# Pointwise (a.e) evaluation of $\sum_{n \geq 0}(u,w_n)_{L^2}w_n$ and equalities in $L^2$

Let $w_n$ be a orthonormal basis of $L^2(\Omega)$. Given $u \in L^2$ we can write $$u=\sum_{n \geq 0}(u,w_n)_{L^2}w_n.$$ Suppose $w_n$ are the eigenfunctions of the Neumann Laplacian. We can write $$\nabla u = \sum_{n \geq 0}(u,w_n)_{L^2}\nabla w_n \tag{1}$$ whenever $u \in H^1$. Both of these equalities are equalities in the sense that the partial sums converge to the LHS in the $L^2$ norm.

Is it possible for me to say that $$\nabla u(x) = \sum_{n \geq 0}(u,w_n)_{L^2}\nabla w_n(x)$$ for a.e. $x$?

Remember that although (1) is some sort of equality in $L^2$ it is not a true equality because of how an infinite sum is defined (as limit of partial sums). So it seems not that straightforward to me. I think the limit taken in $L^2$ gives us pointwise a.e. convergences for a subsequence and I'm not sure if that is enough. The reason I ask this questions is I have a sum defined $$v_t(x) := \sum_{n \geq 0} f_n(t)(u,w_n)w_n(x)$$ where each $f_n(t) \to 1$ as $t \to 0$, and I wish to show that $\nabla_x v_t(x) \to \nabla_x u(x)$ pointwise a.e. $x$ as $t \to 0$ by taking the limit inside the sum. But I do not know if really $\nabla_x u(x)$ can be given by the infinite sum (1) evaluated at $x$.