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The question is :

Does there exist any function $f : \mathbb R \longrightarrow \mathbb R$ such that $f(1) = 1$, $f(-1) = -1$ and $|f(x) - f(y)| \leq |x - y|^{\frac {3} {2}}$?

It is clear that $f$ is continuous over $\mathbb R$ by the given condition and hence it attains all the values between $-1$ and $1$ in $(-1,1)$.Now how can I proceed?Please help me.

Thank you in advance.

2 Answers2

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Notice that if $|f(x) - f(y)| \leq {|x - y|}^{\frac {3} {2}}$, then

$$\left|\frac{f(x)-f(y)}{x-y}\right|\leq {|x - y|}^{\frac {1} {2}}$$

and hence $f$ is differentiable... but $f'=0$ everywhere. In other words, $f$ is constant, so one may not have $f(1)\neq f(-1)$.

Fimpellizzeri
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  • So, it is impossible to construct such a function.Isn't it? –  Dec 21 '16 at 05:41
  • Yes. Functions of this kind are called $\alpha$-Hölder continuous (here, $\alpha$ is the exponent in the RHS). When $\alpha >1$, the trick I used here shows they are all constant. When $\alpha=1$, those are simply the Lipschitz functions. The interesting case is when $0<\alpha<1$. You can read more on these functions here. – Fimpellizzeri Dec 21 '16 at 05:43
  • Thanks @Fimpellizier.It really helps. –  Dec 21 '16 at 06:11
  • You're welcome! – Fimpellizzeri Dec 21 '16 at 06:21
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Another way:

$|f(x)-f(y)| \le \sum_{k=0}^{n-1} | f(y+ {k+1 \over n} (x-y))-f(y+ {k \over n} (x-y)) | \le n ({1 \over n}(x-y))^{3 \over 2} = {1 \over \sqrt{n} } (x-y)^{3 \over 2}$, and since $n$ is arbitrary, we see that $f(x) = f(y)$ for any $x,y$.

copper.hat
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