How to prove this problem about uniform continuity 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!
 A: 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.
