Cfrgtkky

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).

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 $fnotin C^1_{pw}.$

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.

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