# Pointwise vs. uniform convergence

1. Jul 29, 2010

### ForMyThunder

A function is pointwise bounded on a set $$E$$ if for every $$x\in E$$ there is a finite-valued function $$\phi$$ such that $$|f_n(x)|<\phi(x)$$ for $$n=1,2,...$$.

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

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

2. Jul 29, 2010

### Office_Shredder

Staff Emeritus
If E is the set (0,1) and $$\phi=\frac{1}{x}$$...

I think you can figure out the rest

3. Jul 29, 2010

### ForMyThunder

Oh, thanks. It said finite-valued which I took to mean bounded.

4. Jul 30, 2010

### wayneckm

I think for domain being a finite set, both notions coincide because maximum is indeed supremem. However, when it is either countable or uncountable domain, it is not necessary to have the equivalence between maximum and supremum. And it can turn out that supremum is unbounded depsite boundedness at each x. Office_Shredder showed a nice example.

5. Jul 30, 2010

### jostpuur

Finite-valued must have merely meant that $|\phi(x)| < \infty$ for all $x$.