2
$\begingroup$

I need some help with the following problem :

$1

$\ell^p$ 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 .

1 Answers 1

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 $y_n^k:=2^{p—1}(|x_n^k|+|x^k|)-|x_n^k-x^k|$.
  • 0
    I have seen in many places "convergence of the co-ordinates" but what does this actually mean ?2012-12-03
  • 0
    It means $x_n^k\to x^k$ for all $k$.2012-12-03
  • 0
    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.2012-12-03
  • 0
    @DavidMitra I agree it deserves more details, that I think I've written now.2012-12-03
  • 0
    @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.2012-12-03
  • 0
    Use a particular $f$, namely, the linear functional which maps a sequence to its $k$-th coordinate.2012-12-03
  • 0
    @DavideGiraudo : I think there is a mismatch with the indices in the first few lines .2012-12-06
  • 0
    @Theorem fixed now (I think).2012-12-06
  • 0
    @DavideGiraudo : I think its still not, because $n$ stands for co-ordinates right ? and $k$ stands for the sequence index ?2012-12-06
  • 0
    No, it's the contrary.2012-12-06