Let $\Omega$ be open and bounded, $\ h \in L^2(\Omega)\ $,$\ q \in L^{\infty}(\Omega)$, $\ Lu = -\Delta u + q(x)u$.
If $\lambda _1(- \Delta + q(x)) >0$, then the expression
$(u,v)_q= \int _{\Omega} \nabla u \nabla v \ dx +\int_{\Omega} q(x)uv \ dx$
defines on $H_0^1(\Omega)$ a scalar product that induces a norm equivalent to the standard one.
To show that the two norms are equivalent , we need prove there exists two nonnegative constants $C_1,C_2$ such that $C_1\|u\|_{H_0 ^1} \leq \|u\|_q \leq C_2\|u\|_{H_0 ^1}$
My question is how to prove that there exists a constant $C>0$ such that $\|u\|_q \ge C\|u\|_{H_0 ^1}$.
We want to prove that there is $c>0$ such that $$ \|u\|_{H^1}^2 \le c \int_\Omega |\nabla u_n|^2 + qu^2 dx \quad \forall u\in H^1_0(\Omega). $$ Let me prove the inequality by contradiction. Assume the claim does not hold. Then for every $n$ there is $u_n \in H^1_0(\Omega)$ such that $$ \|u_n\|_{H^1}^2 > n \int_\Omega |\nabla u_n|^2 + qu_n^2 dx. $$ The right-hand side is $\ge \lambda_1 \|u_n\|_{L^2}^2$, this implies $u_n \ne0$. We can replace $u_n$ by $\frac1{\|u_n\|_{H^1}}u_n$, and thus we can assume $\|u_n\|_{H^1} =1$. The inequality $$ 1 = \|u_n\|_{H^1}^2 > n \int_\Omega |\nabla u_n|^2 + qu_n^2 dx \ge n\lambda_1 \|u_n\|_{L^2}^2 $$ implies $u_n \to 0$ in $L^2(\Omega)$, which in turn implies $\int_\Omega qu_n^2 dx \to 0$ by Hoelder inequality. In addition, we have $$ \frac 1n - \int_\Omega qu_n^2 dx> \int_\Omega |\nabla u_n|^2 dx \ge 0, $$ and hence $\nabla u_n \to 0$ in $L^2(\Omega)$, which implies $u_n \to 0$ in $H^1(\Omega)$. This is a contradiction to $1 = \|u_n\|_{H^1}^2$.