# Uniform is “Stronger” than Pointwise

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.

• 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

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.