Characterisation of one-dimensional Sobolev space

Question

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 $$\mathcal A=\{v\in C^2([a,b]):v(a)=v(b)=0\}.$$

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• $H^1_0((a,b))\subseteq X.$

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:

1. $u_h\rightarrow u$ in $L^2$;
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.

Answer 1

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.

I- $H_0^1((a,b))\subset X$.

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 $u\in X$.

$X\subset H_0^1((a,b))$.

$X$ 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.

Answer 2

Let $Lf = -if'$ be defined on the linear space $\mathcal{D}(L)$ of absolutely continuous functions $f \in L^{2}[a,b]$ for which $f(a)=f(b)$ and $f' \in L^{2}[a,b]$. $L$ is symmetric on its domain, i.e., $(Lf,g)=(f,Lg)$ for all $f,g\in\mathcal{D}(L)$. It is not hard to show that $(L-\lambda I)$ is surjective for $\lambda \ne n\frac{b-a}{2\pi}$ for $n=0,\pm 1,\pm 2,\cdots$, which is verified by directly solving the the following ODE for arbitrary $g \in L^{2}$: $$ -if'-\lambda f = g,\\ f(a)=f(b). $$ This ODE has an explicit classical solution. Because $L\pm iI$ are surjective, then $L$ is densely-defined and selfadjoint. In particular, $L$ is closed. And, $$ \|(L+iI)f\|^{2}_{L^{2}}=\|Lf\|^{2}_{L^{2}}+\|f\|^{2}_{L^{2}}=\|f'\|^{2}_{L^{2}}+\|f\|^{2}_{L^{2}},\;\; f \in \mathcal{D}(L). $$ The restriction $L_{0}$ of $L$ to $\mathcal{D}(L_{0})=\{ f \in \mathcal{D}(L) : f(a)=f(b)=0 \}$ has a graph which is of co-dimension $1$ in the graph of $L$. Because $L$ has a closed graph, then the same is true of $L_{0}$.

Now, assuming $\{ u_{h} \} \subset \mathcal{A}$ with $u_{h}\rightarrow u$ in $L^{2}$ and $u_{h}'\rightarrow w$ in $L^{2}$, then the ordered pair $(u,-iw)$ must be in the closure of the graph of $L_{0}$, which is the same as the graph of $L_{0}$ because $L_{0}$ is closed. That guarantees that $u$ is absolutely continuous with $u'\in L^{2}$ and $u(a)=0=u(b)$. Therefore, $H^{1}_{0}\subseteq \mathcal{D}(L_{0})$. The opposite inclusion follows from the norm identity stated above.