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Statement of the problem:

Prove: If a sequence of complex functions $s_n$ on a set $X$ converges uniformly to a complex function $s$, then the sequence of Cesaro means $\sigma_N$ also converges uniformly to $s$.

I have already shown that a sequence of complex numbers $\{s_n\}$ which converges to $s$ has Cesaro means which also converge to $s$, but I'm not sure how to use that to prove this statement. My first instinct was to say that, for each $x$ in the domain, we just have a sequence of complex numbers, so each $\sigma_N(x)$ converges to $s(x)$, but this only establishes pointwise converges (right?). I don't know how to deal with the uniform condition.

Edit: According to my professor, this exercise he assigned is actually false. Can someone come up with an explanation why?

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  • $\begingroup$ Does writing $\sigma_N - s = \frac1N\sum_{n=1}^N (s_n-s)$ help ? $\endgroup$
    – Sary
    Mar 2, 2015 at 1:56
  • $\begingroup$ @Sary I used that to prove the statement about the sequence complex numbers. Is that still a valid way to prove it for functions when there are infinitely many possibilities for $x$? I figure that each sequence of complex numbers for each point $x$ (denote it $S_x$) and any $\epsilon > 0$ has an associated $N_{x,\epsilon}$ where $n > N_x \Rightarrow |\sigma_N(x) - s(x)| < \epsilon$. While this proves pointwise convergence, does it prove uniform? If it helps, I can post my proof for the sequence of complex numbers statement. $\endgroup$
    – MT_
    Mar 2, 2015 at 2:00
  • $\begingroup$ Then write $\delta_n := \sup_x|s_n(x)-s(x)|$. What do you know about that sequence of numbers ? Can you give an upper bound for $\sup_x |\sigma_N(x) - s(x)|$ in terms of the $\delta_n$'s ? $\endgroup$
    – Sary
    Mar 2, 2015 at 2:09
  • $\begingroup$ @Sary OK, how do I know that $\delta_n$ exists and is finite? That is essentially my main confusion. If I know why it exists, the rest of the proof I can do. $\endgroup$
    – MT_
    Mar 2, 2015 at 2:11
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    $\begingroup$ @Sary see my answer. It doesn't work because $\delta_n$ might not exist for small values of $n$. $\endgroup$
    – MT_
    Mar 2, 2015 at 15:25

1 Answer 1

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Turns out that the premise is incorrect, as a recent email by my professor has notified.

On $(0,1]$ let $s_1(x) = \frac{1}{x}$ and $s_n = 0$ for $n > 1$, then $s_n \to 0$ uniformly but $\sigma_N \to 0$ pointwise.

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  • $\begingroup$ The theorem is true when every $s_n$ is bounded and the proof is almost identical to the usual proof for sequences $\endgroup$
    – janes
    Dec 31, 2021 at 9:36

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