Convergence in $\ell^p$ norm provided it weakly converges.

Question

I need some help with the following problem :

$1<p < \infty$ , let $x_n$ be a sequence in $\ell^p$ and also $x\in \ell^p$ . I am interested in showing $$\lim_{n\to \infty} \|x_n-x\|_p\to0$$ if and only if $x_n$ converges to $x$ in weak topology and $$\lim_{n\to \infty} \|x_n\|=\|x\|_p$$

I need someone to show me the footsteps that I should follow and think .

Thank you .

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

It's a general fact that strong convergence implies weak convergence and convergence of the norms.

To see the converse, fix $\varepsilon>0$. There is an integer $N$ such that $\sum_{k\geqslant N}|x^k|^p<\varepsilon$. Weak convergence gives convergence of the coordinates (i.e. $x_n^k\to x^k$ for each $k$, using the linear functional $x\mapsto x^k$), so we also have for $n$ large enough that $\sum_{k\geqslant N}|x_n^k|^p<\varepsilon$. Indeed, we have $\sum_{k=1}^{N-1}|x_n^k|^p\to \sum_{k=1}^{N-1}|x^k|^p$ by weak convergence, and $\sum_{k\geqslant 1}|x_n^k|^p\to \sum_{k\geqslant 1}|x^k|^p$, which gives $\sum_{k\geqslant N}|x_n^k|^p\to \sum_{k\geqslant N}|x^k|^p$.

This gives, by the inequality $|a+b|^p\leqslant 2^{p-1}(|a|^p+|b|^p)$, $$\lVert x_n-x\rVert_p^p\leqslant \sum_{j=1}^{N-1}|x_n^j-x^j|^p+2^{p-1}\cdot 2\varepsilon.$$ Taking the $\limsup_{n\to +\infty}$, and using the fact that $\varepsilon$ is arbitrary, we get the wanted result.

Note that the result also holds when $p=1$, and actually, we just need weak converge. It's done in this thread.
We can use Fatou's lemma applied to counting measure and the sequence $y_n^k:=2^{p—1}(|x_n^k|+|x^k|)-|x_n^k-x^k|$.

edited Apr 13 '17 at 12:20

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answered Dec 03 '12 at 21:39

Davide Giraudo

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I have seen in many places "convergence of the co-ordinates" but what does this actually mean ? – Theorem Dec 03 '12 at 21:47
It means $x_n^k\to x^k$ for all $k$. – Davide Giraudo Dec 03 '12 at 21:48
I think it might be worthwhile to expand on why the second sentence in the second paragraph is true; as it's the hardest part of the proof and since it's where the hypothesis that $\Vert x_n\Vert \rightarrow\Vert x\Vert$ is needed. – David Mitra Dec 03 '12 at 21:57
@DavidMitra I agree it deserves more details, that I think I've written now. – Davide Giraudo Dec 03 '12 at 22:04
@DavideGiraudo I was trying to apply the definition of weak convergence ie $f(x_n)→f(x)∀f∈lp′$, but i cannot see how i can connect it with the solution you gave. – Theorem Dec 03 '12 at 22:08
Use a particular $f$, namely, the linear functional which maps a sequence to its $k$-th coordinate. – Davide Giraudo Dec 03 '12 at 22:12
@DavideGiraudo : I think there is a mismatch with the indices in the first few lines . – Theorem Dec 06 '12 at 20:38
@Theorem fixed now (I think). – Davide Giraudo Dec 06 '12 at 20:41
@DavideGiraudo : I think its still not, because $n$ stands for co-ordinates right ? and $k$ stands for the sequence index ? – Theorem Dec 06 '12 at 20:50
No, it's the contrary. – Davide Giraudo Dec 06 '12 at 20:51

Convergence in $\ell^p$ norm provided it weakly converges.

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