Let $f_n:\mathbb{R}\to\mathbb{R}$ be a sequence of uniformly continuous functions. Assume that $f_n$ converges uniformly on all bounded intervals $[a,b]$ to a function $f$, i.e. $\displaystyle\lim_{n\to \infty}\sup_{a\leq x\leq b}|f_n(x)-f(x)|=0$ for all $a<b$. Is the limit $f$ also uniformly continuous ?
In the case where the convergence is uniform on the whole real line, I can prove that the limit is also uniformly continuous.
No, in general, the locally uniform limit of uniformly continuous functions is not uniformly continuous. For an example, consider
$$h_n(x) = \begin{cases} -n &, x \leqslant -n \\ x &, -n < x < n\\ n &, x \geqslant n\end{cases}$$
and
$$f_n(x) = h_n(x)\cdot x.$$
Then all $f_n$ are uniformly continuous, and the convergence is uniform on all compact subsets of $\mathbb{R}$, but the limit function $f(x) = x^2$ is not uniformly continuous.
You get a uniformly continuous limit if the sequence is uniformly equicontinuous, for example. (The uniform convergence of $f_n$ on all of $\mathbb{R}$ implies the uniform equicontinuity of $(f_n)$, so that is a special case of this sufficient criterion.)