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The statement of the theorem is:

Let $E$ and $F$ be two reflexive Banach spaces. Let $ A:D(A) \subset E \rightarrow F$ be an unbounded linear operator that is densely defined and closed. Then $D(A')$ is dense in $F'$. Thus $A''$ is well defined ( $A'': D(A'')\subset E'' \rightarrow F''$) and it may also be viewed as an unbounded operator from E into F. Then we have $A''=A$.

At the end of the proof, in order to prove $A''=A$ it is stated:

$I[ G(A')]=G(A)^{\bot}$, $I[G(A'')]=G(A')^{\bot}$ $\implies$ $G(A'')=G(A)^{\bot \bot}=G(A)$

Where $I:F'\times E' \rightarrow E' \times F'$ is defined by $I([v,f])=[-f,v]$. I suppose that when he says $I[G(A'')]$ he is considering $I$ with the same form but defined on $E''\times F''$.

The last equality should be ok cause $G(A) \subset E \times F$ is closed, so $\overline{G(A)}=G(A)$ (is this correct?). I am trying to understand why it is true that $G(A'')=G(A)^{\bot \bot}$ . I have proved that $I[G(A')]=G(A)^{\bot}$, is it also true that $I[G(A')^{\bot}]=G(A)$?

Could anyone help me to prove it? Thank you all in advance.

F. Buseghin
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1 Answers1

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There is a less tricky way of proving this. Let $J_E, J_F$ be the canonical injections from $E, F$ to $E'', F''$, respectively.

Then, if $x \in D(A)$ is such that $J_Ex \in D(A'')$, then

$$ \langle A''J_E x, f \rangle = \langle J_E x, A' f\rangle = \langle A'f, x \rangle = \langle f, Ax \rangle = \langle J_F Ax, f \rangle $$

for all $f \in D(A')$. So $A''J_Ex$ and $J_FAx$ agree on $D(A')$ and thus $A''J_E x = J_F Ax$, since these are continuous and $D(A')$ is dense in $F'$. Now notice that \begin{align*} D(A'') &= \{p \in E'': \exists C \ge 0 \text{ with } |\langle p, A'f \rangle| \le C\|f\| \forall f \in D(A')\}\\ &= \{J_Ex \in E'': x \in E \text{ and } \exists C \ge 0 \text{ with } |\langle J_Ex, A'f \rangle| \le C\|f\| \forall f \in D(A')\}\\ &= \{J_Ex \in E'': x \in E \text{ and } \exists C \ge 0 \text{ with } |\langle A'f, x \rangle| \le C\|f\| \forall f \in D(A')\}\\ &\supseteq \{J_Ex \in E'': x \in D(A) \text{ and } \exists C \ge 0 \text{ with } |\langle A'f, x \rangle| \le C\|f\| \forall f \in D(A')\}\\ &= J_E\{x \in D(A): \exists C \ge 0 \text{ with } |\langle f, Ax \rangle| \le C\|f\| \forall f \in D(A')\}\\ &= J_E(D(A)), \end{align*}

hence the requirement above that $J_Ex \in D(A'')$ is superfluous, and we se that $A''J_E$ and $J_FA$ agree on $D(A)$, that is,

$$ A''J_E = J_F A. $$

Since $E$ and $F$ are reflexive, one identifies $J_E$ and $J_F$ with the identities and the result follows.

WeakestTopology
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