Let $(X, ||\ ||)$ be a normed space. $dim X = \infty$. $L(X)$ is a space of continuous (which is equialent to bounded in a normed space) linear operators with strong topology. Strong topology is defined by zero neighborhoods: $U(x, W) = \{A \in L(X) | Ax \in W\}$, where $x \in X, 0 \in W \in T_X, T_X$ is topology on $X$, and these neighborhoods finite intersections. Convergence $A_n \rightarrow A (A_n, A \in L(X))$ is equialent to convergence $A_n x \rightarrow Ax$ over $X$, $\forall x \in X$. Show that multiplication operation of End(X) (сomposition of operators) isn't continuous.
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1I think that you have to define things properly. What is $X$? Is it a Banach space? What is $\text{End},(X)$? All endomorphisms, or the bounded ones? What is the strong topology? Pointwise convergence. – Martin Argerami Dec 09 '17 at 23:05
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@MartinArgerami thank you for your remark. Question updated. – Pennywise Dec 10 '17 at 16:30
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Is the considered multiplication the composition? – Christian Dec 10 '17 at 16:46
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@Christian yes. – Pennywise Dec 10 '17 at 16:51
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@Christian in Hilbert space I can proof that composition of an operator and hermitian self adjoint operator is compact. Using this, I tried to prove by contradiction, but it did not work. Since I can find an operator such as $x_0 = Ax \not\in \hbox{span}(X)$ If it is that exist operator B such that $||Bx_0|| = 1$ – Pennywise Dec 10 '17 at 20:25
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@Pennywise: are you looking for an example of an $X$, or to show that for every $X$ continuity fails? – Martin Argerami Dec 10 '17 at 20:28
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1@MartinArgerami for every $X$ continuity fails – Pennywise Dec 10 '17 at 20:29
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1by "multiplication operation of $ End(X) $(сomposition of operators) isn't continuous" do you mean for each $u\in L(X)$, the map $$M_u:L(X)\to L(X), v\mapsto u\circ v$$ isn't strongly continuous? – C. Ding Dec 13 '17 at 00:40
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@C.Ding I think yes – Pennywise Dec 13 '17 at 23:01
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1I assume the question refers to joint strong continuity. Multiplication by a fixed elements is trivially strongly continuous. – Martin Argerami Dec 14 '17 at 03:26
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@MartinArgerami I think so yes. – Pennywise Dec 20 '17 at 09:21