I want to prove the following:

If we have a sequence of functions $\{f_n\}$ that converges uniformly on $E\subset X$ in a metric space X, and each $f_n$ is continuous on E, is the limit function $f(x)=\lim_{n\to\infty}f_n(x)$ also continuous?

Here is what I've tried:

Let's fix $\epsilon>0$. Since we know that the sequence of functions $\{f_n\}$ is uniformly convergent on E, we can find an integer N so that $n\geq N $ implies

$$\mid f_n(x) - f(x) \mid < \frac{\epsilon}{3}$$

for all $x\in E$.

Additionally, we know that each of these functions $f_n$ is continuous, so for all $\epsilon>0$ we can find a $\delta>0$ so that

$$\mid f_n(x) - f_n(p)\mid<\frac{\epsilon}{3}$$

if $\mid x -p\mid < \delta$ for all $p$ in $E$.

Then the following inequality $$\mid f(x)-f(p)\mid ~\leq ~\mid f(x) - f_n(x)\mid + \mid f_n(x)-f_n(p)\mid + \mid f_n(p) - f(p)\mid $$

shows that for $n \geq N$, $$\mid f(x)-f(p)\mid ~<~ \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3 ~}=~ \epsilon$$ whenever $\mid x - p \mid < \delta$.

Therefore, we conclude that the limit function $f(x)=\lim_{n\to\infty}f_n(x)$ is continuous on $E$ given the following conditions.

Is this correct? Thanks in advance

  • $\begingroup$ It's (correct, and) the first occurence of the $\epsilon / 3$ type argument, of which you will see a few more as you go across your mathematical education. $\endgroup$ Commented May 19, 2018 at 5:11

1 Answer 1


The core ideas are correct, but there are some mistakes.

First off, $|f(x)-f(p)|$ is meaningless on a metric space (unless you are defining that symbol to mean $d(f(x),f(p))$, which I don't think is the case and is also a contrived thing to do). That said, changing this is almost irrelevant to the proof. It is not clear if the image is in $\mathbb{R}$ (or a normed space), which would make this valid, or a general metric space. Nevertheless, you say $|x-p|$ later in the proof, which suffers from the same issue and $x,p$ are definitely in a metric space not assumed to be $\mathbb{R}$ or a normed space by your hypotheses.

Second, it would be good to state, where it is appropriate, that you have fixed a $p \in E$. You don't do that, and it makes the fragment

(...) so for all $\epsilon>0$ we can find a $\delta>0$ so that

$$\mid f_n(x) - f_n(p)\mid<\frac{\epsilon}{3}$$

if $\mid x -p\mid < \delta$ for all $p$ in $E$.

come out a little strangely written. Take note also that the $\delta$ depends on $n$ and $p$ here. It would be better if you started the proof by

Let $p \in E$ be a fixed arbitrary point of $E$. We need to prove that $f$ is continuous at $p$. Therefore, let $\epsilon >0$. (...)

This dependence of $\delta$ with respect to $n$ haunts you later when you want to use the triple triangle inequality. This is relatively minor, but important to keep track of and can cause mistakes in other contexts.

You want $\delta$ to depend only on $p$, of course. So what you can do is not let that $n$ be free. Pick a specific $N$ such that $d(f_N(x),f(x))<\epsilon/3$ for all $x \in E$. For that specific $N$, let $\delta>0$ be such that $d(f_N(x),f_N(p))<\epsilon/3$ for all $x$ which are $\delta$-close to $p$. And then we have that \begin{align*} d(f(x),f(p)) &\leq d(f(x),f_N(x))+d(f_N(x),f_N(p))+d(f_N(p),f(p)) \\ &< \epsilon/3+\epsilon/3+\epsilon/3 \\ &=\epsilon, \end{align*} for all $x$ satisfying $d(x,p)<\delta$. $$ $$


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