# Cesàro Mean of Convergent Subsequence Converges?

Let $$X$$ be a compact and convex subset of $$\mathbb R^n$$ and let $${(x_{t})}_{t}$$ be a sequence in $$X$$ such that $$||x_{t+1}-x_{t}||_{\infty}\leq \frac{1}{t+1}$$ for all $$t=0,1,...$$.

Let $${(x_{t_n})}_{t_n}$$ be a convergent subsequence of $$(x_t)_t$$ such that $$x_{t_n}\to \bar x$$.

Is it the case that $$\lim_{n\to \infty}\frac{1}{t_n+1}\sum_{\ell=0}^{t_n}x_{\ell} = \bar x$$?

I am aware that Cesàro mean of a convergent subsequence converge to the limit of the subsequence when the average is taken only with respect to elements of the subsequence. Here I am averaging over all terms up to $$t_n$$ and taking the limit over indices in the subsequence.

EDIT:

The claim is false and a counterexample can be constructed using the sequence $$\frac{0}{2},\frac{1}{2},\frac{2}{2},\frac{2}{3},\frac{1}{3},\frac{0}{3},\frac{1}{4},...$$ and picking the subsequence $$(x_{t_n})_{t_n}$$ such that $$x_{t_n}=0$$. Then $$\bar x=0$$, but $$\lim_{n\to \infty}\frac{1}{t_n+1}\sum_{\ell=0}^{t_n}x_{\ell}=\frac{1}{2}$$. In fact, it seems like this is the Cesàro mean regardless of the subsequence.

So my new questions are:

Let $$X$$ be a compact and convex subset of $$\mathbb R^n$$ and let $${(x_{t})}_{t}$$ be a sequence in $$X$$ such that $$||x_{t+1}-x_{t}||_{\infty}\leq \frac{1}{t+1}$$ for all $$t=0,1,...$$.

(1) **Is there always a convergent subsequence $$(x_{t_n})$$ with convergent mean $$\lim_{n\to \infty}\frac{1}{t_n+1}\sum_{\ell=0}^{t_n}x_{\ell}=\bar x$$? **

(2) **If $$\lim_{n\to \infty}\frac{1}{t_n+1}\sum_{\ell=0}^{t_n}x_{\ell}=\bar x$$, is it the case that $$\lim_{t\to \infty}\frac{1}{t+1}\sum_{\ell=0}^{t}x_{\ell}=\bar x$$? **

• Oh, for (1), are you specifically asking about this sequence that you described? If so, then yes, just take the subsequence of terms $1/2, 2/4, 3/6, \dots$ – Calvin Khor Apr 20 '20 at 5:23
• My questions are the following: $x_t$ may not converge, but it has a convergent subsequence $x_{t_n}$. (1) Is there a convergent subsequence of $x_t$ such that the running average converges, i.e. $\lim_{n\to \infty}\frac{1}{t_n}\sum_{\ell=0}^{t_n}x_{\ell}=\bar x$? (2) Is it the case that if $\lim_{n\to \infty}\frac{1}{t_n}\sum_{\ell=0}^{t_n}x_{\ell}=\bar x$, then $\lim_{t\to \infty}\frac{1}{t}\sum_{\ell=0}^{t}x_{\ell}=\bar x$? – user_newbie10 Apr 20 '20 at 5:24
• but the subsequence and the running average are completely unrelated, the running average always converges to the Cesaro mean if the sequence is Cesaro sumamble. If you want more examples, given any $L$, the sequence consisting of only $L$s and zeros can be made to have Cesaro average $L$ by putting the zeros far apart, and then skipping the zeros gives you a convergent (constant) subsequence – Calvin Khor Apr 20 '20 at 5:27
• – user_newbie10 Apr 20 '20 at 5:30