Let $E,F$ be two Banach spaces and $A$ an unbouded operator from $D(A) \subset E$ to $F$.
We want to prove the following lemma :
Let $A$ be a closed operator. The following are equivalent :
- $A$ is continuous in $D(A)$ (i.e $\|Au\|<C\|u\|$).
- $D(A)$ is closed in $E$.
(2 $\Rightarrow 1$) if $D(A)$ is closed in $E$ then it is a Banach space, since $A$ is closed then by the Closed graph theorem $A$ is continuous.
But how to prove the other direction ?
Best Answer
Suppose that $A$ is continuous. Suppose that the sequence $(x_n) \subset D(A)$ converges with $x_n \to x \in E$; we want to show that $x \in D(A)$.
The continuity of $A$ implies that since $x_n$ is a (convergent and therefore) Cauchy sequence, $Ax_n$ must also be a Cauchy sequence in the Banach space $F$, which means that $Ax_n$ has a limit $y \in F$. Because $A$ is a closed operator, the fact that $x_n \to x$ and $Ax_n \to y$ implies that we must have $Ax = y$, which means that $x$ is an element of $D(A)$.
Thus, we conclude that $D(A)$ must be closed.