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Lecture Notes in Differential Equations - Bruce E. Shapiro

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218 LESSON 26. GENERAL EXISTENCE THEORY*<br />

Figure 26.1: A sufficient condition for a bounded cont<strong>in</strong>uous function to<br />

have a fixed po<strong>in</strong>t is that the range be a subset of the doma<strong>in</strong>. A fixed<br />

po<strong>in</strong>t occurs whenever the curve of f(t) <strong>in</strong>tersects the l<strong>in</strong>e y = t.<br />

b<br />

S<br />

a<br />

a<br />

b<br />

Theorem 26.2 (Sufficient condition for fixed po<strong>in</strong>t). Suppose that<br />

f(t) is a cont<strong>in</strong>uous function that maps its doma<strong>in</strong> <strong>in</strong>to a subset of itself,<br />

i.e.,<br />

f(t) : [a, b] ↦→ S ⊂ [a, b] (26.1)<br />

Then f(t) has a fixed po<strong>in</strong>t <strong>in</strong> [a, b].<br />

Proof. If f(a) = a or f(b) = b then there is a fixed po<strong>in</strong>t at either a or b.<br />

So assume that both f(a) ≠ a and f(b) ≠ b. By assumption, f(t) : [a, b] ↦→<br />

S ⊂ [a, b], so that<br />

f(a) ≥ a and f(b) ≤ b (26.2)<br />

S<strong>in</strong>ce both f(a) ≠ a and f(b) ≠ b, this means that<br />

f(a) > a and f(b) < b (26.3)<br />

Let g(t) = f(t) − t. Then g is cont<strong>in</strong>uous because f is cont<strong>in</strong>uous, and<br />

furthermore,<br />

g(a) = f(a) − a > 0 (26.4)<br />

g(b) = f(b) − b < 0 (26.5)<br />

Hence by the <strong>in</strong>termediate value theorem, g has a root r ∈ (a, b), where<br />

g(r) = 0. Then<br />

0 = g(r) = f(r) − r =⇒ f(r) = r (26.6)<br />

i.e., r is a fixed po<strong>in</strong>t of f.

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