1

In an answer to the question in this post Uniform Convergence Implies $L^2$ Convergence and $L^2$ Convergence Implies $L^1$ Convergence Silvia Ghinassi uses the fact that uniform convergence of $f_n$ to $f$ on a bounded interval $I$ implies that $\sup_I |f_n-f|\rightarrow 0$ on $I$. I was wondering, is the converse of this true? That is, can the condition $\sup_I |f_n-f|\rightarrow 0$ be used as an alternate definition of uniform convergence? I think the answer is yes, but I was hoping someone could verify this. If so, could someone help me see why this is equivalent to the standard definition? I see it intuitively, but the details are a bit fuzzy.

More generally, I was wondering if someone could provide additional characterizations of uniform convergence aside from the ordinary definition (i.e. $f_n\rightarrow f$ uniformly if for each $\epsilon>0$ there is a $N\in\mathbb{N}$ such that for all $n\ge N$ and all $x$ in the domain $|f_n(x)-f(x)|<\epsilon$) and the uniform Cauchy criterion (i.e. for each $\epsilon>0$ there exists an $N\in \mathbb{N}$ such that for all $n,m>N$ and all $x$ in the domain $|f_n(x)-f_m(x)|<\epsilon$ )

Thanks!

Community
  • 1
user140776
  • 1,974
  • I think one uses the fact that $\epsilon$ is an upper bound over the set $I$, so its supremum must also be bounded by $\epsilon$ – qqo Sep 29 '16 at 19:48
  • That makes sense! So then this seems to work for any set, right? It is not necessary that $I$ be a bounded interval, or even an interval? – user140776 Sep 29 '16 at 19:53

0 Answers0