Counter example which violate uniform continuity...

Question

For a given continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$, define a sequence of function $\{f_n\}_{n\in \mathbb{N}}$ by $$f_n(x):=f\Big(x+\frac{1}{n}\Big).$$ Now if $f$ is uniformly continuous, then I can show that {$f_n$} converges to $f$ uniformly. At this point, I was thinking about the converse, i,e; if {$f_n$} defined above converges to $f$ uniformly, does that imply the uniform continuity of $f$.
I think the answer is negative, but I am not getting any counter example for that. Does anyone have any counter example?

Answer 1

Try $f:x\mapsto\sqrt{|x|}$. It is not uniformly continuous, having an unbounded derivative at $0$. But the $f_n$ do converge uniformly to $f$.

Answer 2

Take $$f_n (x)=x^2+\frac {1}{n+1} $$

$(f_n)$ converge uniformly to $$f:x\mapsto x^2$$ at $\Bbb R $ but $f $ is not uniformly continuous at $\Bbb R $ since

$$\lim_{n\to+\infty}f (n+\frac{1}{n})-f (n)\ne 0 $$

If $f_n $ are continuous and if the convergence is uniform at a compact $[a,b] $, then the limit function is Uniformly continuous at $[a,b] $.