$\frac{n}{2}\int_{x-1/n}^{x+1/n} f(t)dt$ converges uniformly to $f$ where $f:\Bbb R\to \Bbb R$ is continuous

Question

Suppose $f$ is a continuous function $\Bbb R\to \Bbb R$, and for each $n\in \Bbb N$ define $f_n:\Bbb R\to \Bbb R$ by $f_n(x)=\frac{n}{2}\int_{x-1/n}^{x+1/n} f(t)dt$. It is clear that $f_n$ converges to $f$ pointwise, by the continuity of $f$. But does $f_n$ converges to $f$ also uniformly? The answer would be positive if $f$ is uniformly continuous, but I can't see whether the answer is still positive in the general case.

Answer 1

You can explicitly compute the value of this expression when $f(t)=e^{t}$ and see that the convergence is not uniform on $\mathbb R$.

Note that $e^{x} a_n$ can never tend to $0$ uniformly on $\mathbb R$ for sequence of positive numbers $a_n$. In this example $f_n(x)-f(x)$ has this form.

PS: I just noticed that $f(t)=t^{k}$ is also a counter-example for every $k >2$.

Answer 2

A correct counterexample is $f(x)=e^x$. Note $$e^x - n/2(\exp(x+1/n)-\exp(x-1/n))=e^x(1-n/2(\exp(1/n) - \exp(-1/n)))$$

and $1 - n/2(\exp(1/n) - \exp(-1/n)) \to 0$ as $n \to \infty$. Since $e^x \to \infty$ as $x \to \infty$, the $N$ required to force $e^x(1-n/2(\exp(1/n) - \exp(-1/n)))<\epsilon$ for $n \geq N$ very much depends on $x$.