[proofplan] We equip the graph of $A$ with the graph norm and show it is a Banach space (since the graph is closed). The coordinate projection $\pi_X$ from the graph to $X$ is a bounded bijective linear operator between Banach spaces, so the Bounded Inverse Theorem (a consequence of the [Open Mapping Theorem](/theorems/631)) gives a bounded inverse. The bound on $\pi_X^{-1}$ directly yields the bound on $A$. [/proofplan] [step:Construct a Banach space from the closed graph of $A$] Equip the Cartesian product $X \times Y$ with the norm \begin{align*} \|(u, v)\|_{X \times Y} := \|u\|_X + \|v\|_Y. \end{align*} Since $X$ and $Y$ are Banach spaces, $X \times Y$ is a Banach space under this norm (completeness follows from component-wise limits). The graph of $A$, \begin{align*} \operatorname{Graph}(A) := \{(u, Au) : u \in X\} \subseteq X \times Y, \end{align*} is a linear subspace of $X \times Y$. The hypothesis that $A$ is closed means precisely that $\operatorname{Graph}(A)$ is a closed subspace of $X \times Y$. A closed subspace of a Banach space is itself a Banach space (under the restricted norm), so $\operatorname{Graph}(A)$ is a Banach space with norm $\|(u, Au)\|_{X \times Y} = \|u\|_X + \|Au\|_Y$. [/step] [step:Define the coordinate projection and verify it is a bounded bijection] Define the projection onto the first coordinate: \begin{align*} \pi_X: \operatorname{Graph}(A) &\to X \\ (u, Au) &\mapsto u. \end{align*} **Linearity:** For $(u_1, Au_1), (u_2, Au_2) \in \operatorname{Graph}(A)$ and $\alpha \in \mathbb{R}$, $\pi_X(\alpha(u_1, Au_1) + (u_2, Au_2)) = \alpha u_1 + u_2 = \alpha \pi_X(u_1, Au_1) + \pi_X(u_2, Au_2)$. **Boundedness:** For any $(u, Au) \in \operatorname{Graph}(A)$: \begin{align*} \|\pi_X(u, Au)\|_X = \|u\|_X \leq \|u\|_X + \|Au\|_Y = \|(u, Au)\|_{X \times Y}, \end{align*} so $\|\pi_X\| \leq 1$. **Bijectivity:** Since $D(A) = X$, every $u \in X$ arises as the first component of $(u, Au) \in \operatorname{Graph}(A)$, so $\pi_X$ is surjective. The pair $(u, Au)$ is uniquely determined by $u$ (since $A$ is a function), so $\pi_X$ is injective. [/step] [step:Apply the Bounded Inverse Theorem to obtain a bounded inverse] The map $\pi_X: \operatorname{Graph}(A) \to X$ is a bounded bijective linear operator between Banach spaces. By the Bounded Inverse Theorem (which follows from the [Open Mapping Theorem](/theorems/631) applied to the surjective bounded operator $\pi_X$), the inverse \begin{align*} \pi_X^{-1}: X &\to \operatorname{Graph}(A) \\ u &\mapsto (u, Au) \end{align*} is bounded. That is, there exists a constant $C > 0$ such that \begin{align*} \|(u, Au)\|_{X \times Y} \leq C\|u\|_X \quad \text{for every } u \in X. \end{align*} [guided] The Bounded Inverse Theorem states that a bijective bounded linear operator between [Banach spaces](/page/Banach%20Space) has a bounded inverse. This is a deep result: in incomplete normed spaces, a bijective bounded operator can have an unbounded inverse. Why does this follow from the [Open Mapping Theorem](/theorems/631)? The Open Mapping Theorem says that a surjective bounded linear operator between Banach spaces maps open sets to open sets. Applied to $\pi_X: \operatorname{Graph}(A) \to X$, this means $\pi_X$ is an open map. An open bijection has a [continuous](/page/Continuity) inverse: for any open set $U \subseteq X$, $(\pi_X^{-1})^{-1}(U) = \pi_X(U)$ is open, so $\pi_X^{-1}$ is continuous. The bound $\|\pi_X^{-1}\| \leq C$ means $\|(u, Au)\|_{X \times Y} \leq C\|u\|_X$ for all $u \in X$. Since \begin{align*} \|Au\|_Y \leq \|u\|_X + \|Au\|_Y = \|(u, Au)\|_{X \times Y} \leq C\|u\|_X, \end{align*} the graph norm bound gives $A \in \mathcal{L}(X, Y)$ with $\|A\| \leq C$. [/guided] [/step] [step:Extract the bound on $A$ from the graph norm estimate] For every $u \in X$, the graph norm bound gives \begin{align*} \|Au\|_Y \leq \|u\|_X + \|Au\|_Y = \|(u, Au)\|_{X \times Y} \leq C\|u\|_X. \end{align*} Hence $A$ is bounded with $\|A\|_{\mathcal{L}(X,Y)} \leq C$. [/step]