Let $(X, \mu)$ be a measure space with $\mu(x)<\infty$ and let $T:L^{\infty}(X)\to \mathbb{R}$, $V\mapsto \int_{X}{Vd\mu}$. Then $T$ is clearly continuous, but what does it mean that $T$ is "weakly continuous"?
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It means that for any $f_n$ converging weakly to $f$, we also have that $Tf_n$ converges weakly to $Tf$.
We say that $f_n$ converges weakly to $f$ in $E$, if for any continuous linear map $\phi\in E'$ (so any element of the dual space of $E$), $\phi(f_n)$ converges to $\phi(f)$.
Jakobian
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And in my example, is it weakly continuous? – Jul 07 '19 at 12:00
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@ShaqAttack1337 any continuous operator is weakly continuous – Jakobian Jul 07 '19 at 12:02
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Look here for reference – Jakobian Jul 07 '19 at 12:04
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$T$ is itself a continuous linear functional on $L^{\infty}$. So if $f_n \to f$ weakly in $L^{\infty}$ then $Tf_n \to Tf$. [ In a normed linear space $x_n \to x$ weakly iff $x^{*}(x_n) \to x^{*}(x)$ for every $x^{*} \in X^{*}$].
It is also true generally that any norm-norm continuous linear map is weak-weak continuous.
Kavi Rama Murthy
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