Weak convergence in $L^2[0,1]$

Question

In $\ell^2$, a sequence converges weakly if it is bounded and each component converges. My question is: How can I think about weak convergence in $L^2[0,1]$?

Answer 1

As $\ell_2(\mathbb N)$ is isometrically isomorphic to any separable infinite-dimensional Hilbert space $\mathcal H$ (such as $L^2[0,1]$), the $\ell_2$-condition for weak convergence of a bounded sequence transfers to every such $\mathcal H$ where

$$ \text{"each component"} $$

gets replaced by

$$ \text{"each }(\langle e_k,x_n\rangle)_{n\in\mathbb N}\text{ where }(e_k)_{k\in\mathbb N}\text{ is any orthonormal basis of }\mathcal H", $$

e.g. see here (Lemma 8). Then, you can take any orthonormal basis of $L^2[0,1]$ (e.g. $\exp(2\pi inx)$, or the Legendre polynomials on $[0,1]$, or $\ldots$) you like to characterize weak convergence in it.

Answer 2

A sequence $(u_n)$ in $L^2[0,1]$ converges weakly iff it is bounded and for all intervals $(a,b)\subset [0,1]$ the integrals $\int_a^b u_n$ are converging. Instead of 'each component' converges, we have now each integral on an arbitrary interval converges.

Answer 3

The space $L^2([0,1])$ equipped with the inner product $$\langle f,g\rangle:=\int^1_0f(x)\overline{g(x)}\,dx$$ is a Hilbert space. By definition of weak convergence we say a sequence $(f_n)\subset L^2([0,1])$ converges weakly to $f\in L^2([0,1])$ whenever $$\lim_n\langle f_n,g\rangle=\langle f,g\rangle\hspace{0.2cm}\text{for all}\hspace{0.2cm}g\in L^2([0,1])$$ On the other hand the spaces $l^2(\mathbb{N})$ and $L^2([0,1])$ are isometric isomorphic to each other. To see this note that if $(e_n)\subset L^2([0,1])$ is a complete orthonormal basis for $L^2([0,1])$ then every $f\in L^2([0,1])$ admits a representation of the form $$f=\sum_{n=1}^{\infty}\langle f,e_n\rangle e_n\Rightarrow ||f||^2_{L^2}=\int_0^1f(x)\overline{f(x)}\,dx=\langle f,f\rangle=\sum^{\infty}_{n=1}|\langle f,e_n\rangle|^2:=||f_n||^2_{l^2}$$ implies $(f_n)\in l^2(\mathbb{N})$ where $f_n:=\langle f,e_n\rangle$. Clearly the operator $A:L^2([0,1])\mapsto l^2(\mathbb{N})$ defined by $Af:=(f_n)$ is a linear bounded operator which preserves the norms (as we can see for the above relation). Now to come back to the initial statement that if $(f^k_n)$ is a bounded sequence for each $k$ and for each $n$ we have $|f^k_n-f_n|\to 0$ as $k\to\infty$ then $(f^k_n)$ converges weakly to $(f_n)\in l^2(\mathbb{N})$. Again by definition $(f^k_n)$ converges weakly to $(f_n)$ whenever $$\lim_k\langle f^k,g\rangle=\langle f,g\rangle$$ where the inner product for $l^2(\mathbb{N})$ is defined by $$\langle f,g\rangle:=\sum_nf_n\overline{g_n}$$ Boundedness of each $(f^k_n)$ implies $(f^k_n)\in l^2(\mathbb{N})$ for each $k$ so their inner product with any other element $g$ of $l^2(\mathbb{N})$ is finite and well defined. On the other hand component wise convergence, $|f^k_n-f_n|\to 0$ as $k\to\infty$ for each $n$, implies $$|\langle f^k,g\rangle-\langle f,g\rangle|=|\langle f^k-f,g\rangle=\Big|\sum_n(f^k_n-f_n)\overline{g_n}\Big|\leqslant \sum_n|f^k_n-f_n||\overline{g_n}|\to 0$$ as $k\to\infty$. Therefore the same definition for both spaces is valid.