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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:

  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.

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  • $\begingroup$ Are $a,b$ finite? $\endgroup$ – Tomás Aug 20 '14 at 15:57
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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.

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  • $\begingroup$ Thank you for the answer! For the second inclusion also our professor suggested us to use mollifiers, but for the first one he suggested us to use the relations I've written in my question. The actual problem is that I'm not so sure why we can pass to the limit under the integral. $\endgroup$ – batman Aug 21 '14 at 7:43
  • $\begingroup$ If $u\in L^2_{loc}((a,b))$ then $\in L^1_{loc}((a,b))$ by Holder inequality. $\endgroup$ – Tomás Aug 21 '14 at 12:02
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    $\begingroup$ @Tomás Since a question asking about this answer was posted on the site, I thought it would be a good idea to notify you about that question. $\endgroup$ – Martin Sleziak Aug 28 '14 at 9:50
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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.

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