My CVFunctional Analysis Homework 11.15
Continuous linear operator $T$ is not necessarily a closed operator, and a closed operator is not necessarily a continuous linear operator. (Here continuity is only defined on $\mathcal{D}(T)$.)
Proof:
On the space of continuous functions $C[a,b]$, the differentiation operator is a closed operator, but it is not a continuous linear operator. Verification can be found in the lecture notes or textbook p.104.
Now consider the spaces $L^1[a,b]$ (integrable functions) and $C[a,b]$. Clearly, the latter is a proper subset of the former. From real analysis, we know that ${f_n}_{n=1}^\infty \subset C[a,b]$ can approximate functions in $L^1[a,b]$ that are not continuous. Consider the identity operator $I$. Since its domain is not closed, even though the identity operator is obviously continuous on its own domain, it is in fact a non-closed operator (the failure is due to the non-closed domain).
14. Suppose $A,B$ are linear operators defined everywhere on a Hilbert space $H$, and
\((Ax,y)=(x,By),\quad \forall x,y\in H.\)
Prove: $A,B$ are bounded, and $B=A^*$.
Proof:
We prove that $A$ is a closed operator and then apply the Closed Graph Theorem to deduce boundedness.
Let ${x_n}\to x_0$ and $Ax_n\to y_0$. By the assumption,
\[\lim_{n\to\infty}(Ax_n,y)=(y_0,y)=(x_0,By)=(Ax_0,y).\]Thus $\lim_{n\to\infty}Ax_n=Ax_0=y_0$. This shows that $A$ is a closed operator defined everywhere on $H$. By the Closed Graph Theorem, $A$ is bounded.
Similarly, $B$ is also bounded. From $(Ax,y)=(x,A^\star y)=(x,By)$ we conclude that $A^\star=B$.
15. Suppose $X,Y$ are Banach spaces, and $T\in L(X,Y)$. If $T$ is injective, then $T^{-1}$ is a closed operator.
Clearly, $T^{-1}(R(T))$ is both injective and surjective. We have
\[Tx_0=T\left(\lim_{n\to\infty}T^{-1}y_n\right)=\lim_{n\to\infty}TT^{-1}y_n=\lim_{n\to\infty}y_n=y_0.\]This shows $y_0\in R(T)$ and $T^{-1}y_0=x_0$. Hence $T^{-1}$ is a closed operator.
16. Prove: If $T$ is a closed operator, then its graph $G(T)$ is closed.
Take any Cauchy sequence ${x_n}\to x$ in $G(T)$. By the definition of a closed operator, $Tx_n\to Tx=y\in R(T)$. Then
\[\| \langle x_n,Tx_n\rangle - \langle x,Tx\rangle \| = \|x_n-x\|+\|Tx_n-Tx\|\to 0, \quad \text{as } n\to\infty.\]This shows that the Cauchy sequence converges, and $\langle x,Tx\rangle\in G(T)$. Therefore $G(T)$ is complete and hence closed.