Let $(M^n,g)$ be a compact Riemannian manifold with boundary, with $\operatorname{dim} M = n \geq 3$. Denote by $\Delta$ the Laplacian operator acting on functions as $\Delta f = -\operatorname{div} \nabla f$. Let $\lambda_1$ be the first eigenvalue of the Laplacian on $M$, with Dirichlet boundary condition.
Does there exist a smooth function $f : M \to \mathbb{R}$ satisfying the following boundary problem?
$$\cases{\Delta f = -\left(\frac{4 \lambda_1}{n-2}\right) f \quad \text{on } M \\f = -\frac{n}{\lambda_1} \quad \text{on } \partial M}$$
My thoughts: Let $\{u_j\}_{j \geq 1}$ be a complete orthonormal set in $L^2(M)$ of eigenfunctions of $\Delta$ with Dirichlet boundary condition, associated to the eigenvalues $\lambda_j$. Let $f \in C^2(M)$ satisfy the above boundary problem. Write $f = \sum_{j=1}^\infty a_j u_j$, with convergence in $L^2(M)$. Then, formally, we have:
$$\Delta \left( \sum_{j=1}^\infty a_j u_j \right) = -\left(\frac{4 \lambda_1}{n-2}\right) \left( \sum_{j=1}^\infty a_j u_j \right) \\ \implies \sum_{j=1}^\infty a_j \lambda_j u_j = -\left(\frac{4 \lambda_1}{n-2}\right) \left( \sum_{j=1}^\infty a_j u_j \right) \\ \implies \sum_{j=1}^\infty a_j\left( \lambda_j + \frac{4 \lambda_1}{n-2} \right)u_j = 0$$
Since $\lambda_j > 0$ for every $j \geq 1$, it follows that $a_j = 0$ for every $j \geq 1$. So, $f \equiv 0$, which contradicts the fact that $f$ is negative on the boundary.
Question: How do I justify that the sum comutes with the Laplacian in the above calculations?