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Let $f:\mathbb{R}\to\mathbb{R}$ be a function satisfying $$\vert f(x)-f(y)\vert\le\frac{1}{2}\vert x+y\vert$$ for any $x,y\in\mathbb{R}$.

One can easily deduce that $f$ is continuous uniformly on $\mathbb{R}$.

How about the differentiability of $f$ on $\mathbb{R}$ or some subset of $\mathbb{R}$?

Can anyone give some insight or advice? Thank you!

AnonyMath
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  • Are you sure about the (uniform) continuity? – Martin R Sep 25 '19 at 13:58
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    Or did you mean $\vert f(x)-f(y)\vert\le\frac{1}{2}\vert x-y\vert$? – Martin R Sep 25 '19 at 14:08
  • Assuming the previous suggestion by @MartinR, just compute $ \frac{f(x+h) - f(x)}{h}$ for some $h > 0$ and prove that the limit as $h \to 0$ exists (which is the basic definition of being differentiable). – Gâteau-Gallois Sep 25 '19 at 14:11
  • If it is indeed $|x+y|$ on the right-hand side, you might still be able to obtain something on some subset of $\mathbb{R}$ from the inequality $| |x| - |y| | \leq |x - y|$ – Gâteau-Gallois Sep 25 '19 at 14:14
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    @MartinR Yes. Put in $y=-x$. Then, $f(x)=f(-x)$ for all $x\in\mathbb{R}$. From this, we have $\vert f(x)-f(y)\vert=\vert f(x)-f(-y)\vert\le\tfrac{1}{2}\vert{x+(-y)}\vert=\frac{1}{2}\vert{x-y}\vert$ for all $x,y\in\mathbb{R}$. So, $f$ is continuous uniformly on $\mathbb{R}$ since it satisfies the Lipschitz condition. – AnonyMath Sep 25 '19 at 15:38
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    I see. Then it is differentiable almost everywhere: https://math.stackexchange.com/questions/2609765/is-a-lipschitz-function-differentiable – Martin R Sep 25 '19 at 16:04

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The limit $$\lim_{x\rightarrow y}\frac{f(x)-f(y)}{x-y}$$ need not exist everywhere (all that you know is that this is uniformly bounded), so differentiability is not guaranteed. You can't generally prove that contractions are differentiable on the whole real line, such as $\frac{1}{2}|x|.$ It really boils down to differentiability of Lipschitz functions, which is true almost everywhere. This is the content of Rademacher's theorem: https://en.wikipedia.org/wiki/Rademacher%27s_theorem.

cmk
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