Wrong with the following proof that seems to prove pointwise convergence implies uniform convergence

functional-analysismetric-spacesreal-analysisuniform-convergence

I know uniform convergence implies pointwise convergence and not the other way around. Then what is wrong with the following proof that seems to prove that pointwise convergence implies uniform convergence?

Lef $\{f_n(x)\}_n$ be a pointwise convergent sequence of functions in $C[a,b]$. Let the pointwise limit be f(t) for every t. Then $\{f_n(x)\}_n$ is Cauchy that is $\forall \epsilon >0 ,\exists N$ such that if $n,m>N$, $|f_n(x)-f_m(x)|<\epsilon/2$

Passing to the sup:
$\text{Sup}_{x \in [a,b]}|f_n(x)-f_m(x)|\le\epsilon/2$

Taking the limit $m\to \infty$ in the last inequality : $\text{Sup}_{x \in [a,b]}|f_n(x)-f(x)|\le\epsilon/2<\epsilon $ Which is the definition of uniform convergence.

The definitions I am using:

$f_n:I\subset \mathbb{R}\to \mathbb{R}$,$f:A\subset I\to \mathbb{R}$

  1. $f_n \to f$, that is $f_n$ converges pointwise to f if $\lim_{n\to\infty}f_n(x)=f(x) \forall x \in A$
    $\iff \forall x \in A \forall \epsilon>0 \exists N $such that $|f_n(x)-f(x)|<\epsilon , \forall n>N$
  2. $f_n$ converges uniformly to $f$ if $\lim_{n\to\infty}\text{Sup}_{x \in A}|f_n(x)-f(x)|=0 $
    $\iff \forall \epsilon>0 \exists N $such that $\text{Sup}_{x \in A}|f_n(x)-f(x)|<\epsilon, \forall n>N$.

I got this idea from (**) below, in the proof of the completeness of $C[a,b]$ with respect to the uniform metric that my professor did. There is a step in which she took the limit $m\to \infty$ of the inequality: $\text{Sup}_{t \in [a,b]}|f_n(t)-f_m(t)|< \epsilon $.

The proof that my professor of the completeness of $C[a,b]$ with respect to the uniform metric did was:

Let $\{f_n(t)\}_n$ be a Cauchy sequence in $C[a,b]$ Then $\forall \epsilon >0 ,\exists N$ such that if $n,m>N$, $|f_n(t)-f_m(t)|<Sup|f_n(t)-f_m(t)|<\epsilon$ …(*)

For each fixed $t\in[a,b]$: $\{f_n(t)\}_n$ is a Cauchy sequence in $\mathbb{R}$.
Because $\mathbb{R}$ is complete, $\exists \lim_{n\to\infty}f_n(t) = f(t) \in \mathbb{R}$

So we have built a function $f:[a,b]\to \mathbb{R}$, now we have to see that it is the limit of the f_n with respect to the uniform distance defined on $C[a,b]$ and that it is continuous.

In order to find that the function f(x) built as the pointwise limit of f_n(x) for each x, is also the limit with respect to the uniform distance, (that is that $\sup_{x \in [a,b]}|f_n(x)-f(x)| \to 0)$we take the limit $m\to \infty$ in (*):$\sup_{x \in [a,b]}|f_n(x)-f_m(x)| )<\epsilon$ : $\text{Sup}_{x \in [a,b]}|f_n(x)-f(x)|\le \epsilon$..(**)

That is we have $\forall \epsilon >0 ,\exists N$ such that if $n,m>N$, $\text{Sup}_{x \in [a,b]}|f_n(x)-f(x)|\le \epsilon$ (this can be made into <\epsilon working with \epsilon/2 at the beginning), which means $f$ is the limit of the f_n with respect to the uniform distance defined on $C[a,b]$.

To show $f \in C[a,b]$:
$|f(t)-f(t_0)|<|f(t)-f_n(t)|+|f_n(t)-f_n(t_0)|+|f_n(t_0)-f(t_0)| <3\epsilon$, where $|f(t)-f_n(t)|<\epsilon $and $|f(t_0)-f_n(t_0)|<\epsilon$ because of the pointwise convergence of $f_n$ and $|f_n(t)-f_n(t_0)|<\epsilon $ because of the continuity of $f_n$ at $t_0$

Best Answer

We can make a counterexample graphically. Define a function $f_n(x)$ on the interval $[0,1]$ by plotting $y = f_n(x),$ where the graph of this function consists of a straight line segment from $(0,0)$ to $(2^{-n},1),$ a straight line segment from $(2^{-n},1)$ to $(2(2^{-n}),0),$ and finally a straight line segment along the $x$ axis connecting $(2(2^{-n}),0)$ to $(1,0).$

For large $n$ this function is a constant $0$ over most of the interval, but there is a sharp triangular "bump" at the left end of the interval whose highest point is at $y=1.$

Now at $x=0,$ we always have $f_n(x) = f_n(0) = 0.$ The sequence $0,0,0,\ldots$ converges to zero, so we have convergence at $x=0.$

For any $x > 0,$ you can always find $N$ such that $2(2^{-N}) < x.$ Then for any $n > N$ you have $f_n(x) = 0,$ so we have convergence at that value of $x.$

So we have pointwise convergence to the function $f(x) = 0$ on $[0,1].$

But no matter how large you make $n,$ there is still a triangular bump at the left end of the graph and the highest point on that bump still hits the line $y=1.$

So consider $\epsilon = \frac12.$ For what $n$ is it the case that $\sup_{x \in [0,1]} \lvert f_n(x)-f(x)\rvert < \epsilon = \frac12$?

The answer is there is no such $n,$ because for every $n,$

$$ \sup_{x \in [0,1]} \lvert f_n(x)-f(x)\rvert = 1. $$

Therefore we do not have uniform convergence.

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