uniform convergence of the composition of two sequence of functions

Question

Let $(f_{n}:\mathbb{R}\to\mathbb{R})$ be a sequence of continuous functions with uniform limit $f:\mathbb{R}\to\mathbb{R}$. Now, for each $n\in\mathbb{N}$, define $g_{n}:\mathbb{R}\to\mathbb{R}$ by $$g_{n}(x)=f_{n}\left(x+\frac{1}{n}\right).$$

I proved that $g_{n}\to f$ pointwise. Is it true that $g_{n}\to f$ uniformly?

I don't think so, but i failed to create the counterexample.

Give some advice! Thank you!

Answer 1

Not even true in a simple case. Take $f_n(x)=x^2$ then $$|g_n(x)-f(x)|=\left|2\frac{x}{n}+\frac{1}{n^2}\right|$$ For $n>1$ one has: $$\Vert g_n(x)-f(x)\Vert_\infty\geq \left|2\frac{n^2}{n}+\frac{1}{n^2}\right|\geq 2n >2 $$ So uniform convergence can be forgotten.

Answer 2

no, this is not true. Consider $f(x) = \exp(x^2)$ and $f_n(x) = f(x) + \frac{1}{n}$ Then

\begin{eqnarray}f_n(x+1/n) -f(x)& =& \exp(x^2 + 2x/n + 1/n^2) + \frac{1}{n}- \exp(x^2)\\ &=& \exp(x^2) (\exp(2x/n + 1/n^2)-1) + \frac{1}{n}\end{eqnarray}

For each $n$ you will find $x$ such that this is larger than $1$ in absolute value.