Let $\Omega$ be an open domain in $\mathbb {R^n}$ and $f \in C^{\infty}(\Omega)$ then how can we prove there is a weak solution $u \in W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) \cap L^{\infty}(\Omega) $ for the problem \begin{cases} \Delta^2u=f & in\hspace{.2cm} \Omega \\ u>0 & in \hspace{.2cm} \Omega \\ u=\Delta u =0 & on\hspace{.2cm} \partial \Omega \end{cases}
my attempt:
In this case $u$ must satisfy
$$\int_{\Omega} (-\Delta u )(-\Delta \phi ) \mathrm{d}x = \int_{\Omega} f \phi \mathrm{d}x $$ for all $$\phi \in W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$$
I know that $W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$ is a Hilbert space with inner product $$\langle u,v\rangle=\int_{\Omega} \Delta u \Delta v dx $$ and I know $$\Delta ^2: W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) \to W_0^{-2,2}(\Omega) $$ is coercive.
So by the Lax-Milgram Theorem I can prove existence of $u \in W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) $ .
But my question is why also we have that $$u \in L^{\infty}(\Omega)$$ $$ u>0 $$