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Consider the following subsets of $C([0,1],\|\cdot\|_{\sup}).$ $$A_1=\{f\in C([0,1])\mid |f(x)|\leq 1, \forall x\in[0,1]\}$$ $$A_2=\{f\in C^1([0,1])\mid |f'(x)|\leq 2, \forall x\in[0,1]\},$$ and $$A_3=\{f\in C([0,1])\mid f\in \mathbb{R}[x]\}.$$

Now define $B_1=A_1\cap A_2$, $B_2=A_1\cap A_3$ and $B_3=A_1\cap A_2\cap A_3.$

I have shown that $B_1$ and $B_2$ are both precompact using Arzela-Ascoli.

However, I'm not sure how to show either are compact. I know that is suffices to show that both $B_1$ and $B_2$ are closed, i.e. if $f_n\in B_1$, $f_n\to f\in B_1$.

How do I actually go about showing closedness in these cases?

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$B_1$ is not closed because there are sequences of $C^1$ functions with derivative bounded by $2$ and $\sup$ norm less than $1$ converging uniformly to functions that are not $C^1$, e.g. $f_n(x)=|x-\frac12|^{1+\frac1n}$.

$B_2$ is not closed by the Weierstrass approximation theorem. Take any continuous nonpolynomial with $\sup$ norm strictly less than $1$, and you can find a sequence of in $B_2$ converging uniformly to it.

$B_2$ is also not precompact, as for example $(x^n)$ has no uniformly convergent subsequence.

An example to show that $B_3$ is not closed would be the sequence of Taylor polynomials of $\sin(x)$, which also shows that $B_2$ is not closed.

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  • $\begingroup$ "$B_2$ is not closed by the Weierstrass approximation theorem. Take any continuous nonpolynomial with $\sup$ norm strictly less than $1$, and you can find a sequence in B_2 converging uniformly to it." Can you elaborate on this? If you can find such a sequence, doesn't that not imply $B_2$ is closed? $\endgroup$ – user225477 Nov 1 '17 at 2:18
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    $\begingroup$ @Zermelo's_Choice: $B_2$ is a subset of $A_3$, so by showing that $B_2$ contains sequences that converge uniformly to nonpolynomials, you are showing that $B_2$ contains sequences converging to things outside of $A_3$, hence outside of $B_2$, hence $B_2$ is not closed. $\endgroup$ – Jonas Meyer Nov 1 '17 at 2:22

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