In today’s Monday Math post, I stated that one could see from the Hessian matrix that a harmonic function cannot have local extrema. To clarify, if the function has a local maximum at a point , , then the Hessian matrix at that point is negative semidefinite. Similarly, if is a local minimum, then the Hessian matrix at that point is positive semidefinite. (See here).
Now, the eigenvalues of a negative semidefinite matrix are all non-positive, and the eigenvalues of a positive semidefinite matrix are all non-negative. Now, consider the minimum case, with a positive semidefinite matrix. The eigenvalues all obey λi≥0. If they are all zero, then the Hessian matrix (as it is a symmetric matrix) is then the zero matrix, and our function is a constant function. If at least one eigenvalue is nonzero, then their sum is positive.
The sum of the diagonal elements of a square matrix is called the trace, and is equal to the sum of the eigenvalues of the matrix. Thus, the trace of the Hessian at a local minimum of a non-constant function must be positive. However, note that the trace of the Hessian is the Laplacian of the function, and therefore must be zero for a harmonic function. Thus, a harmonic function cannot have local minima. An analogous argument shows that local maxima are forbidden as well, so that a harmonic function cannot have local extrema.
Addendum to Monday Math 49