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Question:

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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    I hope no one is offended by $\ell^1$ instead of $l^1$.2011-10-30

2 Answers 2

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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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    @Martin: I saw your post. I will try to convince myself later.2011-10-29
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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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    @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 $n$.2011-10-30