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Does uniform boundedness imply pointwise boundedness? I take Walter Rudin's definition shown below. enter image description here

If yes, we can also find a converging subsequence for $\{f_n\}$ if it's uniformly bounded, contradicting to Rudin's statement that "there need not exist a subsequence which converges pointwise on E"

I can prove pointwise boundedness doesn't imply uniform boundedness by setting $f_n(x)=x$ and $\phi (x)=2x$ for $x\in(0,+\infty)$.

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    $\begingroup$ Unless you use non-standard definitions, uniform boundedness implies pointwise boundedness. $\endgroup$ Nov 18 '17 at 15:59
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    $\begingroup$ Uniform boundedness implies pointwise boundedness, take $\phi(x) = M$ for all $x$. That doesn't contradict what Rudin writes. We can find a subsequence that is pointwise convergent on the countable subset $E_1$. But that doesn't imply that it is pointwise convergent on all of $E$. $\endgroup$ Nov 18 '17 at 16:25
  • $\begingroup$ I see, thanks! @DanielFischer $\endgroup$
    – Hank
    Nov 18 '17 at 17:52
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Certainly uniform boundedness implies pointwise boundedness (in the notation of the definitions you cite, define $\phi(x)=M$ for all $x$).

Say $(f_n)$ is uniformly bounded. This does not imply there is a pointise convergent subsequence. The error you're probably making is this: Yes, for every $x$ there exists a subsequence $(f_{n_j})$ such that $f_{n_j}(x)$ is convergent. That does not say there is a pointwise convergent subsequence, because the subsequence you have depends on $x$, while saying there is a pointwise convergent subsequence says there is one subsequence that works for every $x$.

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