Let $X$ and $Y$ be Banach spaces and $A:D(A)\subset X \to Y$ a densily defined linear operator. Suppose the graph of $A$ is closed. Then the follwing are equivalent:
- $D(A)=X$;
- $A$ is bounded;
- $D(A^*)=Y^*$;
- $A^*$ is bounded.
I figured out the equivalence of the first three sentences. And I showed that the third implies the fourth. But I couldn't show that the fourth implies any of the others. I'm suspecting that this last implication is false. But I have no idea about how to build a counterexample (if there is one). So, is it true?
If $A^*$ is defined as $A\left(f\right)\left(x\right)=f\left(A\left(x\right)\right)$ for all $x\in X$ and all $f\in Y'$, then we have $$\|A^*\|_{\mathcal{L}\left(Y';X'\right)} =\sup_{\|f\|_{Y'}=1}\|A^*\left(f\right)\|_{X'} =\sup_{\|f\|_{Y'}=1}\sup_{\|x\|_{X}=1}|f\left(A\left(x\right)\right)|$$ $$=\sup_{\|x\|_{X}=1}\sup_{\|f\|_{Y'}=1}|f\left(A\left(x\right)\right)| =\sup_{\|x\|_{X}=1}\|A\left(x\right)\|_Y =\|A\|_{\mathcal{L}\left(X;Y\right)}$$
then $A$ is bounded iff $A^*$ is bounded. Note: the above operator norms exist iff $A$ and $A^*$ are bounded, so writing them without knowing the last conditions is an abuse, but it was easier to write it like that.