Let $(X, \|\cdot\|_X)$, $(Y, \|\cdot\|_Y)$ be two Banach spaces and $T: X \to Y$ a linear operator. Then:
$T$ is continuous $\iff$ $T$ is closed
where "$T$ closed" means its graph $G(T)\equiv\{(x, Tx) \in X \oplus Y\}$ is a closed subspace of $(X \oplus Y, \|\cdot\|_{X\oplus Y } := \max\{\|\cdot\|_X, \|\cdot\|_Y \})$, or, equivalently, $\forall$ converging sequence $\{x_n\}_{n \in \mathbb{N}}$ such that $\{Tx_n\}_{n \in \mathbb{N}}$ also converges, we have $\displaystyle \lim_n T(x_n) = T (\lim_n x_n)$
I have some trouble proving $\Longleftarrow$ since I never use the fact that $T$ is closed.
The projection maps
- $P_X : G(T) \to X$
- $P_Y : G(T) \to Y$
are linear bounded and onto, therefore open maps. Moreover, $P_X$ is injective, for $x_1 \neq x_2 \implies (x_1, Tx_1) \neq (x_2, Tx_2)$, even in the worst case of having $Tx_1 = Tx_2$. This means the inverse $P_X^{-1} : X \to G(T)$ is also (linear and) bounded.
Hence $T = P_Y \circ P_X^{-1}$ is continuous, being the composition of two continuous operators.
I think I overlooked something… any help would be appreciated.
Best Answer
You said that $P_X$ is linear bounded and onto, therefore open. This requires the domain to be complete.