> [!theorem] > > Let $\cx$, $\cy$ be [[Banach Space|Banach spaces]] and $T: \cx \to \cy$ be a [[Closed Map|closed linear map]], then $T \in L(\cx, \cy)$ is [[Bounded Linear Map|bounded]]. > > *Proof*. First use the fact that the product norm ensures that projection maps are continuous. Build a correspondence the graph itself and $\cx$ via a projection. Since the graph is closed and $\cx$ is complete, this correspondence is an isomorphism by the Open Mapping Theorem. Reversing the projection to $\cx$ and composing it with the projection to $\cy$ yields the original function as a composition of continuous linear maps. > > Let > $ > \pi_x: \Gamma(T) \to \cx \quad (x, Tx) \mapsto x > $ > and > $ > \pi_y: \Gamma(T) \to \cy \quad (x, Tx) \mapsto Tx > $ > be projection maps, then both of them are continuous as > $ > \norm{(x_1 - x_2, Tx_1 - Tx_2)} < \varepsilon \Rightarrow \norm{x_1 - x_2}, \norm{Tx_1 - Tx_2} < \varepsilon > $ > > As $\cx, \cy$ are both complete, so is $\cx \times \cy$. Since $\Gamma(T)$ is closed in $\cx \times \cy$, it is also complete. Now, each point on the graph has a unique $x$ value, meaning that $\pi_x \in L(\Gamma(T), \cx)$ is a bijection. By the [[Open Mapping Theorem]], $\pi_x$ is an isomorphism and $\pi_x^{-1} \in L(\cx, \Gamma(T))$ is a continuous linear map. > > Composing the two functions yields > $ > \pi_y \circ \pi_x^{-1}(x) = \pi_y(x, Tx) = Tx > $ > $T$ as a composition of continuous linear maps, which is also continuous.