This is a self-posed question, so I do not know the answer and I would like to know what you think about.

Let $f,f_n:[0,1]\to \mathbb R$ be continuous functions. Assume that for every sequence $(x_n)_{n\in \mathbb N}$ converging to a point $x\in [0,1]$, we have $$\lim_n f_n(x_n)=f(x).$$ Can we say that $f_n$ converges uniformly to $f$ in $[0,1]$?

First of all, do you think the question is well-posed? I'm stuck in thinking about it and I don't manage to prove it neither to find a counterexample. Thanks for your help.

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    $\begingroup$ The result is true and this might be one of the few situations where going for a proof by contradiction really helps. Hence, let us assume that $(f_n)$ does not converge uniformly to $f$. This means that... $\endgroup$
    – Did
    Commented Aug 13, 2012 at 22:29
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    $\begingroup$ The mode of convergence you ask about is well-established and is called continuous convergence and I think @did's hint should be very helpful (don't forget to use compactness of $[0,1]$!) $\endgroup$
    – t.b.
    Commented Aug 13, 2012 at 22:41
  • $\begingroup$ @did: I tried to follow your advice but is kind of stuck. $\endgroup$ Commented Aug 13, 2012 at 23:11
  • $\begingroup$ How did you write down the hypothesis that $(f_n)$ does not converge uniformly to $f$? If you have a problem with this, you may first write down the hypothesis that $(f_n)$ does converge uniformly to $f$, then negate this. So: what does it mean to say that $(f_n)$ converges uniformly to $f$? $\endgroup$
    – Did
    Commented Aug 13, 2012 at 23:31
  • $\begingroup$ Hopefully fixed by now, but may still have problems. $\endgroup$ Commented Aug 13, 2012 at 23:33

2 Answers 2


Yes, this is true:

Suppose that the convergence $f_n \to f$ is not uniform but that $f(x) = \lim_{n\to\infty}f_n(x_n)$ for every convergent sequence $x_n\to x$ in $[0,1]$.

After passing to a subsequence of the $f_n$ we may assume that there is $\varepsilon \gt 0$ such that for all $n \in \mathbb{N}$ we have $\sup_{x \in [0,1]} \lvert f(x) - f_n(x) \rvert \gt \varepsilon$. This implies that there is a sequence of points $x_n \in [0,1]$ such that $\lvert f(x_n) - f_n(x_n) \rvert \gt \varepsilon$ for each $n$. After passing to further subsequences, we may assume that $x_n \to x \in [0,1]$ by compactness of $[0,1]$. But then we're in trouble: our hypotheses tell us that $$\tag{$\ast$} \lim_{n\to \infty} \lvert f(x) - f_n(x_n)\rvert = 0, $$ which is incompatible with $f(x_n) \to f(x)$ and $\lvert f_n(x_n) - f(x)\rvert \gt \varepsilon$.

More formally: continuity of $f$ ensures together with $x_n \to x$ that there is $N$ such that $\lvert f(x) - f(x_n)\rvert \lt \varepsilon/2$ for all $n \geq N$ so that the triangle inequality gives us for all $n \geq N$ that $$ \begin{align*} \lvert f_n(x_n) - f(x) \rvert &= \lvert [f_n(x_n) - f(x_n)] - [f(x) - f(x_n)]\rvert \\ & \geq \underbrace{\lvert f_n(x_n) - f(x_n)\rvert}_{\gt \varepsilon} - \underbrace{\lvert f(x) - f(x_n)\rvert}_{\lt \varepsilon/2} \\& \gt \varepsilon/2 \gt 0, \end{align*} $$ which is absurd in view of $(\ast)$.


  1. The mode of convergence $f_n(x_n) \to f(x)$ for all $x_n \to x$ is called continuous convergence. Continuity of the $f_n$ is never used in the above argument. In fact, $f_n(x_n) \to f(x)$ for all $x_n \to x$ ensures continuity of $f$ independently of the continuity of the $f_n$ (exercise).

  2. You can find a generalization of the above result and a more detailed discussion in the section Compact and continuous convergence (Chapter 3, §1, section 5) starting on page 98 in R. Remmert's Theory of complex functions. The precise result reads that $f_n$ converges to $f$ continuously if and only if $f$ is continuous and $f_n \to f$ uniformly on compact sets. More details in loc. cit. (don't miss the historical notes preceding and following that section!).

  • $\begingroup$ I posted this answer simultaneously with did's hint but I then deleted it to give the OP some time to think about it. I hope it's fine to show it now. $\endgroup$
    – t.b.
    Commented Aug 14, 2012 at 6:53
  • $\begingroup$ Basically, I have no opinion here. Let me mention that Romeo probably did not even see my two comments (the first one addressed to him and the second one answering @user32240's query), nicely completed by your own comment (mentioning compactness). $\endgroup$
    – Did
    Commented Aug 14, 2012 at 8:15
  • $\begingroup$ This is a very clear proof. Thank you. $\endgroup$ Commented Aug 14, 2012 at 12:07
  • $\begingroup$ Is the converse true? That is, if we are given that $ f_n (x_n)$ converges UNIFORMLY to $f(x)$ and the sequence of points $(x_n) \rightarrow x$, then does $f_n(x_n) \rightarrow f (x)$ [pointwise convergence]? If so, can anyone help me out by providing a proof. I am having a hard time thinking that the converse is true. $\endgroup$
    – user43901
    Commented Mar 13, 2013 at 0:17
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    $\begingroup$ @t.b.: $\{f_n\}$ does not converge to $f$ implies $\exists \epsilon >0$ such that for some subsequence $\{f_{n_k}\}$ of $\{f_n\}$ , there is a sequence $\{x_k\}$ in domain such that $|f_{n_k}(x_k)-f(x_k)| \ge \epsilon$ ; but how does one ensure that $|f_n(x_n)-f(x_n)|\ge \epsilon$ , as used in the answer ? $\endgroup$
    – user228168
    Commented Jul 12, 2015 at 15:03

There is a classical theorem by Dini that says this is possible when $f_{i}$ are monotonically increasing.

We now assume $f_{n}$ converges to $f$ pointwise and $f,f_{n}$ are all continuous but the limit is not uniform. Thus there is some $\epsilon$ such that $\forall N$ there is some $n\ge N$ such that $$|f_{n}(x)-f(x)|> \epsilon$$ for some $x\in [0,1]$.

Now since $f_{i},f$ are uniformly continuous on $[0,1]$, for $\epsilon_{1}=\frac{\epsilon}{3}$ we have a list of $\delta_{i}$s such that $0< \delta_{i}\le 1$ and $|f_{i}(x)-f_{i}(y)|\le \epsilon_{1}$ for $|x-y|\le \delta_{i}$. Consider the interval $U_{i}=[0,\delta_{i}]$, I claim $\lim_{i\rightarrow \infty} \bigcap U_{i}\not=\emptyset$. Suppose $\cap_{i=N} U_{i}=\emptyset$, then $f_{i}$ would have nontrivial variation $$\lim_{x \rightarrow y} |f(x)-f(y)|=0$$ a fact contradicting they are being continuous. Thus let $U=[0,\delta_{A}]$ be the intersection, and let $\delta_{B}$ be the corresponding value for $f$. We choose $\delta=\min[\delta_{A},\delta_{B}]$.

We now assume the counterexample in the beginning. For fixed $y$ such that $|x-y|\le \delta$ we can choose $N$ large enough such that $$|f(y)-f_{n}(y)|\le \epsilon/3$$ for any $n\ge N$. If it turns out $N$ is greater than the initial $n$ we chose, we may switch to a different $n$ guaranteed to exist by the hypothesis.

We have the estimate that $$|f_{n}(x)-f(x)|\le |f_{n}(x)-f_{n}(y)|+|f_{n}(y)-f(y)|+|f_{n}(y)-f_{n}(x)\le \epsilon$$ which contradicts the hypothesis we had earlier. So the convergence is uniform afterall.

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    $\begingroup$ In fact the desired result in the OP is true. $\endgroup$
    – t.b.
    Commented Aug 13, 2012 at 22:11
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    $\begingroup$ My example has serious problem at point 1. Let me try to modify it. $\endgroup$ Commented Aug 13, 2012 at 22:14
  • $\begingroup$ @t.b.:can you check if my 'proof' is correct? $\endgroup$ Commented Aug 13, 2012 at 23:35
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    $\begingroup$ No, this doesn't really work this way. 1) the OP assumes something much stronger than pointwise convergence. 2) By your definitions you certainly have $0 \in \bigcap_{i=N}^\infty [0,\delta_i]$ and you probably try to argue that there is $\varepsilon \gt 0$ such that $[0,\varepsilon]$ is in this intersection: so you're trying to establish equicontinuity of the $f_n$, right? 3) I don't understand what the sentence including "nontrivial variation" is supposed to mean. 4) Sorry I am lost in the last two paragraphs: what is the $x$ doing? How do you get from $|x-y|\lt \delta$ to uniform conv.? $\endgroup$
    – t.b.
    Commented Aug 14, 2012 at 6:07
  • $\begingroup$ See my answer for a detailed argument. $\endgroup$
    – t.b.
    Commented Aug 14, 2012 at 6:57

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