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.

  • 5
    $\begingroup$ When mathematicians say "X is a stronger condition than Y" they typically mean "X implies Y but not vice versa". $\endgroup$ – 79037662 Sep 24 '19 at 19:41
  • $\begingroup$ 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 $X \implies Y$. $\endgroup$ – Jesse Madnick Sep 24 '19 at 19:46
  • $\begingroup$ 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? $\endgroup$ – zhw. Sep 24 '19 at 19:54

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.