Weak convergence under linear operators

Asked Jan 18 '18 at 02:10

Active Jan 18 '18 at 03:39

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Consider the linear bounded operator $A:X\to Y$. Assume a sequence $x_n$ weakly converges to $x_0$ in $X$. It can be shown that the sequence $y_n:=Ax_n$ weakly converges to some element $y_0$ in $Y$.

Question: Under what conditions do we have $y_0=A x_0$?

edited Jan 18 '18 at 03:39

asked Jan 18 '18 at 02:10

Saj_Eda

https://math.stackexchange.com/questions/536101/weak-convergence-together-with-convergence-of-norms-implies-strong-convergence – Tsemo Aristide Jan 18 '18 at 02:37
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@TsemoAristide How is that related to this problem? – Saj_Eda Jan 18 '18 at 03:38
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Yes, and the proof follows easily from definitions: let $y^{} \in Y^{}$. Then $x^{} = y^{} A$ is in $X^{}$ so $y^{} (Ax_n)=x^{}(x_n) \to x^{}(x_0)=y^{*}(Ax_0)$. – Kavi Rama Murthy Jan 18 '18 at 06:07
I agree with Kavi. It is enough that $X$ and $Y$ are normed spaces, see Robert E. Megginson's An Introduction to Banach Space Theory. – max_zorn Jan 18 '18 at 08:21
@KaviRamaMurthy: please use the "answer" box for answers. The "comment" box is for comments. – Martin Argerami Jan 18 '18 at 15:16
@KaviRamaMurthy Does this property also hold for a convex nonlinear operator? – Saj_Eda Jan 18 '18 at 16:47
Sorry, I have no idea if this is true for non-linear operators. – Kavi Rama Murthy Jan 19 '18 at 10:16

Weak convergence under linear operators

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