[proofplan] Apply the Lax-Milgram theorem to the Hilbert space $H_0^1(U)$ with bilinear form $B$ and linear functional $f$. [/proofplan] [step:Check the hypotheses] The space $H_0^1(U)$ is a Hilbert space. By assumption, $B$ is bounded on $H_0^1(U)\times H_0^1(U)$ and coercive: there is $\alpha>0$ such that \begin{align*} B[v,v]\ge \alpha\|v\|_{H_0^1(U)}^2 \end{align*} for every $v\in H_0^1(U)$. The element $f\in H^{-1}(U)$ is a bounded linear functional on $H_0^1(U)$. [/step] [step:Apply Lax-Milgram] By the [Lax-Milgram Theorem](/theorems/91), there exists a unique $u\in H_0^1(U)$ such that \begin{align*} B[u,v]=f(v) \end{align*} for every $v\in H_0^1(U)$. [/step] [step:Estimate the solution] Taking $v=u$ gives \begin{align*} \alpha\|u\|_{H_0^1(U)}^2\le B[u,u]=f(u) \le \|f\|_{H^{-1}(U)}\|u\|_{H_0^1(U)}. \end{align*} If $u=0$ the estimate is immediate. Otherwise divide by $\|u\|_{H_0^1(U)}$ to obtain \begin{align*} \|u\|_{H_0^1(U)}\le \alpha^{-1}\|f\|_{H^{-1}(U)}. \end{align*} Thus the asserted estimate holds with $C=\alpha^{-1}$. [/step]