Let $\Omega \subset \mathbb{R}^2$ be open, bounded and symmetric with respect to the origin. Let$ f:\partial \Omega \to \mathbb{R} $ be odd and continuous function. Show that if $u$ is solution of:
$$ \begin{cases} \Delta u(x)&=0,&\text{ if }x \in \Omega \\ \phantom{-}u(x)&=f(x),&\text{ if }x \in \partial\Omega \end{cases} $$
then $u$ is odd.
Hint: study of the Laplacian's invariance in relation to $T(x,y)=(-x,-y)$
I tried to show that $\Delta u= \Delta (u\circ T)=0$, but how to finish?
On
If you already showed that $\Delta u(-x, - y) =0$, take $v=u(x, y) +u(-x, - y) $.
By the fact that f is odd, $v$ is solution for the problem: \begin{equation} \begin{cases} \Delta v=0 \\ v=0, \ \ \ in \ \partial \Omega \end{cases}\end{equation}
So, by the maxima principle $v=0$ in $\Omega $ then $u(x, y)=-u(-x, - y) $ in $\Omega$.
If you show that $\Delta (u\circ T)=0$ in $\Omega$ you're done by uniqueness.
Indeed, first note that since $\Omega$ is bounded and symmetric with respect to the origin then $(x,y)\in\partial\Omega$ iff $(-x,-y)\in\partial\Omega$. Then, using that $f$ is odd and $u$ a solution we have for all $(x,y)\in\partial\Omega$ $$u(x,y)=f(x,y)=-f(-x,-y)=-u(-x,-y)=-(u\circ T)(x,y).$$ Now, as you noted $\Delta(u\circ T)=0$, which follows by symmetry of $\Omega$ and the identities for all $(x,y)\in\Omega$ $$\partial^2_x(u\circ T)(x,y)=\partial^2_xu(-x,-y),\\\partial^2_y(u\circ T)(x,y)=\partial^2_yu(-x,-y).$$ Finally, this implies that both $u$ and $-(u\circ T)$ are solutions to your Dirichelt problem. The fact the $f$ is continuous implies that you only have one solution, hence $$u(x,y)=-(u\circ T)(x,y)=-u(-x,-y)$$ for all $(x,y)\in\mathbb{R}^2$.
Note: While I was writing this, another answer was posted. The solution is essentially the same: maxima principle $\Rightarrow$ uniqueness.