• $H$ be a $\mathbb R$-Hilbert space
  • $D(\mathfrak a)$ be a dense subspace of $H$
  • $\mathfrak a:D(\mathfrak a)\times D(\mathfrak a)\to\mathbb R$ be bounded, i.e. $\exists c\ge 0$ with $$\left|\mathfrak a(u,v)\right|\le c\left\|u\right\|_{\mathfrak a}\left\|v\right\|_{\mathfrak a}\;\;\;\text{for all }u,v\in D(\mathfrak a)$$ where $$\left\|u\right\|_{\mathfrak a}:=\sqrt{\mathfrak a(u,u)+\left\|u\right\|_H^2}\;\;\;\text{for all }u\in D(\mathfrak a)$$

Now, let $$D(A):=\left\{u\in D(\mathfrak a)\mid\exists v\in H:\mathfrak a(\;\cdot\;,u)=\left.\langle\;\cdot\;,v\rangle_H\right|_{D(\mathfrak a)}\right\}\;.\tag 1$$ It's easy to see that $$u\in D(A)\Leftrightarrow u\in D(\mathfrak a)\text{ and }\mathfrak a(\;\cdot\;,u)\text{ is }H\text{-continuous}\;.\tag 2$$

I've seen that many people write $$D(A)=\left\{u\in D(\mathfrak a):\mathfrak a(\;\cdot\;,u)\in H\right\}\;.\tag 3$$ So, my question is: Why are $(1)$ and $(2)$ equivalent?

I guess $(2)$ has to be interpreted in a suitable sense. Clearly, if $u\in D(A)$, we know that $u\in D(\mathfrak a)$ such that $\mathfrak a(\;\cdot\;,u)$ is $H$-continuous. Since $D(\mathfrak a)$ is dense in $H$, there is a unique $F\in H'$ with $$\mathfrak a(\;\cdot\;,u)=\left.F\right|_{D(\mathfrak a)}$$ by the bounded linear transformation theorem. Thus, there is a unique extension of $\mathfrak a(\;\cdot\;,u)$ to $H$. Since $H'\cong H$ we could justify the notation $\mathfrak a(\;\cdot\;,u)\in H$.

For the other direction, $\mathfrak a(\;\cdot\;,u)\in H$ would imply the existence of a unique extension $F\in H'$ of $\mathfrak a(\;\cdot\;,u)$ to $H$. Since $H'\cong H$, there is a unique $v\in H$ with $$F=\langle\;\cdot\;,v\rangle_H$$ and hence $$\mathfrak a(\;\cdot\;,u)=\left.\langle\;\cdot\;,v\rangle_H\right|_{D(\mathfrak a)}\;.$$

Is this the way we need to understand $(3)$?

  • $\begingroup$ can you simplify your questions ? you don't need all those details. $\endgroup$ – reuns Jul 9 '16 at 13:39
  • $\begingroup$ and in general, if $T : D(X) \to Y$ is bounded with $D(X)$ a dense subspace of $X$, then it can be extended to a bounded operator $X \to Y$ by considering for each $x \in X$ a sequence $x_n \in D(X)$ such that $x_n \to x$ $\endgroup$ – reuns Jul 9 '16 at 13:41
  • $\begingroup$ By the way: since $\mathfrak{a}$ is bilinear, your definition of "boundedness" is trivially satisfied by the Cauchy-Schwarz inequality if $\mathfrak{a}$ is non-negative, i.e., $\mathfrak{a}(u,u) \ge 0$ for all $u \in D(\mathfrak{a})$. $\endgroup$ – gerw Jul 9 '16 at 17:04
  • $\begingroup$ @gerw It's the definition of continuity which you can find in Definition 1.4 (3) in Analysis of Heat Equations on Domains by El Maati Ouhabaz. $\endgroup$ – 0xbadf00d Jul 9 '16 at 17:50
  • $\begingroup$ @user1952009 Which "details" are you talking about? The main part of my question is my interpretation of $(3)$. And if I want to ask how I need to interpret $(3)$, I should introduce the necessary objects. $\endgroup$ – 0xbadf00d Jul 9 '16 at 17:53

If you would ask me, it is meant in the sense you mentioned. Since a Hilbert space is isometric isomorphic to its dual space due to the Riesz representation theorem you can identify the functionals in $H'$ with their riesz representers in $H$. I did that frequently in my bachelor thesis. It is not the most formal way but you can give shorter definitions and assumptions by using this identification.


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