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Let $f: \mathbb{R} → \mathbb{R}$. Suppose that for all $x, t ∈ \mathbb{R}$ we have to $|f(x) − f(t)| ≤ |x − t|^{1+α}$ , For α> 0. Show that $f$ is constant

I tried the following way: $0 ≤ |f(x) − f(t)/x − t| ≤ |x − t|^{α}$ And using the sandwich theorem to conclude $|f(x) − f(t)/x − t|=0$ so $f(x) − f(t)/x − t=0$ so $f(x) − f(t)=0$ so $f(x)=f(t)$ so f is constant.. It is well?

Nash
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2 Answers2

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You cannot say for sure that $|\frac{f(x) - f(t)}{x - t}| = 0$.

Nevertheless, you can say, by the sandwich theorem, that $\lim_{x \to t}|\frac{f(x) - f(t)}{x - t}| = 0$.

Now you can use the squeeze theorem once again to prove that

$\lim_{x \to t}|\frac{f(x) - f(t)}{x - t}| = 0 \implies \lim_{x \to t}\frac{f(x) - f(t)}{x - t} = 0$.

Which means, as @DavidMitra points out, that $f'(t)$ exists and is $0$. But the above limit holds for all $t \in \mathbb{R}$. Thus, $f'(t)$ exists everywhere and is identically $0$.

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dubist
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Since $|\dfrac{f(a+h)-f(a)}{h}|\le |h|^\alpha$,so $\dfrac{f(a+h)-f(a)}{h}\to 0$ as $h\to 0$

Similarly $|\dfrac{f(a)-f(a-h)}{h}|\le | h|^\alpha$,so $\dfrac{f(a)-f(a-h)}{h}\to 0$ as $h\to 0$.

Combining both we have $\lim _{h\to 0} \dfrac{f(a+h)-f(a)}{h}=0\implies f^{'}(a)=0$ forall $a\in \Bbb R$.

Thus $f$ is constant.

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