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Let $X$ be a Banach space, $A \in B(X)$ be injective, and $B: X \rightarrow X$ be a linear operator such that for every $f \in X^*$, $fAB \in X^*$. Prove that $B \in B(X)$.

Attempt: To prove that $B \in B(X)$, we need to show that $B$ is bounded, i.e., $\|Bx\| \leq M\|x\|$ for some constant $M$ and for all $x \in X$. I also know:

\begin{aligned} \|Bx\| &= \sup_{\|f\|=1} |f(Bx)| \\ &= \sup_{\|f\|=1} |f(ABx)| \quad (\text{since } fAB \in X^*) \\ &\leq \|ABx\| \quad (\text{by definition of the operator norm}) \\ &\leq \|A\| \|Bx\| \quad (\text{by the property of bounded operators}) \\ \end{aligned}

Are my steps till here fine? How should I proceed?

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We can use the closed graph theorem. Assume $x_n\to x$ and $B(x_n)\to y$. We want to show that $B(x)=y$. Since $A$ is injective, this is equivalent to showing that $AB(x)=Ay$. By a standard corollary of the Hahn-Banach theorem this is equivalent to showing that $fAB(x)=fAy$ for any $f\in X^*$.

So let $f\in X^*$. By hypothesis, $fAB$ is bounded, and since $x_n\to x$, it follows that $fAB(x_n)\to fAB(x)$. On the other hand, $fA$ is bounded, and since $B(x_n)\to y$, it follows that $fA(B(x_n))\to fA(y)$. By the uniqueness of a limit, indeed $fAB(x)=fA(y)$.

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