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I am reviewing the concept of point-wise and uniform convergence in the context of sequence of measurable functions.

I have looked at this post and understand the game semantics answer.

People say the uniform convergence is "stronger" than point-wise convergence. But how do they mean when using the adjective "stronger"?

For the point-wise convergence, the process goes:
1. I choose $x_0\in E$.
2. Partner chooses $\varepsilon>0$.
3. I find $N\in\mathbb{N}$ such that $\forall n>N$, $|f_n(x_0)-f(x_0)|<\varepsilon$.
4. If I am successful with steps #1-3 for all $x\in E$, then the sequence converges point-wise to $f$.

For the uniform converence, the process goes:
1. Partner chooses $\varepsilon>0$.
2. I find $N\in\mathbb{N}$ such that $\forall n>N$, $|f_n(x)-f(x)|<\varepsilon$ for all $x\in E$.

In which step is it apt to say the adjective "stronger" ? Clear explanation would be appreciated.

Frank Swanton
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    When mathematicians say "X is a stronger condition than Y" they typically mean "X implies Y but not vice versa". – 79037662 Sep 24 '19 at 19:41
  • For pointwise, the $N$ in Step 3 depends on both $\epsilon > 0$ and on $x_0$. But for uniform, the $N$ depends only on $\epsilon > 0$ and works for all $x$ simultaneously. So, uniform convergence $\implies$ pointwise convergence.

    As the above commenter says, "X is stronger than Y" just means $X \implies Y$.

     – Jesse Madnick Sep 24 '19 at 19:46
  • In the first case $N$ depends on $x_0.$ In the second case, there is an $N$ that works for all $x.$ Which is better? – zhw. Sep 24 '19 at 19:54

1 Answers1

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If $f_n \to f$ uniformly then also $f_n \to f$ pointwise. So it's stronger in the sense that it implies it. It gives more information, and it's rarer. That's common usage.

Henno Brandsma
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