Let $f$ be continuous on $\mathbb R.$
Define $\{f_n : \mathbb R \to \mathbb R \}_{n=1}^\infty \ $ by $f_n(x)= n \displaystyle\int_0^{1/n} f(x-u) \ du$.
Then, for all $x\in \mathbb R,$ $\displaystyle\lim_{n\to \infty} f_n (x)=f(x)$ holds. (I already proved.)
Then, determine whether $f_n$ converges to $f$ uniformly or not.
I think $f_n$ doesn't uniformly converge to $f$.
A counter example is $f(x)=x^2.$(This is continuous on $\mathbb R.$)
Let $f(x)=x^2$, and suppose $f_n(x)=n \displaystyle\int_0^{1/n} f(x-u) du$ converges to $f$ uniformly .
Then, there is $N \in \mathbb N$ s.t. $$n \geqq N \Rightarrow |f_n (x)-f(x)|<\frac{2}{3} \mathrm{\ for \ all \ } x \in \mathbb R.$$
Thus, $|f_N(N)-f(N)|<\dfrac{2}{3}$ must hold.
However, \begin{align} |f_N(N)-f(N)| &=N \left|\int_0^{1/N} f(N-u)-f(N) du \right| \\ &=N \left|\int_0^{1/N} (N-u)^2-N^2 du \right| \\ &=N \left|\int_0^{1/N} -2Nu+u^2 du \right| \\ &=1-\frac{1}{3N^2}\\ &\geqq 1-\frac{1}{3}\\ &=\frac{2}{3}. \end{align} This is contradiction.
Therefore, I think $f(x)=x^2$ is a counter example.
Is my idea correct ?