 Solution

Let $B$ and $F$ satisfy the hypotheses of the Lax-Milgram theorem. By the Riesz representation theorem, there are a bounded linear operator $T:H\to H$ and $z\in H$ such thatCoercivity and the Cauchy-Schwarz inequality implyso $\Vert Tu{\Vert }_H\ge \alpha \Vert u{\Vert }_H$. Thus $T$ is injective and its range is closed. If $y$ is orthogonal to its range, then $B(x,y)=0$ for every $x$; taking $x=y$ and using coercivity gives $y=0$. The range is therefore dense as well as closed, hence all of $H$. There is a unique $u=T^{-1}z$, and it satisfies $B(u,v)=F(v)$ for all $v$. The lower bound also gives $\Vert u{\Vert }_H\le \Vert F\Vert /\alpha$.