$f(x) = \pi + \frac{1}{2}\sin \left ( \frac{x}{2}\right) $ has a unique fixed point on $[0, 2\pi]$

Question

I am looking for some guidance and help with the following question

Prove that $$f(x) = \pi + \frac{1}{2}\sin \left ( \frac{x}{2}\right) $$ has a unique fixed point on $[0, 2\pi]$

Answer 1

The continuous function $h(x) = \pi + \frac{1}{2}\sin \left ( \frac{x}{2}\right) - x $ is positive at 0 and negative at $2\pi$, the interval being $[ \pi, -\pi ]$, so by the Intermediate Value Theorem there is a point $a \in [0,2\pi]$ with $f(a) - a = 0 $ or $f(a) = a$. This fixed point is unique because h(x) is monotonically decreasing (mentioned by Diger).

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In general, if the intermediate value theorem for {h(x) = f(x) - x} shows that there is a zero at this function, then there is a point a, such that f(a) - a = 0 or f(a) = a. This fixed point is unique because h(x) is monotonic.

Answer 2

You may proceed as follow.

• $f([0,2\pi]) \subset [0,2\pi ] \Rightarrow f$ has a fixpoint according to the fixed point theorem.
• $|f'(x)| =|\frac{1}{4} \cos \frac{x}{2}| \leq \frac{1}{4} \Rightarrow f$ is contractive, so the fixpoint must be unique. To see that assume there were $2$ fixpoints $a, b$ with $a \neq b$. Then, you had $$|a - b| = |f(a) - f(b)| = |f'(\xi)||a-b| < |a-b| \mbox{ Contradiction! } \Rightarrow a = b$$