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If in a Banach Space $X$ we have everywhere point wise convergent sequences of bounded operators $A_n, B_n\rightarrow A,B$ is it true that their composition is also everywhere point wise convergent?

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Yes, the composition is pointwise convergent. For any $x\in X$, one has

$$\|A_n B_n(x) - AB(x)\| \leq \|A_n B_n (x) - A_n B(x) \| + \|A_n B(x) - AB(x)\|.$$

The second term tends to zero as $n\rightarrow \infty$ since $A_n(B(x)) \rightarrow A(B(x))$. For the first term, use uniform boundedness on $(A_n)$ to get that $\alpha:= \sup\|A_n\| < \infty$. Then

$$ \|A_n B_n (x) - A_n B(x) \| \leq \alpha \|B_n(x) - B(x)\| \rightarrow 0,$$

which proves $A_n B_n (x)\rightarrow AB(x)$.

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  • By the principle of uniform boundedness, $M:=\sup_n\lVert B_n\rVert$ is finite.
  • We have \begin{align}\lVert B_nA_nx-BAx\rVert&\leq \lVert B_n(A_n-A)x\rVert+\lVert B_nAx-BAx\rVert\\ &\leq \lVert B_n\rVert\lVert A_nx-Ax\rVert+\lVert B_nAx-BAx\rVert\\ &\leq M\lVert A_nx-Ax\rVert+\lVert B_nAx-BAx\rVert, \end{align} and we are done.
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