Let $X$ be a reflexive Banach space and $(A,D(A))$ a densely defined linear operator in $X$. We define its adjoint by $$D(A'):=\{x'\in X': \sup_{\|x\|\le 1, x\in D(A)}|\langle Ax,x'\rangle|<\infty\}$$ and $\langle x,A'x'\rangle:=\langle Ax,x'\rangle$ for $x'\in D(A')$. Denoting the range of $A'$ by $R(A')$ and its nullspace by $N(A')$, we want to show that $$\overline{R(A')}=N(A)^{\perp}=\{x'\in X':\langle x,x'\rangle=0\forall x\in N(A)\}.$$ ''$\subset$'' is clear to me, but I do not know how to prove the other inclusion: How can one construct a sequence of functionals converging in $X'$ to some given $x'\in N(A)$ from which it is 'only' known that it vanishes on $N(A)$? Can somebody give me a hint? Is reflexivity somewhere involved in this issue? Thank you very much in advance!
Orthogonal complement in Banach spaces
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functional-analysis
banach-spaces
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0Have you tried a separation theorem? Take $x \in N(A)^\perp \setminus \overline{R(A')}$ and separate it from $\overline{R(A')}$. – 2017-02-11
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2@gerw : I think at some point in the proof you need to show that $D(A')$ separates points of $X$, which seems difficult. I don't think it's even true in general. – 2017-02-11