= 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 that
$$
B(u,v)=(Tu,v)_H,
\qquad
F(v)=(z,v)_H.
$$
Coercivity and the <Cauchy-Schwarz inequality> imply
$$
\alpha\lVert u\rVert_H^2\leq(Tu,u)_H
\leq\lVert Tu\rVert_H\lVert u\rVert_H,
$$
so $\lVert Tu\rVert_H\geq\alpha\lVert u\rVert_H$. 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 $\lVert u\rVert_H\leq\lVert F\rVert/\alpha$.
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