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Let $A$ be defined as

$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$

I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?

The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.

But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.

I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.

The answer is NO.

Consider the sequence
$$
f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
$$
Then
$$
|,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
$$
If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.