Even more Analysis (sequences of functions)

[latent sparrow]

So… I have no idea how to even start. Can someone walk me through a solution?

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[proud viper]

this is a hint

[latent sparrow]

I mean I solved (a)

But idk how to generalize it to b

Something like there’s a convergent subsequence of $f_n$ s.t. $\forall r\in\bQ\lim_{k\to\infty}f_{n_k}(r)=f(r)$

Then define $f(x) = \sup_{r\leq x}f(r)$

Then choose two sequences $r_m$, $r’_m$ of rationals s.t. $r_m$ monotonically increases, $r’_m$ monotonically decreases and both converge to $x$, furthermore assume $f$ is continuous at this $x$

Then $$f_{n_k}(r_m)\leq f_{n_k}(x)\leq f_{n_k}(r’m)$$
Take the limit as $k\to\infty$ on both sides and we get $$f(r_m)\leq\lim{k\to\infty}\inf f_{n_k}(x)\leq \lim_{k\to\infty}\sup f_{n_k}(x)\leq f(r’_m)$$

Take the limit as $m\to\infty$ on both sides and we get $$f(x)\leq\lim_{k\to\infty}\inf f_{n_k}(x)$$
$$\lim_{k\to\infty}\sup f_{n_k}(x)\leq f(x)$$

As $f$ is continuous at $x$, and by the above it’s obvious that $$\lim_{k\to\infty}f_{n_k}(x)=f(x)$$ QED

[lime oriole]

ℝανιølι

[latent sparrow]

@proud viper how would I change this to get b??

[proud viper]

use \limsup and \liminf

[proud viper]

it's a refinement, not a more general statement

[proud viper]

As for b)

Suppose f_n to f pointwise and f is continuous and let K be compact, without loss of generality, a closed interval [a,b]

1. Then f is uniformly continuous on [a,b]
2. Partition [a,b] into subintervals a=a_0 < a_1 < .. < a_m = b such that sup |f(s)-f(t)| on each subinterval is at most eps/2 (can replace sup with max as well)
3. By assumpion f_n(a_k) to f(a_k) as n to inf for all k = 0, .., m. This implies there exists N large enough such that n > N implies
   |f_n(a_k) - f(a_k)| \leq eps/2 for all k = 0,..,m.
4. Verify that |f_n(x) - f(x)| \leq eps for every n > N and x in [a,b]

@latent sparrow

[latent sparrow]

Hm alr

That’s quite close to what GPT said, but your argument looks simpler - I’ll try yours first

[latent sparrow]

+close

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