Let $X,Y,Z$ be real Banach spaces.
Let $T:X \to Y$ linear, $J:Y \to Z$ be linear, injective and bounded such that $JT:X\to Z$ is also bounded, too. Prove that T is bounded.
My idea was to write $T=J^{-1}(JT)$ and use the closed graph theorem. Since $JT$ is linear and bounded it is closed (has a closed graph). Moreover $J$ is closed. For unbounded operators I found if $J$ injective and closed $J^{-1}$ is closed (see my post If a linear operator $A$ is closed and injective, then its inverse $A^{-1}$ is also closed) but this doesn't apply here. If I would have surjectivity of $J$ I could use the inverse mapping theorem to show the inverse is linear and bounded and hence closed but I don't think $J$ is surjective.
Anyhow, if I would have $J$ is closed then since a composition of a bounded and closed operator is closed (see Composition of continuous and closed operators is closed) $T$ would be closed and hence bounded.
Best Answer
Using the Closed Graph Theorem should solve the problem.
Take a sequence $(x_n) \subset X$ such that $x_n \to x$ and $T(x_n) \to y$ for limits $x \in X$ and $y \in Y$. We will show $T(x) = y$. Then $T$ is closed, hence bounded.
Note that by continuity of $JT$, $JT(x_n) \to JT(x)$. Meanwhile by continuity of $J$, $J(Tx_n) \to J(y)$. Since $JT(x_n) = J(Tx_n)$, it follows from the uniqueness of limits that $J(y) = J(Tx)$. By injectivity of $J$, $y = Tx$ as required.