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Let $L$ be a Hilbert space and $T$ be a linear densely defined operator, $T: D(T)\subset L \to L$ , $\overline {D(T)}=L$ We can make $D(T)$ , a prehilbert space by defining an inner product $\langle \mathbf{x},\mathbf{y}\rangle_T=\langle \mathbf{x},\mathbf{y}\rangle_L+\langle T\mathbf{x},T\mathbf{y}\rangle_L$ for all $x,y \in D(T)$. Please note, with this new structure (inner product/norm) on $D(T)$, it is continuously embedded to $L$. Now, I like to have a completion of $D(T)$ (considering $D(T)$ as a prehilbert space with new structure and the completion would be some Hilbert space). Suppose we have a completion $W$.

1) Can I have $W \subset L$ (an isometric isomorphism between $W$and a subspace of L)? In other words, do I have a realization of completion of $D(T)$ as a subspace of $L$?

2)Generally, if $ A \subset X$ and $A$ and $X$ have different structure (norm or inner product), when I can realize a completion of A as a subspace of $X$? Is there any good reference for this discussion?

Thanks in advance for your help.

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    $\begingroup$ Yes, I think that Kato's "Linear Operators" is the right source. By the way, for things to work T needs to be closable. $\endgroup$ – Urgje Jul 31 '14 at 18:36
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Question 1 is really not the right question to ask :)

As stated, the answer to question 1 is usually "yes". For instance, if $L$ is a separable infinite-dimensional Hilbert space, then so is $W$, and every infinite-dimensional separable Hilbert space is isometrically isomorphic to $\ell^2$. But you don't really know much about what the isomorphisms are.

Here is the question I think you really want to ask. Consider the inclusion map $i : D(T) \to L$. This is a continuous injective linear operator (though not an isometry unless $T=0$), and so it has a continuous linear extension $j : W \to L$. We might ask whether $j$ is injective or surjective.

Having $j$ be injective is equivalent to having the operator $T$ be closable. A typical example of a non-closable operator is the following.

Let $L = L^2([0,1])$, let $D(T) = C([0,1]) \subset L$ which is dense, and define $T$ by $Tf = f(1) \cdot 1$. Consider the sequence $f_n(x) = x^n$, for which $f_n \to 0$ in $L$. Since $T f_n = 1$ for all $n$, we also have that $\{f_n\}$ is Cauchy in the $D(T)$ graph norm, so it converges in graph norm to some $u \in W$. But since $\|f_n - 0\|_{D(T)} \ge 1$, $f_n$ does not converge to 0 in $D(T)$ norm, so $u \ne 0$. By the continuity of $j$, $ju$ equals the $L$-norm limit of $j f_n = f_n$, which is 0. So $j$ is not injective.

If $j$ is bijective, then by the open mapping theorem its inverse is bounded. It then follows easily that $T$ is also bounded.

(I was trying to work out what it means if $j$ is surjective but not injective, or if that's even possible, but I haven't managed it yet.)

Note also that $T$ is a closed operator iff $D(T)$ is complete in the graph norm (so that $D(T) = W$).

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  • $\begingroup$ For an example where $j\colon W\to L$ is surjective but not injective, consider $L = \ell^2(\mathbb{N})$, $D(T) = \operatorname{span} \{ e_k : k \in \mathbb{N}\}$, and $T\colon e_k \mapsto (k+1)e_0$. The completion of $D(T)$ is effectively $\ell^2(\mathbb{N}) \times \mathbb{C}$, and $j$ the projection onto the first factor with that identification. $\endgroup$ – Daniel Fischer Aug 1 '14 at 14:22
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This is a comment to Nate Eldredge's answer.

The correct interpretation of $D(T)$ with the graph norm is not as a subspace of $L$, but rather as a subspace of the product $L \times L$. Indeed, $D(T)$ can be identified with the graph of $T$, i.e. $G(T) = \{(x,Tx) \mid x \in D(T)\}$ via the map $D(T) \to G(T), x \mapsto (x,Tx)$. The graph norm, then, is the restriction of the product $2$-norm on $L \times L$ to $G(T)$.

The completion $W$ of $G(T)$ is simply its closure as a subspace of $L \times L$. If $W$ happens to be the graph of a partially defined function (necessarily a linear operator) then $T$ is said to be closable. As Nate notes, $W$ is a graph of a partially defined function if and only of the projection $j: W \to L$ to the first coordinate is injective.

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Please correct me if there is any logical flaw. I appreciate your comments.

Suppose T is closable. Then, we can define the closure of T, $\overline{T}:D(\overline{T})\subset L\to L$ with $T\subset \overline{T}$ and $G(\overline{T})=\overline{G(T)}$

My claim is: A completion of $(D(T),\|\cdot\|_T)$ is $(D(\overline{T}),\|\cdot\|_{\overline {T}})$ (the corresponding norms are graph norm)

Proof:

Clearly, $(D(T),\|\cdot\|_T)$ and $(G(T),\|\cdot\|_{L\times L})$ are isometric isomorphism (as user167308 indicated) via the map $D(T) \to G(T), x \mapsto (x,Tx)$.

But, a completion of $(G(T),\|\cdot\|_{L\times L})$ is $(\overline{G(T)},\|\cdot\|_{L\times L})$. So, a completion of $(D(T),\|\cdot\|_T)$ is $(\overline{G(T)},\|\cdot\|_{L\times L})$.

As, $G(\overline{T})=\overline{G(T)}$ and $(D(\overline{T}),\|\cdot\|_\overline{T})$ and $(G(\overline{T}),\|\cdot\|_{L\times L})$ are isometric isomorphism via the map $D(\overline{T}) \to G(\overline{T}), x \mapsto (x,\overline{T}x)$, the claim follows (also,$(D(\overline{T})$ is continuously embedded to $L$ ).

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  • $\begingroup$ Looks good to me. $\endgroup$ – Nate Eldredge Aug 1 '14 at 14:57
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EDIT: Warning - this part is not correct, according to commenters below:

If the embedding $A\subset X$ is continuous, then any $A$-Cauchy sequence is $X$-Cauchy too. So if $X$ is complete then the completion of $A$ is always a subset of $X$.


The model example is the embedding $H^1(\mathbb{R}^n)\subset L^2(\mathbb{R}^n)$, which can be obtained, more or less $^{[1]}$ like you did, by taking $A=C^\infty_c(\mathbb{R}^n)$ and $$ \lVert f\rVert_A= \lVert f\rVert_{L^2}+\langle f, -\Delta f\rangle_{L^2}.$$ Set-theoretically, $H^1(\mathbb{R}^n)$ turns out to be the space of all square-integrable functions whose distributional derivative is square-integrable, and indeed it is usually presented that way.


Side note $^{[1]}$. To obtain $H^1$ exactly like you did above you need not consider the Laplace operator $-\Delta$ but its square root $\sqrt{-\Delta}$, which is the unique operator $T$ with the property that $$\langle Tf, Tf\rangle_{L^2} = \langle f, -\Delta f\rangle_{L^2}.$$ That such an operator exists can be easily proven by means of the Fourier transform.

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  • $\begingroup$ This isn't correct. It's true that an $A$-Cauchy sequence will also be $X$-Cauchy, but its limit in the completion of $A$ might not be the same as its limit in $X$. I am working on an answer with a counterexample. $\endgroup$ – Nate Eldredge Jul 31 '14 at 18:46
  • $\begingroup$ @NateEldredge: You made me wonder about another thing that I usually took for granted. $\endgroup$ – Giuseppe Negro Jul 31 '14 at 19:38
  • $\begingroup$ I hope later to have time to read other answers and improve mine as well. For the moment I have added a warning $\endgroup$ – Giuseppe Negro Aug 1 '14 at 13:44

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