> [!theorem] > > Let $(H, \inp)$ be a [[Hilbert Space|Hilbert space]] and $\inp_L$ be a [[Sesquilinear Form|sesquilinear]] form. If $\inp_L$ is bounded, then it induces a [[Bounded Linear Map|bounded linear map]] $A: H \to H$ such that > $ > \angles{Ax, y} = \angles{x, y}_L \quad \forall x, y \in H > $ > If $\inp_L$ is bounded and coercive, then $A$ is a [[Space of Toplinear Isomorphisms|toplinear isomorphism]]. In particular, this implies that for every $x \in H$, there exists $x_L \in H$ such that $\angles{x, \cdot} = \angles{x_L, \cdot}_L$. > > *Proof*. First take $A$ such that $\angles{x, Ay} = \angles{x, y}_L$ for all $x, y \in H$ by [[Riesz Representation Theorem|Riesz representation]]. If $\inp_L$ is bounded, then by the continuity of [[Riesz Representation Theorem|Riesz representation]], $A$ can be expressed as the composition of two bounded conjugate linear maps, hence $A$ is a bounded linear operator. > > Suppose that $A$ is bounded and coercive, then > $ > c\norm{x}_H^2 \le \abs{\angles{x, x}_L} = \abs{\angles{x, Ax}} \le \norm{x}_H\norm{Ax}_H > $ > Hence $\norm{Ax}_H \ge c\norm{x}_H$ for all $x \in H$, and the range of $A$ is closed. If $A$ is not surjective, then there exists $x \in H$ with $x \perp A(H)$, and $\angles{x, Ax} = 0$, which contradicts the fact that $A$ is coercive. By the [[Open Mapping Theorem|open mapping theorem]], $A$ is a toplinear isomorphism. > > Taking $A^*$ gives the desired map.