The contiuity of $f_n(x)=\int_0^{1/n} n f(x-u) du.$ Let $f:\mathbb R\to \mathbb R$ be continuous. Then, prove that　$f_n(x)=\displaystyle\int_0^{1/n} n f(x-u) du$ is continuous on $\mathbb R$ for each $n\in \mathbb N$.
I tried but I couldn't do well.

Fix $n \in \mathbb N$ and fix $a\in \mathbb R$. I'll prove $f_n$ is continuous at $a$.
Let $\epsilon >0$.
(I have to find some $\delta >0.$)
When $|x-a|<\delta,$
\begin{align}
|f_n (x)-f_n(a)|
&=\left| \int_0^{1/n} n f(x-u)-nf(a-u) du\right| \\
&\leqq n \int_0^{1/n} |f(x-u)-f(a-u)| du.
\end{align}
If I could find $\delta>0 $ s.t. $|x-a|<\delta \Rightarrow |f(x-u)-f(a-u)|< \epsilon,$ I come to conclusion.
I think I have to use the continuity of $f$, but both $f(x-u)$ and $f(a-u)$ includes variable $u$ so I don't know how I should use the continuity of $f$.
Thanks for your help.
 A: Set $g(x,y)=f(x-y)$. Let $a\in\mathbb R$ and $\varepsilon >0$. You have that $g$ is continuous on $[a-1,a+1]\times [0,1/n]$. In particular $g$ is uniformly continuous on $[a,-1,a+1]\times [0,1/n]$, and thus, there is $\delta=\delta (a,\varepsilon ) \in (0,1)$ s.t. for all $|x-a|\leq \delta $ and all $u\in [0,1/n]$,$$|g(x,u)-g(a,u)|<\varepsilon .$$
Therefore, $$|f_n(x)-f_n(a)|\leq \varepsilon ,$$
whenever $|x-a|\leq \delta $ as wished.
A: If $|x-a| <1$ and $u \in (0,\frac  1n)$ then  $x-u$ and $a-u$ both belong to $[a-1,a+1]$. $f$ is uniformly continuous on this interval. Now you can choose your $\delta$ so that $|f(x-u)-f(a-u)| <\epsilon$.
A: In this single-variable case, one can make a simple change of variables $t=x-u$ to get
\begin{align}
f_n(x):=\int_0^{1/n}nf(x-u)\,du=-\int_{x}^{x-\frac{1}{n}} nf(t)\,dt.
\end{align}
So, if one defines $F:\Bbb{R}\to\Bbb{R}$ as $F(x)=\int_0^xf(t)\,dt$, then $f_n(x)=-n[F(x-\frac{1}{n})-F(x)]$. Note that from the fundamental theorem of calculus (which we can apply since $f$ is continuous), $F$ is actually $C^1$ (because $F'=f$ is continuous). Therefore, each $f_n$ is also $C^1$.
