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For $1 \le p < \infty$ and each index $n$, let $e_n \in \ell^p$ have $n$-th component 1 and all other componenets $0$. I want to show that $p>1 \Rightarrow \{e_n\} \to 0$ weakly in $\ell^p$ and that this is not the case in $\ell^1$. Any suggestions?

Davide Giraudo
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  • Suggestions: For $p>1$, try using the definitions, try to show that the definition of weak convergence to zero is satisfied, see if you get stuck anywhere. For $p=\infty$, use the defintion, and try to find a countexample that shows that the definition of weak convergence to zero is not satisfied. – Jonas Meyer Feb 14 '13 at 05:24
  • The issue is that I am not quite sure what the definition is in the case of $\ell^p$. Could you please provide it? –  Feb 14 '13 at 05:25
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    Where did you encounter this problem such that you do not have access to the definitions? You are asking here for the solution to a problem but do not know what it means. Your first question might instead be, "what is the definition of weak convergence in $\ell^p$?" This is a special case of the notion of weak convergence in Banach space, but you will want to use the identification of the dual of $\ell^p$ with $\ell^q$ with $p+q=pq$. – Jonas Meyer Feb 14 '13 at 05:32
  • I am sorry, I am not very precise. In the very beginning of these notes, one has the definition of weak convergence (http://people.math.gatech.edu/~heil/6338/summer08/section9a.pdf). My exact question is: what is the dual in this particular case and how is the inner product defined for this case? –  Feb 14 '13 at 05:37
  • Thank you for that. (Oops I meant $1$ where I wrote $\infty$ in my first comment.) Dual of $\ell^p$ when $1\leq p<\infty$ is $\ell^q$, with $q$ the conjugate exponent of $p$, $pq=p+q$ if $p>1$, $q=\infty$ if $p=1$. The functional on $\ell^p$ corresponding to an element of $\ell^q$ is shown in the Wikipedia link above and in Example 1.9 on page 3 of the pdf you linked to. It seems implicit that the reader is assumed to know about the duals of $\ell^p$ when $1\leq p<\infty$. I do not know what sources you may have for that fact, but it should be found in most introductions to $\ell^p$ spaces. – Jonas Meyer Feb 14 '13 at 05:50

2 Answers2

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The definition of $\ell^p, \ 1\leq p<+\infty$ is $$\ell^p=\left\{x=(x(k))_{k\in\mathbb N}:\|x\|_p:=\sqrt[p]{\displaystyle\sum_{k=1}^\infty |x(k)|^p}<+\infty\right\}$$ and $$\ell^\infty=\left\{x=(x(k))_{k\in\mathbb N}:\|x\|_\infty:=\sup\{|x(k)|:k\in\mathbb N\}<+\infty\right\}.$$ A sequense $(x_n)_{n\in\mathbb N}$ in $\ell^p, \ 1\leq p<+\infty$ , converges weakly to $x\in\ell^p$ (denoted $x_n\rightharpoonup x$ )
if for any $y\in\ell^q$ (where $q$ is defined by $p+q=pq$ for $p\neq1$ and $q=+\infty$ for $p=1$): $$\langle x_n,y \rangle \xrightarrow[n\to+\infty]{}\langle x,y \rangle,$$ where $\langle x,y \rangle :=\displaystyle\sum_{k=1}^{\infty}x(k)y(k)$ for $x=(x(k))_{k\in\mathbb N}\in\ell^p,y=(y(k))_{k\in\mathbb N}\in\ell^q$.

We want to show that for $p>1, \ e_n\rightharpoonup 0$ in $\ell^p$.
Equivalently that $\displaystyle\sum_{k=1}^{\infty}e_n(k)y(k)\xrightarrow[n\to+\infty]{}0$ for all $(y(k))_{k\in\mathbb N}\in\ell^q$.
Note that $e_n(k)y(k)=\delta_{nk}y(k) \ \forall n,k \in\mathbb N$; from $\displaystyle \sum_{k=1}^\infty|y(k)|<+\infty$ it follows that $$\displaystyle\sum_{k=1}^{\infty}e_n(k)y(k)=y(n)\xrightarrow[n\to\infty]{}0.$$

Now for $p=1$ find an element $(y(k))_{k\in\mathbb N}\in\ell^\infty$
s.t. $\displaystyle\sum_{k=1}^{\infty}e_n(k)y(k) \not\rightarrow0$ (hint: find one with $\displaystyle\sum_{k=1}^{\infty}e_n(k)y(k)=1, \ \forall n\in\mathbb N $).

Cortizol
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P..
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  • How do we know that $\sum_{k=1}^\infty |y(k)| < \infty$ for an arbitrary $y \in \ell^q$? – mtg Sep 25 '21 at 18:52
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Following the same steps as here, we can establish the following weak convergence conditions.

Let $1<p<\infty$ and $\{x^{(n)}\}\subset \ell^p$. This sequences converges weakly to $0$ in $\ell^p$ is and only if the following two conditions are satisfied:

  1. We have $\lim_{n\to +\infty}x^{(n)}_N= 0$ for all $N\in\Bbb N$.
  2. The sequence $\{\lVert x^{(n)}\rVert_p\}\subset\Bbb R$ is bounded.

To see that $\{e_n\}$ doesn't converge to $0$ in $\ell^1$, consider $L\colon\ell^1\to \Bbb R$, $L(x):=\sum_{j=0}^{+\infty}x_j$. Actually, strong and weak convergence of sequences are the same thing in $\ell^1$.

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Davide Giraudo
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