If $f$ is differentiable ,and $ \forall x \in \mathbb R $ and $\forall h >0 $ hold $\mid {f(x+h)-f(x-h)}\mid <h^2$.Prove that $f$ is constant.I tried to use Lagrange theorem.
If $f$ is differentiable and $\forall x \in \mathbb R $ and $\forall h >0 ( \mid {f(x+h)-f(x-h)}\mid
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2Add your scratch work! – G_D Jan 10 '18 at 23:33
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The result holds even if we don't assume differentiability but the proof is non obvious. See https://math.stackexchange.com/a/2533349/72031 – Paramanand Singh Jan 11 '18 at 05:14
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Hint: divide the inequality by $h$, can you recognize the left hand side? Remember also that this "new" inequality still holds for every $h > 0$.
GaC
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You mean to do that and get$ 2 *$derivative of x and then use limes when h aproach 0? – Jan 10 '18 at 23:41
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We get $\mid {f(x+h)-f(x-h)}\mid <h^2\iff\left|\frac{ {f(x+h)-f(x)}}{h}+\frac{f(x-h)-f(x)}{-h}\right|<h$ for all $x\in\mathbb{R}$ and $h>0$, which means that $2|f'(x)|=\left|\lim_{h\downarrow 0}\frac{ {f(x+h)-f(x)}}{h}+\lim_{h\uparrow 0}\frac{{f(x+h)-f(x)}}{h}\right|=\left|\lim_{h\downarrow 0}\frac{ {f(x+h)-f(x)}}{h}+\lim_{h\downarrow 0}\frac{f(x-h)-f(x)}{-h}\right|=\lim_{h\downarrow 0}\left|\frac{ {f(x+h)-f(x)}}{h}+\frac{f(x-h)-f(x)}{-h}\right|\leq \lim_{h\downarrow 0}h=0$ because by differentiability this right limit equals the left limit. So $f'(x)=0$ for all $x\in\mathbb{R}$, which means that $f$ is constant.
The Phenotype
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Your $y$ depends on $h$ also so that the derivative definition does not apply in the manner you have written. - 1 – Paramanand Singh Jan 10 '18 at 23:55
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1it would indeed be better to rewrite it $\frac{f(x+h)-f(x)+f(x)-f(x-h)}{h}\to 2f'(x)$. – zwim Jan 11 '18 at 00:14
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1@ParamanandSingh, zwim Thanks for pointing the mistake out and your suggestion! – The Phenotype Jan 11 '18 at 00:26
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A more direct proof.
Take $x_1$ and $x_2 > x_1$ arbitrarily. Let $h = (x_2 - x_1)/(2N)$ and divide the interval $(x_1 , x_2)$ into $N $ equally long intervals. We then have
$$ |f(x_1) - f(x_2 )| < N * h^2 = (x_2 - x_1)/ (4N ) $$
for arbitrary $ N$. As $N \rightarrow \infty $, we get $f(x_1 ) = f(x_2 )$.
Beamer
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