# Which one is subharmonic?

The Laplace operator Δ of a function of n variables is defined by

If Δ f = 0 in some region Ω, f is said to be harmonic on Ω.  In that case f takes on its maximum and minimum over Ω at locations on the boundary ∂Ω of Ω. Here is an example of a harmonic function over a square which clearly takes on its maximum on two sides of the boundary and its minimum on the other two sides.

The theorem above can be split into two theorems and generalized:

If Δ f ≥ 0, then f takes on its maximum on ∂Ω.

If Δ f ≤ 0, then f takes on its minimum on ∂Ω.

These two theorems are called the maximum principle and minimum principle respectively.

Now just as functions with Δf equal to zero are called harmonic, functions with Δf non-negative or non-positive are called subharmonic and superharmonic. Or is it the other way around?

If Δ f ≥ 0 in Ω, then f is called subharmonic in Ω. And if Δ f ≤ 0 then f is called superharmonic. Equivalently, f is superharmonic if −f is subharmonic.

The names subharmonic and superharmonic may seem backward: the theorem with the greater than sign is for subharmonic functions, and the theorem with the less than sign is for superharmonic functions. Shouldn’t the sub-thing be less than something and the super-thing greater?

Indeed they are, but you have to look f and not Δf. That’s the key.

If a function f is subharmonic on Ω, then f is below the harmonic function interpolating f from ∂Ω into the interior of Ω. That is, if g satisfies Laplace’s equation

then fg on Ω.

For example, let f(x) = ||x|| and let Ω be the unit ball in ℝn. Then Δ f ≥ 0 and so f is subharmonic. (The norm function f has a singularity at the origin, but this example can be made rigorous.) Now f is constantly 1 on the boundary of the ball, and the constant function 1 is the unique solution of Laplace’s equation on the unit ball with such boundary condition, and clearly f is less than 1 on the interior of the ball.

## One thought on “Which one is subharmonic?”

1. Anjali

What about if f is subharmonic and g is superharmonic on a domain G , with (f-g)(x)<0 near boundary of a domain G. Then what can we say about (f-g) on whole of G. Does it remains negative?