Weak convergence definition

Question

I have a following question, for example: let $f_n\in L^2$ be a bounded sequence of real functions. Then we know that there is some $f$ (in $L^2$?) such that $f_n$ converges weakly to $f$ in $L^2$ i.e..... and now I do not know if I understand it properly. Do weak convergence means:

a) for all $g\in L^2$ scalar product $(f_n,g)=\int f_n(x)g(x)\,dx\rightarrow \int f(x)g(x)\,dx=(f,g)$

or

b) for all $x$, $(f_n(x),g(x))=f_n(x)\cdot g(x)\rightarrow f(x)g(x)=(f(x),g(x))$?

After writing this, I see that I really do not understand this...

Answer 1

In general, if $X$ is a normed space then $x_n\to x$ weakly if for every bounded linear functional $\varphi\in X^*$ we have $\varphi(x_n)\to \varphi(x)$.

Now, $L^2$ is a Hilbert space, so by Riesz representation theorem all the bounded linear functionals have the form $f\to (f,g)=\int_{\mathbb{R}}fg$ for some $g\in L^2$. So here $f_n\to f$ weakly means $(f_n,g)\to (f,g)$ for all $g\in L^2$.

Answer 2

Weak convergence means convergence with respect to all bounded linear functionals hence (a) should be your correct interpretation (as $(L^2)^* \cong L^2)$.

Moreover the second statement is not necessarily true as first of all "for all $x$" shouldn't be the case, it's more like "a.e. $x$". But even in this a.e. atmosphere, the sequence not necessarily has the desired convergence property, you must find a suitable subsequence...

So it's definitely (a)