Linear operator from space to subspace is bounded

Question

Let $(X,||\cdot||_X),(Y,||\cdot ||_Y)$ be Banach spaces with $Y$ a subset of $X$, and assume the obvious map $(Y,||\cdot ||_Y)\to (Y,||\cdot||_X)$ is continuous. Let $T: X\to X$ be a continuous linear operator such that $Tx\in Y\forall x\in X$. How could we prove that $T:(X,||\cdot||_X)\to (Y,||\cdot||_Y)$ is bounded?

Answer 1

By the Closed Graph Theorem, it suffices to prove that if $x_n\to x$ in $X$ and $Tx_n \to y$ in $Y$, then $y=Tx$. By the hypothesis, $||z||_X\le C||z||_Y\forall z\in Y$ for some $C>0$. Since $||Tx_n-y||_Y\to 0$, this inequality now implies $||Tx_n-y||_X\to 0$ and thus $Tx_n\to y$ in $X$. But by continuity of $T$ we know $Tx_n\to Tx$ in $X$. Since limits are unique in Banach spaces, this implies $Tx=y$, as desired.