A function is pointwise bounded on a set [tex]E[/tex] if for every [tex]x\in E[/tex] there is a finite-valued function [tex]\phi[/tex] such that [tex]|f_n(x)|<\phi(x)[/tex] for [tex]n=1,2,...[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

A function is uniformly bounded on [tex]E[/tex] if there is a number [tex]M[/tex] such that [tex]|f_n(x)|<M[/tex] for all [tex]x\in E, n=1,2,...[/tex].

I understand that in uniform boundedness, the bound is independent of [tex]x[/tex] and in pointwise convergence it is dependent. My question is this: if we take [tex]M=\max\phi(x)[/tex], then since [tex]\phi[/tex] is finite-valued, wouldn't this make every pointwise bounded function a uniformly bounded function? I don't understand.

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# Pointwise vs. uniform convergence

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