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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 $t_0$

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  • $\begingroup$ When you say : "passing to the sup : $\sup_{x\in [a,b]}|f_n(x)-f_m(x)|\leq \varepsilon /2"$. This is not true because the $N$ before depend on $x$. $\endgroup$
    – Surb
    Jan 14 at 10:27
  • $\begingroup$ @Surb $|f(x)-f_m(x)|<\epsilon/2$ is true for each x, then it sure must be true for the greatest $|f_n(x)-f_m(x)| $which is the Sup, I don't see why$ N$ depending on $x$ may influence that, perhaps you could provide an explicit example? And isn't it the same my professor did in ( * ), (**) below? $\endgroup$
    – J.C.VegaO
    Jan 14 at 10:33
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    $\begingroup$ Fix $\varepsilon >0$ and $x\in \mathbb R$. Then $(f_n(x))$ is Cauchy, therefore, there is $N=N(x,\varepsilon )$ s.t. $|f_n(x)-f_m(x)|<\varepsilon /2$ whenever $n,m\geq N$. Now, if you change the $x$, you change the $N$, therefore, you don't have $|f_n(t)-f_m(t)|<\varepsilon /2$ for all $t\in [a,b]$ whenever $n,m\geq N$, and thus you can't take the supremum. For $(*)$ and $(**)$ it's not clear what you try to prove, but it's not true if you only suppose that $(f_n)$ converges pointwise... $\endgroup$
    – Surb
    Jan 14 at 10:52
  • $\begingroup$ $|f(x)-f_m(x)|<\dfrac{\varepsilon}{2}$ may not be true for each $x$ if you fix $m$. $\endgroup$
    – Zerox
    Jan 14 at 10:58
  • $\begingroup$ In the answer to this question I encountered counterexample: "The sequence of functions $f_n(x)=\min(1,n|x-1/n|)$ converges to $1$ pointwise on the interval $[0,1]$, yet the convergence is not uniform." $\endgroup$
    – drhab
    Jan 14 at 10:59
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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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  • $\begingroup$ I wanted to see how beginning with $|f_n(x)-f_(x)|<\epsilon$ and taking an N that depends on x,$Sup|f_n(x)-f_(x)|<\epsilon$ is false. You already showed that $Sup|f_n(x)-f_(x)|<\epsilon$ is false Now, before that I needed $|f_n(x)-f_(x)|<\epsilon$ for n>N to be true. Is it? and where is the dependence of N on x?. I got the graphical picture, still not seeing how my N is dependent on x $\endgroup$
    – J.C.VegaO
    Jan 14 at 13:15
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    $\begingroup$ As I wrote, "For any $x>0$, you can always find $N$." But you have to choose $x$ first before you decide what $N$ is. The smaller you make $x$, the larger $N$ needs to be. That's how $N$ depends on $x.$ If you choose $N$ first, then there's always an $x$ for which this $N$ is no good, for example $x = 2^{-N-1}$, where if $n = N+1 > N$ then $f_n(x) = 1.$ $\endgroup$
    – David K
    Jan 14 at 14:09
  • $\begingroup$ I think I got it, the descending line of the triangular bump has to intersect the line $x=x_N$ for all of the inequality to be true for all subsequent $n>N$. This descending line has as equation $ y=-2^Nx +2$, whose intersection with $y=\epsilon$ yields $x_N=2^{-N}(2-\epsilon)$. Inverting it $N=log((2-\epsilon)/x_N)/log2$, which shows an explicit dependence of N on both x and $\epsilon$ $\endgroup$
    – J.C.VegaO
    Jan 14 at 14:17
  • $\begingroup$ Just something else, in (*)the professor takes the limit $m\to \infty$ in $\text{Sup}_{x \in [a,b]}|f_n(x)-f_m(x)|< \epsilon$ so that $\text{Sup}_{x \in [a,b]}|f_n(x)-f_(x)|\le \epsilon$. How does the limit gets to enter inside the sup? I don't recall any explicit theorem about this and why does the inequality becomes non strict? $\endgroup$
    – J.C.VegaO
    Jan 14 at 14:31
  • $\begingroup$ @J.C.VegaO: You can only do that if $f_n\to f$ uniformly (it's not so clear what your teacher tries to prove, could you please write precisely the statement ?)... So, if it's the case, notice that $\|f\|_\infty := \sup_{x\in [a,b]}|f(x)|$ is a norm on $\mathcal C[a,b]$ (prove it if it's not clear). Therefore, by triangle inequality, $|\|f_n-f_m\|_\infty -\|f_n-f\|_\infty |\leq \|f-f_m\|_\infty $, and thus taking $m\to \infty $ gives the wished result. $\endgroup$
    – Surb
    Jan 14 at 15:22
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Passing to the sup: $\text{Sup}_{x \in [a,b]}|f_n(x)-f_m(x)|\le\epsilon/2$

Here you mean of course that this is satisfied for all $n, m \ge N$.

This is not true. You know that $\forall \epsilon >0$ $\exists N$ such that if $n,m>N$, $|f_n(x)-f_m(x)|<\epsilon/2$, but $N$ depends on $x$.

As an example take $[a,b] = [0,1]$ and $f_n(x) = x^n$. Then we get $$\lvert f_n(x) - f_m(x) \rvert = \lvert x^n- x^m \rvert.$$

Let $\delta < 1$ and $x \in (0,1)$. We want to find an index $N$ such that $$\lvert x^n- x^m \rvert < \delta \text{ for all } n,m \ge N \tag{1}$$

Clearly we must have $\lvert x^N- x^m \rvert < \delta$ for all $m \ge N$. Since $x^m \to 0$ as $m \to \infty$, this requires $x^N < \delta$. But if $x^N < \delta$, then $\lvert x^n- x^m \rvert = x^N \lvert x^{n-N}- x^{m-N} \rvert \le x^N < \delta$. Thus $(1)$ is satisfied iff $$x^N < \delta . \tag{2}$$ But $(2)$ means $N\ln x < \ln \delta$. Because $\ln x < 0$ for $x \in (0,1)$, we get $$N > r(x) = \ln \delta / \ln x . \tag{3}$$ But $r(x) > 0$ because $\ln \delta < 0$. Since $\ln x \to 0$ as $x \to 1$, we see that $r(x) \to \infty$ as $x \to 1$. This shows that $N$ depends on $x$. In fact, $N(x) \to \infty$ as $x \to 1$.

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  • $\begingroup$ any explicit example of how taking an N that depends on x, makes the Sup inequality false? I am no very convinced of this $\endgroup$
    – J.C.VegaO
    Jan 14 at 11:03
  • $\begingroup$ @J.C.VegaO: see my last comment under your original post. $\endgroup$
    – Surb
    Jan 14 at 11:49
  • $\begingroup$ How does it follow from this that $\sup_{x \in [0,1]} \lvert f_n(x)-f(x)\rvert < \epsilon$ is false? $\endgroup$
    – J.C.VegaO
    Jan 14 at 13:40
  • $\begingroup$ @J.C.VegaO In this particular example, where $f_n(x) = x^n$, we find that $f(1) = 1$ and $f(x) = 0$ everywhere in the interval $[0,1).$ So $\lvert f_n(x)-f(x)\rvert = 0$ when $x = 1$ and $\lvert f_n(x)-f(x)\rvert = x^n$ when $0\leq x<1.$ Now the limit of $x^n$ as $x\to 1$ is $1,$ so $\sup_{x \in [0,1]} \lvert f_n(x)-f(x)\rvert = 1.$ $\endgroup$
    – David K
    Jan 14 at 14:17
  • $\begingroup$ @J.C.VegaO It shows that there cannot exist an index $N$ such that $\text{Sup}_{x \in [a,b]}|f_n(x)-f_m(x)|\le\epsilon/2$ for all $n, m \ge N$. The latter would imply that $\lvert f_n(x)-f_m(x)|\le\epsilon/2$ for all $n, m \ge N$ and $x \in [0,1]$. But for any given $N$ we have $r(x) > N$ for $x$ sufficiently close to $1$ - although we need $N > r(x)$ for these $x$. $\endgroup$ Jan 14 at 15:43

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