Suppose \( H \) is a Hilbert space . For every \( \varphi \in H^* \), there exists a vector \( h_\varphi \in H \) such that
\[ \varphi = \langle \cdot , h_\varphi \rangle. \]The vector \( h_\varphi \) is called the Riesz representation of \( \varphi \), and it satisfies
\begin{equation}\label{eq:norm_eq} \|\varphi\| = \|h_\varphi\|. \end{equation}- The vector \( h_\varphi \) must appear in the second argument, since the second argument of the scalar product is antilinear.
Uniqueness and equation \eqref{eq:norm_eq} correspond to the injectivity and isometry of the canonical map from \( H \) to \( H^* \) (see (0x68e72ea1) ). Let \( \varphi \in H^* \).
We prove the existence of a Riesz representation of \( \varphi \). Assume \( \ker \varphi \neq H \); otherwise \( h_\varphi = 0 \). Then, since \( H \) is complete and \( K = \ker(\varphi)=\varphi^{-1}(\{0\}) \) is closed, it follows by (0x68e89b9a)
\[ H = K \oplus K^\perp. \]Hence, since \(\ker \varphi \neq H\), there exists a vector \( p \in K^\perp \) with \( \|p\| = 1 \). For \( h \in H \),
\[ (\varphi h)p - (\varphi p)h \in K. \]By the identity
\[ \langle p, (h - \varphi(h)p) \rangle = 0, \]we deduce
\[ \langle h,\overline{\varphi(p)} p \rangle = \varphi(h). \]Thus, the vector \( \overline{\varphi(p)} p \) is the Riesz representation of \( \varphi \).