Is $ C ^ 1 [ 0 , 1 ] $ compact? Is the closure of the unit ball of $ C ^ 1 [ 0 , 1 ] $ in $ C [ 0 , 1 ] $ compact?

Question

I was reading about compactness. I faced two problems.

Is $ C ^ 1 [ 0 , 1 ] $ compact?

$$ \| f - g \| = \max _ { x \in [0, 1] } | f ( x ) - g ( x ) | $$ This norm is given with the two spaces - $ C ^ 1 [ 0 , 1 ] $ and the unit ball of $ C ^ 1 [ 0 , 1 ] $. $ C ^ 1 [ 0 , 1 ] $ is the space of all differentiable functions on $ [ 0 , 1 ] $.

Is the closure of the unit ball of $ C ^ 1 [ 0 , 1 ] $ in $ C [ 0 , 1 ] $ compact?

I think $ C ^ 1 [ 0 , 1 ] $ is not compact as $ f _ n ( x ) = \sqrt { \left( x - \frac 1 2 \right) ^ 2 + \frac 1 n } $ uniformly convergent to $ \left| x - \frac 1 2 \right| $ which is not differentiable at $ \frac 1 2 $.

Is my argument correct?

I have no idea about the closure of the unit ball of $ C ^ 1 [ 0 , 1 ] $. Is this a compact space?

Answer 1

You are making things complicated. $C^{1} [0,1]$ is not compact because the sequence of constant functions $\{1,2,\cdots\}$ has no convergent subsequence. [No normed linear space is compact because on unboundedness]. The closure of the unit ball in $C^{1}[0,1]$ w.r.t the sup norm is nothing but the closed unit ball of $C[0,1]$, thanks to Weierstrass Approximation Theorem. The unit ball of $C[0,1]$ is not compact because $\{x^{n}\}$ has no convergent subsequence.

Answer 2

Your argument is basically correct: if $f_n(x)=\sqrt{\left(x-\frac12\right)^2+\frac1n}$, then, in $C[0,1]$, $\lim_{n\in\mathbb N}=f$, with $f(x)=\left\lvert x-\frac12\right\rvert$. But $f\notin C^1[0,1]$. Therefore, the unit ball of $C^1[0,1]$ is not a closed subset of $C[0,1]$. But, if it was compact, it would be a closed subset.

It follows that $C^1[0,1]$ is not compact either.

Answer 3

Your example works fine to prove that $C^1[0,1]$ is not compact: it is a sequence of $C^1$ functions which converges (uniformly) to a function which is not $C^1$, meaning the sequence has no convergent subsequences in $C^1$. (I think the author meant for you to pick an easier sequence like $f_n=n$ or something, but that's not really important.)

Now note that $\|f_n\|=\sqrt{\frac14+\frac1n}$. This means that from $n=2$ on, these functions are in the unit ball. So, setting $g_n=f_{n+1}$, we have that $g_n$ is a sequence of functions in the unit ball.

The convergence properties of $g_n$ are exactly the same as those of $f_n$, meaning it has no convergent subsequence in $C^1[0,1]$, and therefore no convergent subsequence in the unit ball, proving that the unit ball isn't compact.