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I have a question about weak convergence.

Let $(S,\Sigma,m)$ be a measure space. $(f_{t})_{t>0}$ be a family of square integrable functions. (i.e. for every $t>0$, $f_{t} \in L^{2}(S;m)$) and $f$ be a square integrable function.

Question

Let $(t_{n})_{n \in \mathbb{N}}$ be a sequence with each $t_{n}>0$ and $t_{n} \searrow 0$ as $n \to \infty$. I know the definition $f_{t_{n}} \to f$ weakly in $L^{2}(S;m)$. $f_{t_{n}} \to f$ weakly in $L^{2}(S;m)$ means \begin{align*} \forall g \in L^{2}(S;m),\int_{S}f_{t_{n}}gdm \to \int_{S} fg dm\quad{\rm as\,}n\to\infty \end{align*} But I don't know $f_{t} \to f$ weakly in $L^{2}(S;m)$ as $t \searrow 0$. What is the definition of this?

My opinion

I guess $f_{t} \to f$ weakly in $L^{2}(S;m)$ as $t \searrow 0$ means that for every sequence $(t_{n})_{n \in \mathbb{N}}$ with each $t_{n}>0$ and $t_{n} \searrow 0$ as $n \to \infty$, $f_{t_{n}} \to f$ weakly in $L^{2}(S;m)$. I'm wondering what you think about that.

Thank you in advance.

J.R.
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sharpe
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    You can say it thus. You can also say that $f_t \to f$ weakly as $t \searrow 0$ if and only if for every $\varepsilon > 0$ and $g \in L^2$ there is a $\delta > 0$ such that $\lvert \langle f_t - f, g\rangle\rvert < \varepsilon$ for all $t < \delta$. – Daniel Fischer Jul 28 '15 at 10:25
  • Thanks for your reply. I understood. – sharpe Jul 29 '15 at 04:35

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