Prove that a series converges in $L^2(\Omega)$ Let $\Omega$ be a bounded smooth domain and let $\varphi_k$ be eigenfunctions of the Neumann Laplacian  with eigenvalues $\lambda_k$.
Let $u \in H^1(\Omega)$. I want to show that for all $y \in (0,\infty)$, that the following sum converges in $L^2(\Omega)$:
$$f(x,y) = \sum_{k=1}^\infty e^{-y\sqrt{\lambda_k}}(u,\varphi_k)_{L^2}\varphi_k(x).$$
How do I show this? Doesn't it also hold for $y=0$? 
Of course I know the definition of convergence in $L^2$ (the partial sums converge in the $L^2$ norm), the problem is that I cannot assume a limit holds in order to show that the sum converges to that limit.
 A: Fix $y \in [0,\infty)$. Let $f_N(x,y) := \sum_{k=1}^N e^{-y\sqrt{\lambda_k}} (u,\varphi_k) \varphi_k(x)$ be the partial sum, and let $g_N := f-f_N$ be the tail of the series. Consider the integral$\int_{\Omega } g_N^2$: writing $g_N^2$ as a product of two sums and multiplying out the terms, then only the diagonal terms survive after integrating because the $\varphi_k$'s are orthonormal. It follows that
\begin{align*}
\int_{\Omega} g_N^2 &= \sum_{k=N+1}^{\infty} \int_{\Omega} e^{-2y \sqrt{\lambda_k}} (u, \varphi_k)^2 \varphi_k(x)^2 dx\\
&= \sum_{k=N+1}^{\infty} e^{-2y\sqrt{\lambda_k}}(u,\varphi_k)^2 
\end{align*}
It is a standard fact that the Neumann eigenvalues are non-decreasing, i.e. $\lambda_k \geq \lambda_1 = 0$ for all $k$. Substituting this into the above equality, we get that
$$
\int_{\Omega} g_N^2 \leq e^{-2y\sqrt{\lambda_1}} \sum_{k=N+1}^{\infty} (u,\varphi_k)^2 = \sum_{k=N+1}^{\infty} (u,\varphi_k)^2 \to 0 \textrm{ as $N \to \infty$.}
$$
This last sum goes to zero as $N \to \infty$ because it is a tail of the convergent series $\| u \|_{L^2}^2 = \sum_{k=1}^{\infty} (u,\varphi_k)^2 < \infty$ (and this $L^2$-norm is indeed finite since the Sobolev space $H^1(\Omega)$ embeds in $L^2(\Omega)$). Therefore, as the partial sums $f_N$ converge to $f$ in the $L^2$-norm, the series $f(\cdot, y)$ converges in $L^2(\Omega)$ for each fixed $y \in [0,\infty)$.
