(Uniform limit theorem) Let $f_n:X\to Y$ be a sequence of continuous functions from the topological space $X$ to the metric space $Y$. If $(f_n)$ converges uniformly to $f$, then $f$ is continuous.

I tried to prove this theorem by my own for past two days. Eventually I gave up(saw solution).

There are three info given in this theorem. IMO it is “important”, in what order(else one get lost in the forest) one uses these info. In proof of Munkres, (1) use uniform convergence ,(2)use of continuity of $f_n$, it is also imp to use $B_d(f_n(x_0),\epsilon /3)$ nbhd of $f_n(x_0)$ and surprisingly we only use continuity of $f_N$, existence of $N$ depends on (1), (3) use triangle inequality to show $f(U)\subseteq B_d(f_n(x_0, \epsilon)$. One can also think of it as sequence of “idea”.

Before writing proof I knew, I will get existence of nbhd of $x_0$ by continuity of $f_n$ map. First half of the proof, I did the same thing as Munkres proof. But I was taking general $W\in \mathcal{N}_{f_n(x_0)}$, where $n\geq N$. From this point everything becomes super fuzzy/perplexing. I didn’t notice to use $d(f(x),f(x_0))\leq d(f(x),f_N(x))+ d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))$ fact. However, even after watching this inequality, after few min, I can’t “precisely” write the inequality! One can’t switch stuff in that inequality, for instance like $d(f(x),f_N(x_0))+d(f_N(x_0),f_N(x))$ instead of $d(f(x),f_N(x))+d(f_N(x),f_N(x_0))$. Writing each term “precisely” is important. But writing that inequality exactly the way given in Munkres proof is really really though provoking to me. I mean there are lots of indices.

If you have proved this theorem, you can share your thought process.

  • $\begingroup$ I have reopened your question. $\endgroup$ Feb 10, 2022 at 11:44
  • $\begingroup$ @JoséCarlosSantos Can you please share your thought(parameters you had in mind) when you decided to close this post? $\endgroup$
    – user264745
    Feb 10, 2022 at 11:46
  • $\begingroup$ It looked as if you were asking for a proof of the theorem that states that the uniform limit of a sequence of continuous functions is continuous. $\endgroup$ Feb 10, 2022 at 11:48
  • $\begingroup$ @JoséCarlosSantos The whole essence of this post is to extract main idea from the proof and try to understand pattern of the proof. Because pattern used in this theorem is very similar to proof of Ex 8, sec 21. In this post I discuss, one of the step in proof is very strange and hard(to me) to come up with. So how to make it seem natural. Proof of this theorem is connected with lots of exercise, that’s why I wanted to understand proof at a more deeper level. $\endgroup$
    – user264745
    Feb 10, 2022 at 11:58

1 Answer 1


You want to show that the uniform limit $f$ of a sequence $f_n$ of continuous functions is continuous.

We choose to prove continuity at some arbitrary point $x_0 \in X$ and we'll be done. So we're given $\varepsilon>0$ and we have to find a neighbourhood $U$ of $x_0$ so that

$$\forall x: x \in U \to d(f(x), f(x_0) < \varepsilon$$

We know nothing about $f$ except that it's the uniform limit of the $f_n$ so we know

$$\forall \varepsilon>0: \exists N: \forall n \ge N: \forall x \in X: d(f_n(x), f(x)) < \varepsilon\tag{1}$$

which gives is a way to get $d(f_n(x_0), f(x_0)$ as small as we need for lots of values of $n$.

We want to use the triple triangle inequality that you also gave as the "the idea", for some fixed index $N$ and an arbitrary $x \in X$:

$$d(f(x_0), f(x)) \le d(f(x_0), f_N(x_0)) + d(f_N(x_0), f_N(x)) + d(f_N(x), f(x)) \tag{2}$$

This is just from choosing $f_N(x)$ and $f_N(x_0)$ as intermediate points from $f(x_0)$ to $f(x)$ (note that these points repeat in the consecutive terms).

Now, the first and the last term of the RHS can be made arbitrarily small (at the same time!) by applying $(1)$. The middle term $d(f_N(x_0), f_N(x))$ can be made small by choosing $x$ close to $x_0$ and applying the continuity of the single function $f_N$ at $x_0$. So that allows us to finish the proof:

we pick $N$ so that $(1)$ holds for $\frac{\varepsilon}{3}$ (we have three terms whose total must become $<\varepsilon$ so that's fairly logical to do, IMO).

Finally we note that $f_N$ (now a definite function) is continuous (as all $f_n$ are; note that we acutally only need infinitely many $f_n$ to be continuous so we always have one with index large enough, but that's a minor remark) and so we can apply the continuity at $x_0$ to get $U$ so that

$$\forall x: x \in U \to d(f_N(x), f_N(x_0) < \frac{\varepsilon}{3}\tag{3}$$

And now for that $U$ and the chosen $N$ we combine $(1)$ (just for $N$, not all $n \ge N$) with $(3)$ into $(2)$ to get

$$\forall x \in U: d(f(x), f(x_0) < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$$ which is as required!


I hope my build up of the proof (what can we control, and in what way etc.) shows how to come up with such proofs by yourself. The "creative" step is to use some $f_N$ values as the intermediate points in the triangle inequality. Once you have that, there is only one way forward.

  • $\begingroup$ Thank you for the answer. In the end of chapter I always get excited(to do next chapter), and mess up things. Yup, existence of $N$ depends on (1). Perhaps in triangle inequality we can take $n\in \Bbb{N}$(fix it) s.t. $n\geq N$. We have two intermediate points $f_N(x)$ and $f_N(x_0)$. As I said in Que section, one can’t switch first term(for instance) on RHS by $d(f(x_0),f_N(x))$ instead of $d(f(x_0),f_N(x_0))$. By reading your post, there is no shortcut to write triangular inequality. One should have presence of mind. IMO in this kind of problem, one gets easily lost in the forest(of info). $\endgroup$
    – user264745
    Feb 11, 2022 at 7:50
  • 1
    $\begingroup$ @user264745 the $n$ in the triangle inequality must be for two fixed points, so that’s why you pick the fixed index $N$. We know that particular index will work as $N \ge N$. $\endgroup$ Feb 11, 2022 at 11:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .