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I am trying to figure out why completeness is necessary in this theorem. And I was given the following task:

Consider $C[0,1]$ with norm $$\|x\| =\int_{0}^{1}| x(t)| dt$$ and operators $$A_nx = n\int_{0}^{1/n} x(t) dt$$

So I want to prove that this operator is pointwise bounded but not uniform bounded. I've managed to do the following:

$|n\int_{0}^{1/n} x(t) dt|$ $\le$ $n\int_{0}^{1/n} |x(t)| dt$ $\le$ ($1/n-0)n$ $\|x\|$ $\le$ $\|x\|$

I guess this proves pointwise boundedness (?) But now I don't now how to find operator norm, which will show that uniform unboudedness fails.

whatanhonor
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    Find a sequence $x_n$ with $|x_n|=1$ but $|A_n(x_n)|$ large. – kimchi lover Dec 22 '19 at 15:06
  • Your proof seems wrong, as you've allegedly proven that $\Vert A_n\Vert \leq 1$. You should have used the continuity of $x$ somewhere. $A_n(x)$ should tend to $x(0)$. Unless you also use $\Vert \cdot \Vert$, for a different norm. – Keen-ameteur Dec 22 '19 at 15:10
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    I guess you want to write $$A_n: C[0;1]\rightarrow \mathbb{R}, \ A_n x= n \int_0^\frac{1}{n} x(t) dt$$ also your estimate for pointwise bounded is wrong. What you can show is the following $$\vert A_n x\vert \leq \Vert x\Vert_\infty$$ To see that this family is not uniformly bounded, you should show that $$\Vert A_n \Vert \geq n$$ this you can do by choosing $x$ with support in $[0,1/n]$. – Severin Schraven Dec 22 '19 at 15:11
  • @SeverinSchraven Maybe something like this? $|n\int_{0}^{1/n} x(t) dt|$ $\le$ $n\int_{0}^{1} |x(t)| dt$ = $n |x|$ And then find the least possible $n$ ? – whatanhonor Dec 22 '19 at 15:19
  • I am not sure how your comment should be related to what I wrote – Severin Schraven Dec 22 '19 at 15:23
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    There are two things you want to show. First that the family is poinwise bounded, meaning $$\sup_n \vert A_n x\vert < \infty$$ here you cannot allow something that explodes with $n$. That's why I wrote that you should prove the estimate $$\vert A_n x\vert \leq \Vert x\Vert_\infty := \sup_{t\in [0;1]} \vert x(t)\vert $$ On the other hand you want to show that the operator norms of $A_n $ explode. That was why I suggested to show $\Vert A_n \Vert \geq n$. We are dealing with the operator norm, for every $A_n$ we should find some $x_n$ with $\Vert x_n \Vert =1 $ and $\vert A_n x_n\vert = n$. – Severin Schraven Dec 22 '19 at 15:32
  • @SeverinSchraven thank you! I've managed to do the first part. But can't really think of a proper function for the second part. – whatanhonor Dec 22 '19 at 16:21
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    Just take for $x_n$ any function supported in $[0;1/n]$ such that $\Vert x_n \Vert =1$. – Severin Schraven Dec 22 '19 at 16:30
  • Also I would suggest you to write a complete answer to your answer here. Then you also get feedback and see whether it is really correct (and also will give you some reputation). – Severin Schraven Dec 22 '19 at 22:20

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A simpler example may be of use to you. In space $l_1$ with sup norm we can consider the sequence of functionals $f_n(x)=\sum\limits_{k=n}^{2n+1}\xi_k$.

thing
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