Let be A an open and bounded set in $\mathbb{R}^{n}$ with border $\partial A$. A function $f:A\to \mathbb{R} $ is harmonic in $A$ if $f$ is continuos in $\bar{A}$ and satisfies Laplace ecuation $\displaystyle\sum_{k=1}^{n}{\frac{\partial^2 f}{\partial x_k^2}{(x)}}=0$ for all $x \in A$.
Prove that if $f,g$ are harmonic functions such that $f(x)=g(x)$ for all $x \in \partial A $ then $f(x)=g(x)$ for all $x \in \bar{A}$. Prove that if $f,g$ are harmonic and $f(x)= \psi(x)$ and $g(x)=\phi(x)$ for all $x\in \partial A$, then
$sup_{x\in A}{|{f(x)}-{g(x)}|}$ $=sup_{y\in \partial A}{|{\psi(y)}-\phi(y)|}$
I tried to show that the maximum and the minimum of $f$ and $g$ were equals but i couldn't use that in the proof.
I need your help with some alternative to prove this.
You are in the right track. It suffices to show that if $\triangle f=0$ and $f|_{\partial A}=0$, then $f=0$. The maximum principle says that $$\sup_A f =\sup_{\partial A} f =0,$$so $f\leq 0$ on $\overline{A}$. Since $-f$ is also harmonic and vanishes at $\partial A$, using that $\sup(-f)=-\inf f$ we also conclude that $-f\leq 0$, which is to say $f\geq 0$. So $f=0$. This takes care of the first part.
The second part is also immediate from the maximum principle: the difference $f-g$ is harmonic and equals $\phi-\psi$ on the boundary, so $$\sup_A (f-g) =\sup_{\partial A}(\phi-\psi).$$ Reverse the roles of $f$ and $g$ to get the conclusion with the absolute values.