3
$\begingroup$

I need just a hint please. It seems that I have to prove that $f(x)=mx$ in which $m\in \mathbb{R}.$ But I couldn't handle it.

Problem: We say that a sequence $x_{n}\; , n = 1, 2,\cdots ,$ Cesaro converges to $a,$ if $$ \lim\limits_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}x_{i}=a. $$ A function $f$ is Cesaro continuous at $a$ if $x_{n}\to a$ (in Cesaro mean) implies $f(x_{n})\to a$ (in Cesaro mean). Prove that, if $f : \mathbb{R}\to\mathbb{R}$ is Cesaro continuous at $0$ and $f(0) = 0,$ then $f$ is linear.

$\endgroup$
2
$\begingroup$

I guess you mean to say that $f(x_n)\to f(a)$ whenever $x_n\to a$, all in the Cesaro sense.

If this is the case, let $x\in\mathbb R$ and consider the sequence $x_n=(-1)^nx$. Then $x_n\to 0$ in the Cesaro sense, therefore $$f(x)+f(-x)=\lim_{n\to\infty}\frac{1}{2n}\sum_{k=1}^{2n}f(x_k)=f(0)=0,$$ therefore $f$ is odd.

Note now that, if $x,y\in\mathbb R$ and $(y_n)$ is the sequence $$x+y, -x,-y, x+y, -x, -y,\dots,$$ then $y_n\to 0$ in the Cesaro sense, therefore, as above, $$f(x+y)+f(-x)+f(-y)=\lim_{n\to\infty}\frac{1}{3n}\sum_{k=1}^{3n}f(y_k)=f(0)=0,$$ therefore $$f(x+y)=-f(-x)-f(-y)=f(x)+f(y),$$ since $f$ is odd. This shows that $f(x)=ax$ for some $a$, at least when $x\in\mathbb Q$.

Finally, for $r$ being irrational, let $(x_n)$ be a sequence of rational numbers that converges to $r$, and consider the sequence $(z_n)$, which is defined by $$x_1,-r,x_2, -r,x_3,-r,\dots .$$ Then $z_n\to 0$ in the Cesaro sense, and as above you can show that $f(r)=ar$.

$\endgroup$

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.