# 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 $f \in X^*$, $fAB \in X^*$.

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?

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 $$fAB(x)=fA(y)$$.