# If $(f_n)$ converges pointwise to $f$ on $\mathbb{R}$, does this imply that $(f_n)$ uniformly converges to $f$ on an interval of $\mathbb{R}$?

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?

• Also, this is a vague question. Let me know how to correct it. Commented Dec 18, 2013 at 5:21

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.

• I am not sure whether I understand the function. Isn't $f_n$ just equal to a constant function? Commented Dec 18, 2013 at 5:23
• No. I have elaborated the definition slightly. Commented Dec 18, 2013 at 5:25
• Is the $x\geq n$ condition for the $1$ or the $0$? Commented Dec 18, 2013 at 5:25
• It is for the 1. Commented Dec 18, 2013 at 5:27
• Fix $x$. Then if $n \le x$ we have $f_n(x) = 1$, if $n >x$, we have $f_n(x) = 0$. So, in your example, $f_{1000}(999) = 0$, since $1000>999$. Commented Dec 18, 2013 at 5:33

You are correct that the claim is false: there are sequences of continuous functions that converge to nowhere-continuous functions. An explicit way to do this is to construct a sequence of continuous functions that converge to $\chi_{\mathbb{Q}}$, the function which is $1$ on the rationals and zero otherwise.

What is true, however, is a related idea called Egorov's Theorem.