Let $S,T$ be linear operators. Then $S$ is called a restriction of $T$ if $D(S) \subset D(T)$ (where $D$ denotes the domain of the linear operator) and $Sx = Tx$ for all $x \in D(S)$.

We denote by $G(T) := \{ (x,Tx) | x \in D(T) \}$ the graph of $T$.

$T$ is called closed if $G(T)$ is closed.

$S$ is called closable if there exists a linear operator $\bar{S}$ such that $\overline{G(S)} = G(\bar{S})$. In this case $\bar{S}$ is called closure of $S$.

Now I want to proof the following: If $S \subset T$ where $T$ is closed then $S$ must be closable.

My ideas so far: If $S \subset T$ and $T$ is closed then $\overline{G(S)} \subset \overline{G(T)} = G(T)$. But I am not sure how to use this to show the existence of such an operator $\bar{S}$.

Could you please help me? Thank you.

  • $\begingroup$ $\overline{G(S)}$ is a (closed) linear subspace of $X\times Y$. What conditions must a linear subspace of $X\times Y$ satisfy to be the graph of a linear operator $A \colon D(A) \to Y$, where $D(A)$ is a linear subspace of $X$? Are these conditions a consequence of $\overline{G(S)} \subset G(T)$? $\endgroup$ – Daniel Fischer Dec 2 '17 at 21:01
  • $\begingroup$ The condition would be that the graph has to be $D(A) \times \operatorname{range}(A)$, or what do you mean? I don't know what it means in terms of the second question. $\endgroup$ – Diglett Dec 2 '17 at 21:51
  • $\begingroup$ $D(A) \times \operatorname{range} (A)$ is rarely the graph of a map. What can you tell about $G(A) \cap (\{x\} \times Y)$ for an $x\in X$? $\endgroup$ – Daniel Fischer Dec 2 '17 at 21:56
  • $\begingroup$ Whoops, I see my mistake now! Well, the only element in $G(A) \cap (\{x\} \times Y)$ is $(x,Ax)$, am I right? Of course if $x \in D(A)$. If $x \not\in D(A)$ then the intersection is empty. $\endgroup$ – Diglett Dec 2 '17 at 22:06
  • $\begingroup$ Right. Can you then go on to show "A linear subspace $H$ of $X \times Y$ is the graph of a linear operator $A \colon D(A) \to Y$, where $D(A)$ is a linear subspace of $X$ if and only if $H \cap (\{0\}\times Y) = \{(0,0)\}$."? $\endgroup$ – Daniel Fischer Dec 2 '17 at 22:09

A linear operator $T : \mathcal{D}(T) \subseteq X\rightarrow Y$ is closed iff the graph $$ \mathcal{G}(T)=\{ (x,Tx) \in X\times Y : x\in\mathcal{D}(T) \} $$ is a closed subspace of $X\times Y$. A subspace $\mathcal{M}$ of $X\times Y$ is a graph iff the only element $(0,y)\in \mathcal{M}$ is where $y=0$. A linear operator $T$ is closeable iff the closure of the graph $\mathcal{G}(T)$ in $X\times Y$ is a graph. Because of this, it can be seen that the the restriction of any closed operator is closable.

  • $\begingroup$ Hey, thanks for your response! Could you please explain why the implication $(0,y) \in \mathcal{M} \Rightarrow y=0$ implies that $\mathcal{M}$ is a graph? And I don't understand why your last statement holds. $\endgroup$ – Diglett Dec 3 '17 at 10:37
  • 1
    $\begingroup$ @Diglett : Let $\mathcal{D}(T) = \{ x \in X : (x,y) \in \mathcal{M} \mbox{for some } y \in Y\}$. For every $x\in\mathcal{D}(T)$ there is a unique $y$ such that $(x,y)\in\mathcal{M}$, which then defines $Tx = y$. This is because $(x,y_1),(x,y_2)\in\mathcal{M}$ would give $(0,y_2-y_1)\in\mathcal{M}$, which, by assumption, would give $y_2=y_1$. The linearity of $\mathcal{M}$ then forces the map $T$ to be linear. So a graph is closeable iff it does not pick up $(0,y)$ in the closure for some $y\ne 0$. $\endgroup$ – DisintegratingByParts Dec 3 '17 at 21:13

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