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?

  • $\begingroup$ 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, 2015 at 22:39

1 Answer 1


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.


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