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

Question

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?

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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:

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

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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.

Answer 1

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

Answer 2

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.

Answer 3

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 $0$.

Answer 4

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".