(b) g(x) = ex2
x 4 + 1
Solution. The 2π-periodic extension of g is continuous, and g is differentiable on (−π, π) so the
Fourier series for g converges uniformly to g.