# Show that for any eigenvalue we can find a real-valued eigenfunction (for the laplacian)

Let $$\Omega \in \mathbb{R}^d$$ be an open bounded set with smooth boundary. Consider $$-\bigtriangleup q(x) = \lambda q(x)$$ with either Dirichlet or Neumann boundary conditions $$q(x)=0, x \in \partial \Omega$$or $$\bigtriangledown q(x) \centerdot n(x) = 0, x \in \partial \Omega$$ I need to show that for any eigenvalue, we can find a real-valued eigenfunction. I have shown that the eigenvalues are real by taking the conjugate of the pde, multiplying the conjugate one by $$q$$ and the non-conjugate one by the conjugate of $$q$$ and integrating. I've tried subtracting the conjugate equation from the original and integrating but that didn't get me anywhere. The best I can get from doing any integration is that $$\frac{q(x)}{\bigtriangleup q(x)}$$ is real but that follows immediately from the pde if you know $$\lambda$$ is real. I've been stuck for a while and don't know what to do. Thanks.

• Multiply by $\bar{q}$, integrate over the domain. Then integrate by parts the LHS. That should do the trick. – maxmilgram Apr 13 at 19:15
• @maxmilgram what am I supposed to get from doing that? The boundary term vanishes and the other term is a dot product of two gradients. If i integrate that term by parts I get another vanishing boundary term and the conjugate of what I started with which is the RHS if you substitute the pde back in. – N Dizzle Apr 13 at 20:20
• You get $||\nabla q||_{L^2}^2=\lambda||q||_{L^2}^2$, don't you? – maxmilgram Apr 14 at 6:52
• Yeah that's what I got, I guess I just don't see how that means $q$ is real. – N Dizzle Apr 14 at 18:04
• Oh! Really sorry my bad, i missread that as "eigenvalues". So I gave a proof for eigenvalues being real. I dont think your statement about eigenfunctions is actually true. Take a real eigenfunction $q(x)$. Then $i\cdot q(x)$ is an eigenfunction as well. – maxmilgram Apr 14 at 18:12