Does pointwise convergence to a continuous function on a closed interval imply uniform convergence? [duplicate]

Let's suppose that $(f_n)_{n}$ is a sequence of continuous functions that converges pointwise to a continuous function $f(x)$ on a closed interval $[a, b]$. Is then the convergence uniform, too?

If it is so, how do you prove it? If it isn't, could you give a counterexample, please?

My attempt

I managed to write down the hypothesis in symbolical terms, but could not go beyond that:

• continuity of the sequence ($x_0\in[a, b]$):

$\forall n, \ \forall \epsilon >0 \ \ \exists \delta>0 \ : \ |x-x_0|<\delta \Rightarrow |f_n(x)-f_n(x_0)|<\epsilon$

• continuity of the limit function ($x_0\in[a, b]$):

$\forall \epsilon >0 \ \ \exists \delta>0 \ : \ |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon$

• pointwise convergence:

$\forall x\in[a, b], \ \forall \epsilon >0 \ \ \exists n_\epsilon>0 \ : \ n\geq n_\epsilon \Rightarrow |f_n(x)-f(x)|<\epsilon$

The thesis should be:

$\forall \epsilon >0 \ \ \exists n_\epsilon>0 \ : \ n\geq n_\epsilon \Rightarrow |f_n(x)-f(x)|<\epsilon \ \ \forall x\in[a, b]$

Note

Please do not bring sequences such as that of $x^n (x\geq 0)$ as a counterexample, because they are not counterexamples:

$\text{for } n \rightarrow \infty , \ x^n \rightarrow f(x)= \begin{cases} 0 & \text{if } 0 \leq x<1 \cr 1 & \text{if } x=1 \end{cases}$

so there is a pointwise convergence to $f(x)$ on $[0, 1]$; but $f(x)$ - the limit function - is not continuous, then this example lacks the conditions for the theorem to be applied.

• If the main point is to request feedback on your own proof attempt, you should use the (proof-verification) tag, see the tag-info. Jul 13, 2018 at 18:15
• I've added the proof-verification tag to your question. Please remove it if you feel that this was inappropriate. There are a number of possibly related questions: (1), (2), (3). Jul 13, 2018 at 18:16
• @XanderHenderson The above post does not contain any proof to be verified. Jul 13, 2018 at 21:25
• @Adayah Then this question offers nothing new which has not already been addressed on MSE before, and should be closed as a duplicate. Jul 13, 2018 at 21:26

4 Answers

Check out $f_n(x) = nx e^{-nx}$ on $[0, 1]$.

• I am sorry but in this case the function $f(x)$ to which $f_n(x)$ pointwisely converges in $[0, 1]$ is not continuous, so the theorem cannot be applied: $f(x)= \begin{cases} 0 & \text{if } 0 \leq x<1 \cr e^{-1} & \text{if } x=1 \end{cases}$
– user
Jul 13, 2018 at 17:09
• @MattiaPegolo: $f(1)=0$ not $e^{-1}$. Jul 13, 2018 at 17:20
• @MattiaPegolo . If $x>0$ then $e^{nx}=\sum_{j=0}^{\infty}(nx)^j/j!>(nx)^2/2!.$... An analogous example: Suppose $f_n(x)=0$ for $x\in [0,1/3n]\cup [1/n,1]$ but $f_n(2/3n)=n.$ Then for any $x\in [0,1]$ the set $\{n\in \Bbb N: f_n(x)\ne 0\}$ is finite. Jul 13, 2018 at 17:31
• @AbdelmalekAbdesselam I am sorry, I meant $f(x)= \begin{cases} e^{-1} & \text{if } x=0 \cr 0 & \text{if } 0<x \leq 1 \end{cases} \$ ; but even with this correction I think that I was wrong.
– user
Jul 14, 2018 at 9:29
• @AbdelmalekAbdesselam ...I was wrong, because $f(x)=0$ in $[0, 1]$.
– user
Jul 14, 2018 at 10:50

Another example is $$f_n(x)=\begin{cases} n^2x&0\le x\le\frac{1}{n},\\ 2n-n^2x&\frac{1}{n}\le x\le\frac{2}{n},\\ 0&\frac{2}{n}\le x\le 1 \end{cases}$$ The graph looks like a triangle of height $n$ and base $2/n$ so the $\int_0^1f_n(x)\mathrm{dx}=1.$ If the convergence were uniform, the integrals would go to $0$.

As Adayah's counterexample shows, this does not work without extra hypotheses. However, if the pointwise convergence is monotone, then a classical theorem of Dini shows the convergence is indeed uniform.

• But with the extra hypothesis required by the theorem above and only with those ones (no monotony needed), can it work?
– user
Jul 13, 2018 at 17:17
• @MattiaPegolo: no it doesn't because Adayah's counterexample proves it. Jul 13, 2018 at 17:22
• I think Dini's theorem requires that the functions themselves be monotone increasing. I know the Wikipedia article doesn't list this as a hypothesis, but it is used in the proof given there. Jul 13, 2018 at 17:39
• @saulspatz: No. Dini's theorem also works on compact spaces where the functions themselves being monotone doesn't even make sense. See these notes by my colleague Joel Feldman: math.ubc.ca/~feldman/m321/dini.pdf Jul 13, 2018 at 18:08
• Indeed, in the French system fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8mes_de_Dini this is known as Dini's 2nd theorem, although it seems to be due to Polya. Jul 13, 2018 at 19:36

It might help to think of the negation of uniform convergence. Given a sequence $f_n$ and a limit function $f$, $f_n$ does not converge uniformly to $f$ if there is an $\epsilon$ such that for all $N$, there exist $k$ and $x_k$ such that $k>N$ and $|f_k(x_k)-f(x)|>\epsilon$. That is, to disprove uniform convergence, it suffices to find an infinite sequence of pairs $(n,x_n)$ such that $f_k(x_k)$ does not go to zero. An example would be $f(x) = 0$, $x_k = 2^{-k}$ with $f_k$ chosen such that $f_k(x_k)$ is equal to a constant. To make this converge pointwise to $f$, it suffices that each $x$ has nonzero $f_k(x)$ for only finitely many $k$. Or, we can make it even simpler and have only one such $k$ for each $x$. This can be done by having $f_k$ be zero for all but an interval of length $10^{-k}$ centered at $x_k$. It is then simple to construct continuous functions that satisfy these conditions.

BTW, "converges pointwise" is more standard than "pointwisely converges", and "attempt" rather than "trial".