 Solution

By part (a), $B$ is a bounded bilinear form: $|B(u,v)|\le M\Vert u\Vert \Vert v\Vert$. 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 thatUniqueness of the representing vector makes $L$ linear, and $\Vert Lv\Vert \le M\Vert v\Vert$ makes it a bounded linear operator.

The coercivity assumption implieshence $\Vert Lv\Vert \ge C\Vert v\Vert$. Moreover, using the adjoint operator,so the same argument gives $\Vert L^{\ast }v\Vert \ge C\Vert v\Vert$. 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 $B(u,v_f)=⟨u,L{v}_f⟩=⟨u,f⟩$ 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 the unique vector is $v_f=L^{-1}f$.