Show that a function can not be harmonic

Question

A function $u$ is twice continuously differentiable in $D(0,2)$. It has the property: $u(e^{i\theta})=cos(\theta)$ and $u(0)=1$ Show that $u$ can not be harmonic.

Answer 1

Hint. Note that the average of $u$ along the unit circle is $$\frac{1}{2\pi}\int_0^{2\pi}u(e^{i\theta})d\theta=\frac{1}{2\pi}\int_0^{2\pi}\cos(\theta)d\theta=0$$ which is less than $1=u(0)$. On the other hand, harmonic functions satisfy the mean value property.

Answer 2

Let $K=\{z \in \mathbb C:|z| \le 1\}$. Suppose that $u$ is harmonic. Then

$\max u(K)= \max u( \partial K)= \max \{ cos t : t \in [0,2 \pi]\}=1$

and, since $u(0)=1$, the function $u$ has a maximum in $K^o$. By the Maximum - principle, $u$ is constant on $K$, but this is a contradiction to $ u(e^{i\theta})=cos(\theta)$.