Problem 6

Exercise 2.6 (Weakly converging operators have a bounded limit)   Suppose $A_n : X \to Y$ is a sequence of bounded linear operators converging weakly to $A$ in the sense that for all $\phi\in Y^*$ and $x\in X$ the following limit holds

$$\lim_{n\to\infty} \phi(A_n x) = \phi(A x).$$ (2.26)

Then $\sup \Vert A_n\Vert<\infty$ and $A$ is bounded.

Proof. Define a few linear maps

$$A_n^* : Y^* \to X^* \quad \phi \mapsto \phi \circ A_n$$ (2.27)

$$T_n^x : Y^* \to \mathbb{R} \quad \phi \mapsto \phi(A_n x)$$ (2.28)

$$J_n : X \to Y^{**} \quad x \mapsto T_n^x$$ (2.29)

Fixing $x\in X$, we know

$$\lim_{n\to\infty} \phi(A_n x) = \phi(A x) \implies \sup_{n\geq 1}... ...vert = \sup_{n\geq 1} \vert T_n^x(\phi)\vert < \infty \quad \forall \phi\in Y^*$$ (2.30)

Uniform boundedness implies that $\sup_{n\geq 1}\Vert T_n^x\Vert<\infty$. Since $x$ was fixed, this is true for any $x$, so that uniform boundedness can be applied again on

$$\sup_{n\geq 1}\Vert T_n^x\Vert = \sup_{n\geq 1}\Vert J_n(x)\Vert<\infty \quad \forall x\in X$$ (2.31)

so that $\sup_{n\geq 1}\Vert J_n\Vert<\infty$.

After we show

$$\Vert J_n\Vert = \sup_{\Vert x\Vert=1}\Vert J_n(x)\Vert = \sup_{\Vert x\Vert=1} ... ...ert\phi\Vert=1}\Vert J_n(x)(\phi)\Vert = \sup_\phi \sup_x \Vert\phi(A_n x)\Vert$$ (2.32)

$$= \sup_\phi \Vert\phi\circ A_n\Vert = \Vert A_n^*\Vert = \Vert A_n\Vert$$ (2.33)

it is true that $\sup\Vert A_n\Vert = \sup\Vert J_n\Vert<\infty$.

Now we are ready to show $A$ is bounded, working in the double dual.

$$\Vert A^*\Vert = \sup_{\Vert\phi\Vert=1} \Vert A^*(\phi)\Vert = \sup_\phi \sup_x \vert\phi(A x)\vert = \sup_\phi \sup_x \lim_{n\to\infty} \vert\phi(A_n x)\vert$$ (2.34)

$$\leq \liminf_{n\to\infty} \sup_\phi \sup_x \vert\phi(A_n x)\vert ... ...\circ A_n\Vert \leq \liminf_{n\to\infty} \sup_\phi \Vert\phi\Vert\Vert A_n\Vert$$ (2.35)

$$\leq \liminf_{n\to\infty} \Vert A_n\Vert \leq \sup_{n\geq 1} \Vert A_n\Vert < \infty$$ (2.36)

Therefore, $\Vert A^*\Vert$ is bounded, proving that $\Vert A\Vert$ is bounded. $\qedsymbol$