I've got some doubts proving that $$H^1_0((a,b))=\{u\in AC([a,b]): u'\in L^2 \text{ and } u(a)=u(b)=0\}:=X.$$Let $$\mathcal A=\{v\in C^2([a,b]):v(a)=v(b)=0\}.$$
- $H^1_0((a,b))\subseteq X.$
Let $u\in H^1_0(a,b)$. We want to use the Fundamental theorem of calculus for Lebesgue's integral to prove that $u\in X$.
Since $u\in H^1_0(a,b)$, $u\in L^2$ and there exists a sequence $(u_h)_h\subset \mathcal A$ and a function $w\in L^2(a,b)$ such that:
- $u_h\rightarrow u$ in $L^2$;
- $u'_h\rightarrow w$ in $L^2$.
Now we know that $$ (*)\qquad u_h(x)=u_h(x)-u_h(a)=\int_a^x u'_h(t)dt, \text{ for each } x\in [a,b], h\in \Bbb N,$$
and we want to infer that $$ (**) \qquad u(x)=\int_a^x w(t)dt.$$
My problems start here. I think we should pass to the limit for $h\rightarrow \infty$ in $(*)$, using 1., 2. and Lebesgue's dominated convergence theorem.
I can't justify precisely why $(**)$ holds.
Concerning the application of Lebesgue's dominated convergence theorem, we only know that a subsequence of $u'_h$ pointwise converges to $w$, let's say $({u'_h}_k)_k$. We have to show that there exists $g\in L^1$ such that $|{u'_h}_k (x)|\leq |g(x)|$ for each $x\in[a,b],$ for each $k\in \Bbb N$. But who is this $g$?
If we had $g$, then $$\int_a^x {u'_h}_k(t)dt\rightarrow \int_a^x w(t)dt.$$
On the other hand, if $\lim_{k\to \infty} {u_h}_k (x)=u(x)$ (for the same $k$) we would have finished, but I'm not sure that $\lim_{k\to \infty} {u_h}_k (x)=u(x)$ is true.
Can someone help me, please? Any suggestion would be very appreciated.
Best Answer
This proof cover all cases: $a,b$ finite or not. In the case that $a,b$ are not finite, $u(a)$ is understood as $\lim_{x\to-\infty}u(x)$. Analogous for $b$. Also, if $a,b$ are not finite, then we weill consider locally things, i.e. $BV_{loc}((a,b))$.
I am also assuming that $H_0^1((a,b))$ is the closure of $C_0^1((a,b))$ with respect to the $H^1((a,b))$ norm.
Take $u\in H_0^1((a,b))$ and let $\eta_\delta$ be the standard mollifier sequence. Let $u_\delta=\eta_\delta\star u$ and note that for any $c\in (a,b)$ $$|u_\delta(x)-u_\epsilon(x)|\le \int_c^x |u'_\delta (t)-u'_\epsilon(t)|dt+|u_\delta (c)-u_\epsilon(c)|\tag{1}.$$
If $c$ is a Lebesgue point of $u$, we conclude from $(1)$ that $u_\delta\to u$ uniformaly in compact sets of $(a,b)$.
Therefore, once $$u_\delta (x)=\int_c^x u'_\delta (t)dt +u_\delta(c),\tag{2}$$
and $u'_\delta \to u'$ in $L^2_{loc}((a,b))$, we must conclude that $$u(x)=\int_c^x u'(t)dt+u(c)\tag{3}$$
If, for example $a$ is finite, then $u_\delta$ converge to $u$ uniformly in every compact set of the form $[a,y]$, hence, noting that $u_\delta (a)=0$ for all $\delta$, we have from $(2)$ that $u(a)=0$. The same happens for $b$, if it is finite.
On the other hand, if for example, $a=\infty$ then $lim_{x\to -\infty} u(x)=a$ (prove it). We conclude from the above that $u\in X$.
$X$ is obviously contained in $H^1((a,b))$, so it only remains to show that $u$ can be approximated by a sequence $u_\delta$ in $C_0^1((a,b))$. Let $u_\delta=\eta_\delta\star u$. This sequence satisfies $u_\delta (a)=u_\delta(b)=0$ for all $\delta$, so it is possible to aproximate each $u_\delta$ by a function in $C_0^1((a,b))$.
You can try to do it by yourself or you can take a look in theorem 8.12 of Brezis book.