Conversation started Dec 20, 2011 at 8:03.
Matt
Matt
Dec 20, 2011 08:03
How do you see that $\sup_{x \in X} \lim_{m \to \infty} | f_n(x) - f_m(x)| \leq \lim \inf_{m \to \infty} \sup_{x \in X} |f_n(x) - f_m (x)|$?
I thought $\sup_{x \in X} \lim_{m \to \infty} | f_n(x) - f_m(x)| = \sup_{x \in X} \lim \inf_{m \to \infty} | f_n(x) - f_m(x)|$ but then I'm not sure how to justify swapping the $\sup$ and the limit...
t.b.
t.b.
@Matt Can you say a word about the hypotheses on $f_n$?
Matt
Matt
@tb They are continuous and $X$ is compact but otherwise no hypotheses.
It's a step in a proof that $C(X)$ with the sup norm is complete.
t.b.
t.b.
I see, so you assume that $f_n$ is a $\|\cdot\|_\infty$-Cauchy sequence, don't you?
Matt
Matt
Yes, exactly.
t.b.
t.b.
Dec 20, 2011 08:19
You then have already shown that $f_n$ converges pointwise to a function $f: X \to \mathbb{R}$, now you want to show that this convergence is uniform if I understand correctly.
Matt
Matt
Yes.
Skullpatrol
Skullpatrol
@JM "There will always be asshats, in both the Internet and real life." Without asshats the brains couldn't get rid of their waste products.
Matt
Matt
The proof is here.
Skullpatrol
Skullpatrol
The proof is all around us.
t.b.
t.b.
I would do this as follows (a bit less efficiently but more transparently): choose $\varepsilon \gt 0$ and let $N$ be such that for $n,m \geq N$ you have $\|f_n - f_m \|_\infty \leq \varepsilon$. But this means that for all $x$ you have $|f_N (x) - f_m(x)| \leq \varepsilon$ and thus $f(x) \in [f_N(x) - \varepsilon, f_N(x)+\varepsilon]$.
This shows that $\|f_N - f\|_\infty \leq \varepsilon$ and, as $\|f_N - f_n\| \leq \varepsilon$ we also have $$\|f_n - f\| \leq \|f_n - f_N\| + \|f_N - f\| \leq 2\varepsilon$$
 
Conversation ended Dec 20, 2011 at 8:28.