Let $\Omega$ be a smooth bounded domain and $f\in L^p(\Omega)$, $u\in W^{1,1}(\Omega)$ such that $p\geq 1$. Then $$ \int_{\Omega}|fu|\,dx\leq C\,\int_{\omega}\{|f(x)|dx(\int_{\Omega}\frac{|\nabla u(\zeta)|}{|x-\zeta|^{n-1}}\,d\zeta+||f||_{L^1(\Omega)})\}\leq\,C\{(\int_{\Omega}\int_{\Omega}\frac{|f(x)|}{|x-\zeta|^{n-\frac{1}{q}}}d\zeta\,dx)^{1-\frac{1}{n}} (\int_{\Omega}|\nabla\zeta|^n\,d\zeta\int_{\Omega}\frac{|f(x)|\,dx}{|x-\zeta|^\frac{n-1}{q}})^\frac{1}{n}+||u||_{L^1(\Omega)}||f||_{L^1(\Omega)} \} $$ for some positive constant $C$. In the above line, I have used the following Lemma 1, proved in Trudinger 1967 paper which states that for any $u\in W^{1,1}(\Omega)$, we have $$ |u(x)|\leq C(n)\{\int_{\Omega}\frac{|\nabla u(\zeta)|}{|x-\zeta|^{n-1}}\,d\zeta+||u||_{L^1(\Omega)}\}. $$
Please help me. Thank you.
First you use the Lemma to get \begin{align}\int_{\Omega}|fu|\,dx&\leq C\,\int_{\Omega}|f(x)|\left(\int_{\Omega}\frac{|\nabla u(\zeta)|}{|x-\zeta|^{n-1}}\,d\zeta+||u||_{L^1(\Omega)}\right)dx \\&=\int_{\Omega}\int_{\Omega}|f(x)|\frac{|\nabla u(\zeta)|}{|x-\zeta|^{n-1}}\,d\zeta dx+\int_{\Omega}|f(x)|\,dx||u||_{L^1(\Omega)}\end{align} Since $$\frac{n-1}n\left(n-\frac1q\right)+\frac1n\frac{n-1}{q}=n-1$$ you can rewrite the first term on the right-hand side as $$\int_{\Omega}\int_{\Omega}\left(\frac{|f(x)|}{|x-\zeta|^{n-1/q}}\right)^{(n-1)/n}\left(\frac{|f(x)|}{|x-\zeta|^{(n-1)/q}}\right)^{1/n}|\nabla u(\zeta)|\,d\zeta dx $$ and use Holder's inequality with exponent $p=n$ and $p'=\frac{n}{n-1}$ to bound this term from above by $$\left(\int_{\Omega}\int_{\Omega}\frac{|f(x)|}{|x-\zeta|^{n-1/q}}\,d\zeta dx\right)^{(n-1)/n}\left(\int_{\Omega}\int_{\Omega}\frac{|f(x)|}{|x-\zeta|^{(n-1)/q}}|\nabla u(\zeta)|^n\,d\zeta dx\right)^{1/n}. $$ In the last term you can use Fubini's term to separate the integrals.