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Prove that $f$ is constant if and only if for all point pair $x,y \in \mathbb{R} $, $$ | f(x) - f(y) | \leq (x-y)^2 $$

My attempt:

It's a double implication so...

To right:

$f$ is constant $\Rightarrow \forall c \in \mathbb{R}, f(c) = k $, with $ k \in \mathbb{R} \Rightarrow \forall x,y \in \mathbb{R} , | f(x) - f(y) | = 0 \Rightarrow 0 \leq (x-y)^2 $

We observe that $(x-y)^2$ is the square of the difference of two real numbers, hence, is always positive.

To left:

We cant think $(x-y)^2$ as $|x-y|^2$, so:

$$ | f(x) - f(y) | \leq |x-y|^2$$ $$ \Bigg|\frac{f(x) - f(y)}{x-y}\Bigg| \leq |x-y| $$ $$ \lim_{x \to y}{\Bigg|\frac{f(x) - f(y)}{x-y}\Bigg|} \leq \lim_{x \to y}{|x-y|} $$ $$ |f'(y)| \leq 0 $$ We know that the module of a number is always greater or equal to $0$. So $ |f'(y)| = 0 $. $$ |f'(y)| = 0 \Rightarrow f(y) \ \text{is constant} $$

But I'm not sure if I can get into the absolute value with the limit. The statement doesn't mention that $f$ is differentiable.

Spectree
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    See https://math.stackexchange.com/q/537216/42969 or https://math.stackexchange.com/q/954926/42969 – Martin R Dec 29 '20 at 21:44

1 Answers1

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You proof works, the fact that $f$ is not supposed being differentiable is normal. Indeed, you showed that $\left|\frac{f(x)-f(y)}{x-y}\right|\leqslant |x-y|$ for all $x\neq y$, since $\lim\limits_{x\rightarrow y}|x-y|=0$, we also have $\lim\limits_{x\rightarrow y}\frac{f(x)-f(y)}{x-y}=0$. By doing this, you proved that $f$ is differentiable at $y$ and $f'(y)=0$, since $y$ is arbitrary, this means that $f'=0$ and therefore $f$ is constant.

Tuvasbien
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  • that's a very motivational answer, thank you! But, only to be sure, I don't have $ \lim_{x \to y}{\frac{f(x) - f(y)}{x-y}} = 0 $ at all... it's more like $ \lim_{x \to y}{|\frac{f(x) - f(y)}{x-y}|} = 0 $ and I'm not sure if I can apply the definition of differentiable bc it is not exactly the same... I look for a justification of this fact, you know? – Spectree Dec 29 '20 at 22:01
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    If $\lim\limits_{x \rightarrow y}\left|\frac{f(x)-f(y)}{x-y}\right|=0$ then $\lim\limits_{x\rightarrow y}\frac{f(x)-f(y)}{x-y}=0$ because $$ -\left|\frac{f(x)-f(y)}{x-y}\right|\leqslant \frac{f(x)-f(y)}{x-y}\leqslant\left|\frac{f(x)-f(y)}{x-y}\right| $$ and you can use the squeeze theorem. – Tuvasbien Dec 29 '20 at 22:05
  • that's a good one! The sandvitx theorem – Spectree Dec 29 '20 at 22:08
  • One more thing, how do you know the limit exists? I mean $ \lim_{x\to y}{\frac{f(x)-f(y)}{x-y}}= l $, because $f$ is not supposed to be differentiable. – Spectree Jan 01 '21 at 21:22
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    I used the fact that if $|g(x)|\leqslant h(x)$ and $\lim\limits_{x\rightarrow a}h(x)=0$ then $\lim\limits_{x\rightarrow a}g(x)=0$, it gives both the existence and the value of the limit. – Tuvasbien Jan 02 '21 at 06:04