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If $f$ Continuous bounded on $\mathbb{R}$，and $$\lim_{h\to0}\sup_{x\in\mathbb{R}}\left|f(x+h)-2f(x)+f(x-h)\right|=0.$$ prove that $f$ uniformly continuous on $\mathbb{R}$. I try to find a $\delta>0$ to make it arbitrary $x_1,x_2\in \mathbb{R}$and$\left|x_1-x_2\right|<\delta$have$$\left|f(x_1)-f(x_2)\right|<\varepsilon.$$but I don't know how to use this complex limit expression.I hope you can give me some tips, thank you!

Hilbert1994
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    Hint: the complex limit expression = $f(x+h) - f(x) + (f(x-h)-f(x))$, so use triangle inequality on that. – Hank Igoe Jan 09 '21 at 14:24
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    @Hank Igoe Thank you for the tip,and I got $$\sup_{x\in\mathbb{R}}\left|f(x+h)-f(x)\right|+\inf_{x\in\mathbb{R}}\left|f(x-h)-f(x)\right|\leqslant\sup_{x\in\mathbb{R}}*\leqslant\sup_{x\in\mathbb{R}}\left|f(x+h)-f(x)\right|+\sup_{x\in\mathbb{R}}\left|f(x-h)-f(x)\right|$$.But forgive me for being stupid,still can't connect to consistent continuity... – Hilbert1994 Jan 09 '21 at 14:54
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    @Jean Marie Could you please help me with this problem, thank you – Hilbert1994 Jan 10 '21 at 14:29
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    See the same question here asked a day before you asked it for the first time !,10 days ago (https://math.stackexchange.com/q/3966553). – Jean Marie Jan 10 '21 at 18:53
  • @Jean Marie I'm sorry that I just came to MSE to ask that question, so it's against the rules – Hilbert1994 Jan 10 '21 at 19:01
  • @Jean Marie Now that question has been automatically removed by the community – Hilbert1994 Jan 10 '21 at 19:03
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    @Jean Marie I really hope you can guide me. Thank you – Hilbert1994 Jan 10 '21 at 19:12
  • I would llike but I have no true idea. As I see it that Jack d'Aurizio who is a top answerer has erased his answer, it means for me that it is a rather difficult question – Jean Marie Jan 10 '21 at 23:06
  • @Jean Marie Well, there's something wrong with part of his answer,thank you ! – Hilbert1994 Jan 10 '21 at 23:39

1 Answers1

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Let $M>0$ be an upper bound for $|f(x)|$, and $\epsilon > 0$ be given. From the given condition it follows that there is a $\delta > 0$ such that $$ |f(x+h)-2f(x)+f(x-h)| < \epsilon $$ for all $x \in \Bbb R$ and all $h \in (0, \delta)$. We will now show that $$ \tag{*} |f(x) - f(y) | \le 2 \sqrt{M \epsilon} $$ for all $x, y\in \Bbb R$ with $|x-y| < \delta$, which implies that $f$ is uniformly continuous.

In order to prove $(*)$ we use the following result from Prove or disprove $\lim\limits_{n \to \infty}\Delta x_n=0.$ :

Let $(x_n)$ be a sequence of real numbers, and let $\Delta x_n = x_{n+1}-x_n$, $\Delta^2 x_n = \Delta x_{n+1}-\Delta x_n$ be the first and second differences.

If $|x_n| \le M$ and $|\Delta^2 x_n| \le K$ for all $n$ then $|\Delta x_n|^2 \le 4MK$ for all $n$.

For fixed $x, y \in \Bbb R$ with $0 < |x-y| < \delta$ we can apply this result to the sequence $$ f_n = f(x + n(y-x)) $$ which satisfies $|f_n| \le M$ and $|\Delta^2 f_n| \le \epsilon$. It follows that $$ |f(x) -f(y) | = |\Delta f_0| \le 2 \sqrt{M \epsilon} $$ and this completes the proof.

Martin R
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