Is the set of piecewise $C^1([a,b])$ functions complete with $C^1$ norm?

Question

Here's the $C^1$ norm :

$|| f || = \sup | f | + \sup | f '|$

where the supremum is taken on $[a, b]$.

Please, justify your answer (proofs or counterexamples are needed).

Answer 1

A counterexample may be:

$\forall x \in [0;\frac{1}{2}], f_{1}(x)=0$ and $\forall x \in [\frac{1}{2};1], f_{1}(x)=\frac{1}{2^{0}}$

$\forall x \in [0;\frac{2^{n}-1}{2^{n}}], f_{n}(x)=f_{n-1}(x)$ and $\forall x \in [\frac{2^{n}-1}{2^{n}};1], f_{n}(x)=\sum_{i=0}^{n-1}\frac{1}{2^{i}}$

Answer 2

Choose any sequence $a_0=b > a_1 > a_ 2 > \dots \to a.$ Define $f$ on $(a,b]$ by letting $f$ be a tent of height $h_n$ over $[a_n,a_{n-1}], n=1,2,\dots.$ Do this so that the slopes of the tents $\to 0$ in absolute value (i.e., so that $h_n/(a_{n-1}-a_n) \to 0$). Set $f(a) = 0.$ Then $f\notin C^1_{pw}.$

Now define $f_n = \chi_{[a_n,a_0]}\cdot f, n=1,2,\dots $ Then $f_n \in C^1_{pw}$ for each $n.$ Note that $f_n \to f$ pointwise on $[a,b].$ If $m< n,$ then $f_n-f_m$ is just the set of tents that live on $[a_n,a_m],$ with $f_n-f_m= 0$ everywhere else. It follows that $(f_n)$ is Cauchy in the $C^1_{pw}$ norm.

Could $f_n$ converge to some $g$ in $C^1_{pw}?$ Suppose it does. Then $\sup_{[a,b]}|f_n - g| \to 0.$ Thus $g$ is the pointwise limit of the $f_n.$ But we saw earlier that $f$ is the pointwise limit of the $f_n.$ We conclude $f=g$ everywhere, which is a contradiction. Therefore $C^1_{pw}$ is not complete.