Consider the following simple elliptic problem in a bounded domain $\Omega\subset\mathbb{R}^N$: $$ -\Delta u(x) + u(x) = f(x) \qquad \forall x\in \Omega $$ with Neumann boundary conditions in $\partial\Omega$. If $f\in L^2(\Omega)$ and we assume that we have a solution $u\in W^{1,2}(\Omega)$, we can multiply the equation by $u$ and integrate by parts to obtain $$ \int_{\Omega}\left|\nabla u\right|^2+u^2\leq \int_{\Omega}fu\leq \left(\int_\Omega u^2\right)^{1/2}\left(\int_\Omega f^2\right)^{1/2} $$ where we have used Cauchy-Schwarz in the last inequality. So we basically obtain $$ \lVert u \rVert_{H^1(\Omega)}\leq \lVert u \rVert_{L^2(\Omega)} $$ It is possible to obtain higher regularity estimates like $\lVert u \rVert_{H^2(\Omega)}\leq C\lVert u \rVert_{L^2(\Omega)}$ but this involves more complicated arguments, needs regularity on $\partial\Omega$ and the constant $C$ may depend on the domain (see the book of Partial Diff Eqs by Evans).
My question is regarding the $L^p(\Omega)$ spaces. Assume $f\in L^p(\Omega)$. Is it possible to obtain an a priori estimate of the form $\lVert u \rVert_{W^{1,p}(\Omega)}\leq \lVert u \rVert_{L^p(\Omega)}$? I know that there exists a priori estimates of the form $$ \lVert u \rVert_{W^{2,p}(\Omega)}\leq C\lVert u \rVert_{L^p(\Omega)} $$ which in particular imply what I want, but they are really complicated to obtain and the constant depends on the domain (the proof is based on Calderon-Zygmund theory, see Gilbarg-Trudinger Chapter 9). I feel that there may be a simple way (as in the case of $L^2(\Omega)$) to prove that $\lVert u \rVert_{W^{1,p}(\Omega)}\leq C\lVert u \rVert_{L^p(\Omega)}$ with a constant $C$ independent of the domain, but I can not find it.
I tried to do it for a particular case. Let $p=4$ and assume we are in dimension 1. If we multiply the equation by $\varphi=u^3$ and integrate by parts I obtain $$ \int_{\Omega}3(u')^2u^2+u^4\leq \int_{\Omega}fu^3\leq \left(\int_\Omega u^4\right)^{3/4}\left(\int_\Omega f^4\right)^{1/4} $$ so then, in particular, $$ \lVert u \rVert_{L^4(\Omega)}\leq \lVert f \rVert_{L^4(\Omega)} $$ but I do not obtain a bound for $\lVert u' \rVert_{L^4(\Omega)}$ but just for $\int_{\Omega}3\left|u'\right|^2u^2$.
EDITED:
I managed to obtain the estimates in dimension one. Taking $\varphi=u(u')^2$ then $\varphi'=(u')^3+2uu'u''$ and using that $u''=u-f$, we have $\varphi=(u')^3+2u^2u'-2uu'f$. Then, using the variational formulation and an argument similar to the one above one obtains $$ \lVert u' \rVert_{L^4(\Omega)}\leq \sqrt{3}\lVert f \rVert_{L^4(\Omega)} $$ The same can be done for $p>2$. However, I did not managed to make it for higher dimensions.