I am a graduate student.I am studying functional analysis.I am having problem with the following question in my assignment:
Let $X,Y,Z$ be Banach spaces and $A_n$ be a sequence in $B(X,Y)$ and $A_nx\to Ax$ for all $x\in X$ and $B_n$ be a sequence in $B(Y,Z)$ and $B_ny\to By$ for all $y\in Y$.Then show that $B_nA_nx\to BAx$ for all $x\in X$.
I have done a solution but I doubt whether it is correct.The solution is as follows:
$\|B_nA_nx-BAx\|\leq \|B_nA_nx-B_nAx+B_nAx-BAx\|\leq\|B_n(A_nx-Ax)\|+\|B_n(Ax)-B(Ax)\|$
Now,$B_ny\to By$ for all $y\in Y$.So,$\|B_ny\|\to\|By\|$ for all $y\in Y$ and hence $\{\|B_ny\|\}_{n\in \mathbb N}$ is bounded for each $y\in Y$.So,$\|B_ny\|\leq M_y$ for all $n\in \mathbb N$.As $Y$ is Banach,so by uniform boundedness principle $||B_n||\leq M$ for all $n\in \mathbb N$.So, $\|B_nA_nx -BAx\|\leq \|B_n\|\|A_nx-Ax\|+\|B_n(Ax)-B(Ax)\|\leq M\|A_nx-Ax\|+\|B_n(Ax)-B(Ax)\|\to 0$ as $n\to \infty$,for each $x\in X$.This proves the result.
Now can someone please tell me whether it is correct or I have made a mistake?