# Prove or disprove that $\lim_{n→∞}\sup_{x∈R}⁡f_n (x) =\sup_{x∈R}⁡f(x)$

Let $\{f_n\}$ be a sequence of function defined and bounded on $\mathbb R$, and suppose that $\{f_n\}$converges uniformly to $f$ on every finite interval $[a,b]$. Prove or disprove that $$\lim_{n→∞}\sup_{x∈R}⁡f_n (x) =\sup_{x∈R}⁡f(x).$$

I know that $\{f_n\}$converges uniformly to $f$ that mean $\lim_{n\to \infty}\sup |f_n -f| \to0$. But it's not necessary that $\lim_{n→∞}⁡\sup_{x∈R}⁡f_n (x) =\sup_{x∈R}⁡f(x)$.

I don't think this is true, But I can't find counter example either.

• Isn't $f_{n}(x) = I(x=n)$ a counter-example? – madprob Feb 3 '14 at 21:08

Let $$f_n = \left\{ \begin{array}{cc} 1 & \text{ if } x \ge n \\ 0 & \text{ if } x < n \\ \end{array} \right.$$
Given any finite interval $[a,b]$ and any $\epsilon > 0$, just pick $N > b$ and we have that $f_n = 0$ on $[a,b]$ for $n > N$. Thus $f_n \to {\textbf{0}}$ uniformly on $[a,b]$.
• @sos440 thanks :D – 6005 Feb 3 '14 at 21:51