On the very first page of my textbook's Hilbert spaces chapter I've found the following statement:
Let $(V,\langle \cdot ,\cdot \rangle)$ be an inner product space. $(v_j)_{j\in\mathbb N}\subset V$.
$$\lim_{j\rightarrow \infty} v_j=v\iff \lim_{j\rightarrow \infty} \|v_j-v\|=0$$
Now, we know that $L^2_{\mathbb R}$ is a inner product space (a complete one), with norm defined by
$$\|u\|=\|u\|_2=\bigg(\int u^2d\mu\bigg)^{1/2}$$
Hence the previous statement reads
For $(v_j)_{j\in\mathbb N}\subset L^2_{\mathbb R}(\mu)$
$$\lim_{j\rightarrow \infty} v_j=v\iff \lim_{j\rightarrow \infty} \|v_j-v\|_2=0$$
Namely, the sequence converges pointwise iff it converges in $L^2$-sense.
But, if I am not mistaken in general point-wise convergence does not guarantee convergence in $L^2$.
Conversely if the sequence converges in $L^2$ I can say that there's a subsequence converging a.e. in $\mathbb R$.
Is this a particular property of $L^2$ spaces? Am I missing something? Thanks.
Yes. The notation $\lim_{j \to \infty} v_j = v$ says nothing about pointwise convergence. It simply is a notation for the statement that $v_j$ converges to $v$ in $L^2$, which by definition means $\|v_j - v\|_2 \to 0$.