Show that $u(x,y)$ is constant.

Question

Let $u(x,y)$ be a harmonic function on domain s.t all the partial derivatives of $u(x,y)$ vanish at the same point of , then $u(x,y)$ is constant.

Now the thing is if the harmonic conjugate of $u(x,y)$ exists say $v(x,y)$ then $f=u+iv$ is analytic and $f^m(z)$ vanishes for all $z \in D$ then $f(z)$ is const so is $u(x,y)$.

But the question is I know that for a star shaped domain the harmonic conjugates exists not for any domain.. So what will be the solution of this?

Answer 1

For each $z\in D$ there is a star-shaped region $B$ with $z\in B\subset D$. (For example, a small enough open ball centered at $z$.) By your harmonic conjugate reasoning $u$ is constant on $B$. This (and the connectedness of $D$) is enough to show that $u$ is constant on all of $D$.

Answer 2

A basic fact of elliptic regularity theory tells you that a function which is harmonic in a domain $D$ is also a real analytic function in $D$.

Therefore, if all partial derivatives of a harmonic function vanish simoultaneously at a given inner point of $D$, the Taylor expansion of the function centered at that point is null; thus the function itself is everywhere null in $D$ by Principle of Analytic Continuation.