# Non trivial Eigenvectors of the Laplacian in $H^1$

Let $$N$$ be natural and $$\Omega \subset \mathbb R^N$$ be open and bounded. I want to show that $$f-\Delta f=0$$ in $$D'(\Omega)$$ has non trivial solutions in $$H^1$$. So far I've managed to show solutions to this PDE (for general open $$\Omega$$) are orthogonal to $$H^1_0$$ in $$H^1$$ and using Fourier transform I've found the solutions for different $$\Omega \subseteq \mathbb R$$. I'm still not sure how to prove existence for general bounded open $$\Omega$$.

• Why do you expect non trivial solutions to exist? $\Delta$ is a negative operator, it cannot have a positive eigenvalue... Commented Mar 4, 2021 at 13:52
• @LorenzoQuarisa The Laplacian is only a negative operator when restricted to functions with suitable boundary conditions (e.g. Dirichlet, Neumann). If you don't impose such boundary conditions, there can of course be positive eigenvalues. For example, the function $f(x)=e^x$ on $[0,1]$ satisfies $f''=f$. Commented Mar 4, 2021 at 14:07
• Exactly as Mao said. Commented Mar 4, 2021 at 14:46
• @LorenzoQuarisa Because its from a distribution theory exam from 20 years ago. Commented Mar 4, 2021 at 15:27
• Right, I missed that there were no boundary conditions. Commented Mar 4, 2021 at 16:10

$$H_0^1(\Omega)$$ is a proper, closed subspace of $$H^1(\Omega)$$, therefore its orthogonal complement $$H_0^1(\Omega)^{\perp}$$ is non-trivial. Since this is the space of solutions, there exists at least a nontrivial solution.
In fact, we can show that for any $$h\in H^1(\Omega)$$ there exists a uniqe solution to $$\begin{cases}\Delta u= u & \Omega \\ u=h & \partial \Omega. \end{cases}$$
Fix $$h\in H^1(\Omega)\setminus H_0^1(\Omega)$$. Let $$H^1_h(\Omega):=\left\{u\in H^1(\Omega):u-h\in H_0^1(\Omega)\right\}$$ Thus $$0\notin H^1_h(\Omega)$$, and $$H_h^1(\Omega)$$ is closed in $$H^1(\Omega)$$. Since an element $$u\in H^1(\Omega)$$ solves the equation if and only if $$u\in H_0^1(\Omega)^{\perp}$$, we just have to show that $$H^1_h(\Omega)\cap H_0^1(\Omega)^{\perp}\neq \emptyset.$$ An element of this set is the element of $$H_h^1(\Omega)$$ with the smallest $$H^1$$ norm. To show that it is the only element, notice that if $$u_1,u_2\in H_h^1(\Omega)\cap H_0^1(\Omega)^{\perp}$$, then $$u_1-u_2\in H_0^1(\Omega)\cap H_0^1(\Omega)^{\perp}=\left\{0\right\}$$.