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On the space $\ell^1$ for $x=(\alpha_1,\alpha_2,\ldots)\in{\ell^1}$, define $$f(x)=\sum\limits_{n=1}^\infty \alpha_n$$ Prove that $f$ is not continuous with respect to $\|x\|_\infty =\sup_n|\alpha_n|$.

This is my proof:

Since $f$ is a linear map, $f$ is continuous iff $f$ is bounded. Assume for the sake of contradiction that $f$ is bounded. This implies that there exist $k\gt 0$. such that $$\|f(x)\|\le k\|x\|_\infty \text{ for all } x\in \ell^1$$ In particular, $k$ may be $1$.

Now, let $x=(\alpha_1,\alpha_2,\ldots)\in \ell^1$ where $\alpha_i \ge 0$ for all $i$.

$$\|f(x)\|=\|\sum\limits_{n=1}^{\infty}\alpha_{n}\|$$ Since for each $i$ $\alpha_{i}\gt 0$,

$$=\sum\limits_{n=1}^\infty |\alpha_n| \gt\sup_n|\alpha_n|=\|x\|_\infty$$

This implies $$\|f(x)\|\gt\|x\|_\infty$$

Contradicting my first line of proof. Hence $f$ is not bounded, i.e. $f$ is not continuous.

My problem here is that I failed to believe myself, I think something is wrong in the proof. can anyone help me out?

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The line $\|f(x)\| > \|x \|_\infty$ doesn't contradict the line $\forall x\in l^1, \|f(x)\| \leq k \|x\|_\infty$, unless $k$ has become $1$ for some reason. But the idea is right. – user10676 Oct 29 '11 at 17:21
Yes $k$ may be $1$. – Hassan Muhammad Oct 29 '11 at 17:37
But what if $k=2$? What is your contradiction? $k$ may be anything ;) – N. S. Oct 29 '11 at 18:23
$k=2$ is also a good choice. – Hassan Muhammad Oct 29 '11 at 18:52
I hope no one is offended by $\ell^1$ instead of $l^1$. – Michael Hardy Oct 30 '11 at 17:59
up vote 4 down vote accepted

Boundedness means: $$(\exists C)(\forall x) \|f(x)\|\le C\|x\|.$$

Negation of this is $$(\forall C)(\exists x) \|f(x)\|> C\|x\|.$$

Note two important things:

  • You only have shown $C=1$ is not working.
  • You only need to show existence of one such $x$ for a given $C$. (In your attempted proof above you worked with arbitrary $x$.)

So the question is: If a constant $C$ is given, can you find a sequence $x$ such that $\|f(x)\|> C\|x\|$?

EDIT (added after seeing OP's comments):

Maybe this could clarify what's going on.

Problem: Show that the sequence $a_n=2^n$ is bounded.

A sequence is bounded if there exists a constant $C$, such that $|a_n|\le C$ for each $n$. In this case -- since $a_n$ is positive -- this is the same as $a_n \le C$.

Solution 1: Since $a_1=2>1$, the above property fails for $C=1$. So $a_n$ is not bounded.

Solution 2: We can show by induction that $2^n>n$ holds for $n=1,2,\ldots$. Thus for every given $C>1$ we can take $n=\lceil C \rceil$ and we have $C\le n < 2^n=a_n$. Thus the sequence $a_n$ is not bounded by any given constant $C$.

Which of the above solutions is correct? Do you see the similarity with your proof of unboundedness of $f$?

So far I have tried to get you to solve the problem by yourself, however, if there is still a problem, you can have a look at this (Spoiler alert - the solution appears when you move your mouse bellow):

Suppose that $C$ is a positive integer. Then we can choose a sequence $x$ such that $x=(\underset{(C+1)\text{-times}}{\underbrace{1,1,1,\dots,1}},0,0,0,\dots)$. Then $\|x\|=1$ but $f(x)=C+1$, hence $|f(x)|>C\|x\|$. This works for any positive integer $C$, so the function $f$ is not bounded.

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If $c=1$ then $x=(\alpha_{1},\alpha_{2},\cdots)$ where alpha's are positive numbers, I think this $x$ will work? – Hassan Muhammad Oct 29 '11 at 17:40
@Hassan: That's true. What about c=2, c=3, etc.? – Martin Sleziak Oct 29 '11 at 17:42
All I want is to get a contradiction. So if C=1 gives me what I am after then I may relax. – Hassan Muhammad Oct 29 '11 at 17:48
No: to get a real contradiction you have to find a bad vector for every value of $C$ (or $k$). – sibilant Oct 29 '11 at 17:52
@Hassan: I tried to include in the post example of similar reasoning but illustrated on a much simpler problem. – Martin Sleziak Oct 29 '11 at 17:57

You can also argument as follows:

Consider the sequence $x_n\in\ell^1$ given by $$ x_n=\left(1,\frac{1}{2^{1+\frac{1}{n}}},\frac{1}{3^{1+\frac{1}{n}}},\ldots,\frac{1}{k^{1+\frac{1}{n}}},\ldots \right) $$

Note that for all $n\in\mathbb{N}$ we have $\|x_n\|_{\infty}=1$. In other words $x_n$ is a sequence in the unit ball. To show that $f$ is not continuous it is enough to show that $|f(x_n)|\to\infty$. This can be done using the integral test since $$ n=\int_{1}^{\infty} \frac{1}{x^{1+\frac{1}{n}}} \ dx\leq \sum_{k=1}^{\infty}\frac{1}{k^{1+\frac{1}{n}}}=|f(x_n)|. $$

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The evaluation of the integral is not clear to me, can you justify it? I am trying to solve my problem using contradiction, but your solution is a direct proof. – Hassan Muhammad Oct 30 '11 at 5:51
Hi Hassan, this integral can be computed directly using the fundamental theorem of calculus $\int_{1}^{\infty}\frac{1}{x^{1+\frac{1}{n}}}\ dx= \displaystyle\lim_{t\to\infty}-nx^{-\frac{1}{n}}|_{1}^{t}=n$. Note that inequality comes from the comparison between Riemann sums and integrals with the partition of length one. – Leandro Oct 30 '11 at 6:30
I guess the limit of your sequence $x_{n}$ is $1$. For $f$ to be continues $f(x_{n})$ must converge to $1$. $|f(x_{n})|$ converges to $n$. Thus, if $n$ is $1$, then $f$ is continues! Do you see? – Hassan Muhammad Oct 30 '11 at 9:07
Forgive me, I mean $f$ diverges, but I dont want $n=1$. Is this possible? – Hassan Muhammad Oct 30 '11 at 9:31
@HassanMuhammad I did not understood what you typed in the two previous comments. First of all $(x_n)$ is a sequence in $\ell^1$ so there is no possibility for it converges to a real number. Note that $|f(x_n)|$ does not converges to $n$ it is exactly $n$. – Leandro Oct 30 '11 at 18:16

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