Extending uniform convergence of analytic functions on larger domains

Question

Let $f_k, f: ]-\infty , 1 [ \to \mathbb {R}$ be analytic functions. Suppose $f_k $ converges uniformly to $f $ on $]-\infty,0] $. Is it true that $f_k$ converges to $f$ on $]-\infty, \epsilon [$ for some $\epsilon >0$?

I fail to see a counterexample, but I have never seen this before. Thank you for your help.

Answer 1

Define $f\equiv 0,$ and

$$f_n(x) = \frac{1}{(1+(x-1/n)^2)^{n^3}}, \,\,n=1,2,\dots$$

Then $f_n\to 0$ uniformly on $(-\infty,0].$ That's because $f_n(0)\to 0$ and $f_n(0)=\sup _{(-\infty,0]}f_n.$

But note $f_n(1/n)=1$ for all $n,$ which implies uniform convergence to $0$ fails on $(-\infty,\epsilon)$ for every $\epsilon >0.$