Let $f:\mathbb{R}\to \mathbb{R}$ such that $f \in C^2$ and $f''$ is bounded. Show that the sequence of functions defined by $f_n(x)=n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)$ converges uniformly to $f'$.
I would like to know if my proof holds, please and have a feedback on redaction.
First of all as $f''$ is bounded, then $f'$ is Lipschitz on $\mathbb{R}$. This implies that $f'$ is uniformly continuous on $\mathbb{R} $ $\forall \epsilon>0 \ \exists \delta>0 \ \forall x \in \mathbb{R}: |x-y|<\delta \implies |f'(x)-f'(y)|<\epsilon$
Consider now the interval $[x,x+\frac{1}{n}]$. As $f \in C^2$, then $f'$ is differentiable on $\mathbb{R}$ and $f$ is continuous on $\mathbb{R}$ (so on $[x,x+\frac{1}{n}]$) as well. By Mean Value Theorem we have there $\exists c_n\in[x,x+\frac{1}{n}]$: $f'(c_n)=n(f(x+\frac{1}{n})-f(x))$. By inserting RHS and LHS limit, we have: $f'(x)=\lim_{n\to \infty}n(f(x+\frac{1}{n})-f(x))$. The LHS limit is well defined (i.e exists) because $f \in C^2$. So, $f_n \to f'$ pointwise.
To show that $f_n\to f'$ uniformly, we have to show that $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x \in [x,x+\frac{1}{n}]: |f_n(x)-f'(x)|<\epsilon$.
But, $|f_n(x)-f'(x)|=|n(f(x+\frac{1}{n})-f(x))-f'(x)|=|f'(c_n)-f'(x)|<\epsilon$ So the convergence is uniform
Your proof is correct. You can shorten it by using Taylor's theorem: $$ f(x+ \frac 1n) = f(x) + \frac 1n f'(x) + \frac{1}{n^2} \frac{f''(c)}{2} $$ for some $c$ between $x$ and $x+1/n$, which implies $$ \left| f_n(x) - f(x) \right| \le \frac{\max_{c \in \Bbb R} |f''(c)|}{2n} \, . $$ This proves not only the uniform convergence but also provides a concrete estimate on the error term.