= Solution
By part (a), $B$ is a <bounded bilinear form>: $|B(u,v)|\leq M\lVert u\rVert\lVert v\rVert$. For each fixed $v$, the map $u\mapsto B(u,v)$ is consequently a <bounded linear functional>. The <Riesz representation theorem> supplies a unique vector $Lv$ such that
$$
B(u,v)=\langle u,Lv\rangle
\qquad(u\in\mathcal H).
$$
Uniqueness of the representing vector makes $L$ <linear>, and $\lVert Lv\rVert\leq M\lVert v\rVert$ makes it a <bounded linear operator>.
The <coercive bilinear form>[coercivity] assumption implies
$$
C\lVert v\rVert^2\leq B(v,v)=\langle v,Lv\rangle
\leq\lVert v\rVert\lVert Lv\rVert,
$$
hence $\lVert Lv\rVert\geq C\lVert v\rVert$. Moreover, using the <adjoint operator>,
$$
\langle v,L^*v\rangle=\langle Lv,v\rangle=B(v,v),
$$
so the same argument gives $\lVert L^*v\rVert\geq C\lVert v\rVert$. The <invertibility from lower bounds on an operator and its adjoint> therefore shows that $L$ is invertible and $L^{-1}$ is bounded.
Set $v_f=L^{-1}f$. Then $B(u,v_f)=\langle u,Lv_f\rangle=\langle u,f\rangle$ for every $u$. If another $v$ has this property, then $L(v-v_f)=0$, and the injectivity of $L$ gives $v=v_f$. Thus \b[the unique vector is $v_f=L^{-1}f$].
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