# Why is the laplacian positive-definite

Let (M,g) be compact Riemannian manifold (possibly $\partial M\neq\emptyset)$

Now I have read, that "the laplace-beltrami operator is a positive definite operator".

I have shown, if M is a closed manifold or if we consider dirichlet boundary conditions, $\Delta:=-\nabla^2$ is positive definite, i.e. $(\Delta f,f)_{L^2}\geq 0$.

Which setting does the author mean by: "the laplace-beltrami operator is a positive definite operator"?? Is $\Delta$ always positive definite independent of the boundary conditions?

Thank you!

• where did you get this? If the Manifold have boundary,i dont think this is always true. Take look on the book of Eugene Michael Taylor - Partial differential equation volume 1 - page 144 Oct 20, 2012 at 12:00
• I don't know, whether it is always true or not. This is why I asked. Oct 20, 2012 at 15:39

For a very concrete example, consider a compact $1$-manifold with boundary, such as $M=[0,1]$ with the Euclidean metric. Here the L-B operator is just $f\mapsto -f''$, hence $(\Delta f, f)_{L^2}=\int_0^1 (-f'')f =f'(0)\,f(0)-f'(1)\,f(1) +\int_0^1 (f')^2$. If our boundary conditions imply $f'(0)\,f(0) \ge 0$ and $f'(1)\,f(1)\le 0$, we have positive definiteness. On the other hand, the boundary conditions $f(0)=1$ and $f(1)=2$ are satisfied by the function $f(x)=2^x$ which has $(\Delta f, f)_{L^2}<0$.