9
$\begingroup$

Let $H$ be a Hilbert space over $\mathbb{C}$ with inner product $\langle\cdot,\cdot\rangle$, and let $\{x_n\}_{n=1}^\infty\subseteq H$, $x\in H$. I'm using the following definitions:

  • $\{x_n\}_{n=1}^\infty$ norm converges to $x$ $\Longleftrightarrow$ $x_n\rightarrow x$ $\Longleftrightarrow$ $\lim_{n\rightarrow\infty}\|x_n-x\|=0$.
  • $\{x_n\}_{n=1}^\infty$ converges weakly to $x$ $\Longleftrightarrow$ $x_n\rightharpoonup x$ $\Longleftrightarrow$ $\lim_{n\rightarrow\infty}\langle y,x_n\rangle=\langle y,x\rangle$ for all $y\in H$.
  • $\{x_n\}_{n=1}^\infty$ converges in the Cesàro sense to $x$ $\Longleftrightarrow$ $x_n\xrightarrow{Ces} x$ $\Longleftrightarrow$ $\lim_{N\rightarrow\infty}\|\frac{1}{N}\sum_{n=1}^Nx_n-x\|=0$.

If $x_n\rightarrow x$, then $x_n\rightharpoonup x$ and $x_n\xrightarrow{Ces}x$. On the other hand, if $H=\mathbb{C}$ with the standard inner product, then $(-1)^n\xrightarrow{Ces}0$ but $(-1)^n\not\rightharpoonup 0$. I'm having some difficulty, however, finding a sequence that converges weakly but not in the Cesàro sense. Since weak and strong convergence are equivalent in a finite-dimensional space, such a sequence can only exist if $H$ is infinite-dimensional. I'm therefore trying to find an example in the space $\ell^2(\mathbb{N})$ of sequences $\alpha=(\alpha_1,\alpha_2,\ldots)$ such that $\sum_{n=1}^\infty|\alpha_n|^2<\infty$ with the usual inner product $\langle\alpha,\beta\rangle=\sum_{n=1}^\infty\bar{\alpha}_n\beta_n$.

The sequence I'm considering is

$$x_1=\left(1,\frac{1}{2},\frac{1}{3},\ldots\right),$$ $$x_2=\left(0,1,\frac{1}{2},\ldots\right),$$ $$x_3=\left(0,0,1,\ldots\right),$$ $$\vdots$$

i.e, the sequence consisting of repeatedly shifting the harmonic sequence to the right. I've managed to show that $x_n\rightharpoonup 0$, but I haven't been able to show that $\{x_n\}_{n=1}^\infty$ doesn't converge in the Cesàro sense. Am I on the right track? If so, how do I show that Cesàro convergence fails? If not, what would be a sequence that actually works?

$\endgroup$
  • $\begingroup$ Your sequence doesn't work -- you can approximate the sums that occur by integrals that yield logarithms, and the dominant term in the Cesàro norm then becomes $\frac{x\log^2x}{x^2}$, which goes to zero. $\endgroup$ – joriki Sep 3 '15 at 22:39
4
$\begingroup$

I am currently not sure about your sequence, but I will edit the answer if I see something.

Meanwhile, the following example works: Define $$ x_n := e_k \text{ for } 2^k \leq n < 2^{k+1}, $$ where $e_k$ is the $k$th member of the standard basis of $\ell^2$.

Then we have $x_n \rightharpoonup 0$, but $$ \bigg(\frac{1}{2^{k+1} - 1} \sum_{\ell = 1}^{2^{k+1} - 1} x_\ell \bigg)_{k} = \frac{(2^{k+1} - 1) - 2^k + 1}{2^{k+1}-1} = \frac{2^{k}}{2^{k+1} - 1} \rightarrow \frac{1}{2}. $$ In particular, this shows $$ \limsup_k \bigg \Vert \frac{1}{2^{k+1} - 1} \sum_{\ell = 1}^{2^{k+1} - 1} x_\ell \bigg\Vert \geq \limsup_k \bigg| \bigg(\frac{1}{2^{k+1} - 1} \sum_{\ell = 1}^{2^{k+1} - 1} x_\ell \bigg)_{k} \bigg| \geq 1/2 $$ and thus $(x_n)_n$ does not converge Cesaro to $0$.

The intuition here is that if we are at $2^{k+1}$, then still about half of all the sequence members we will have seen are equal to $e_k$.

Note though, that even though Cesaro convergence does not hold in general, Mazur's lemma (https://en.wikipedia.org/wiki/Mazur%27s_lemma) shows that we can always find some convex combination which converges strongly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.