Show that if $\phi \circ T$ is bounded for every $\phi\in Y^*$ then $T$ is bounded

Question

Show that if $T:X \to Y$ is a linear map between Banach spaces and $\phi \circ T$ is bounded for every $\phi\in Y^*$ then $T$ is bounded.

The given hint is to prove the contrapositive. 'Progress':

Suppose $T$ is unbounded. Then there exist $x_n \in X$ such that $\left\lVert x_n\right\rVert=1$ and $\left\lVert Tx_n\right\rVert \geq n$ for all $n$. For each $n$ there is a support functional $\phi_n$ at $Tx_n$: $\phi_n \in Y^*, \left\lVert \phi_n\right\rVert=1, \phi_n(Tx_n)=\left\lVert Tx_n\right\rVert$. I then applied the (contrapositive of the) Principle of Uniform Boundedness to the set of linear maps $\phi_n \circ T$ before realising that this doesn't even work because we don't know that the maps $\phi_n \circ T$ are bounded. So really I've made no progress and this is mostly to show that I've tried. Though I expect the solution will use the PUB somehow. Any small hints would be appreciated (please do not spoil the solution).

Answer 1

Use the Closed Graph Theorem:

Assume that $x_n \to x$ and $Tx_n \to y$. For Arbitrary $\phi \in Y^*$ we have that $\phi \circ T$ is continuous so $(\phi \circ T)(x_n) \to (\phi \circ T)(x) = \phi(Tx)$. On the other hand, we have $\phi(Tx_n) \to \phi(y)$.

Uniqueness of the limit implies $\phi(Tx) = \phi(y)$ for all $\phi \in Y^*$ and hence, as the dual space is separating, $y = Tx$. We conclude that $T$ is bounded.

Answer 2

Define a family of continuous linear maps $\{F_x:x\in X,\|x\|\leq 1\}$ from $Y^{*}$ to $\mathbb R$ (or $\mathbb C$) by $F_x(\phi) =\phi (T(x))$. For fixed $\phi \in Y^{*}$ note that $sup \{|F_x(\phi)|:\|x\|\leq 1\}<\infty$ (by hypothesis). By Uniform Boundedness Principle we get $sup \{|F_x(\phi)|:\|x\|\leq 1, \|\phi \| \leq 1\}<\infty$. This means $T$ is bounded.