If a sequence $(f_n)$ converges pointwise to the same function $f$ on all $\mathbb{R}$, does this imply that $(f_n)$ uniformly converges to $f$ on an interval of $\mathbb{R}$?
From what I understand, this should be false. I know from a theorem that if a sequence of functions converges uniformly to a function $f$, $f$ is continuous. However, the converse is false (I think). In other words, the fact that a sequence of functions converges to a continuous function does not guarantee that the sequence of functions is uniformly convergent.
If $(f_n)$ converges pointwise to the same function $f$ on all $\mathbb{R}$, wouldn't that simply mean that the $f$ is continuous? Using the same logic as the one in the converse of the theorem above, wouldn't that mean that the sequence of functions need not to be uniformly convergent?
I am not quite sure about this though. What are some examples or information that is relevant?
Best Answer
Let $f_n = 1_{[n,\infty)}$ (that is $f_n(x) = \begin{cases} 1, & x \ge n\\0, & \text{otherwise} \end{cases}$).
Then $f_n(x) \to 0$ for all $x$, but $f_n$ does not converge to 0 uniformly.
Another example, let $g_n(x) = \begin{cases} 0, & x < 0 \\ x^n, & x \in [0,1]\\1, & x > 1 \end{cases}$. Then $g_n(x) \to g(x)$, where $g(x) = \begin{cases} 0, & x < 1 \\1, & x \ge 1 \end{cases}$, which is not continuous. Furthermore, the convergence is not uniform.