Problem 6

Exercise 1.6 (Bounding linear maps via elements in the dual space)   Suppose $T : E\to F$ is such that $\phi\circ T\in E^*$ is bounded for every $\phi\in F^*$. Then $T$ is bounded.

Proof. Apply uniform boundedness. Define $T^* : F^* \to E^*$ by sending $\phi \mapsto \phi \circ T$ and define $J : X \to Y^{**}$ by sending $J(x)(\phi) = T^*(\phi)(x) = \phi(Tx)$.

The statement is precisely that

$$\Vert\phi\circ T\Vert = \sup_{\Vert x\Vert=1}\vert\phi(Tx)\vert = \sup_{x} \vert J(x)(\phi)\vert < \infty \quad \forall \phi \in Y^*$$ (1.49)

Uniform boundedness implies

$$\sup_x \Vert J(x)\Vert < \infty$$ (1.50)

Indicating $\Vert J\Vert<\infty$.

Now we show $\Vert T^*\Vert<\infty$.

$$\Vert T^*\Vert = \sup_{\Vert\phi\Vert=1}\Vert T^*(\phi)\Vert = \sup_{\Vert\phi\V... ...i)(x)\Vert = \sup_{\Vert\phi\Vert=1}\sup_{\Vert x\Vert=1} \Vert J(x)(\phi)\Vert$$ (1.51)

$$\leq \sup_{\Vert\phi\Vert=1}\sup_{\Vert x\Vert=1} \Vert J(x)\Vert... ...\Vert=1} \Vert J\Vert\Vert x\Vert\Vert\Vert\phi\Vert \leq \Vert J\Vert < \infty$$ (1.52)

Therefore, $T$ is bounded. $\qedsymbol$