Riesz representation theorem

Remarks

If $\varphi = 0$, take $y = 0$. Otherwise the kernel $N = \ker \varphi$ is a closed subspace different from $H$. By the Hilbert projection theorem, its orthogonal complement $N^\perp$ contains a vector $z \neq 0$, and $N^\perp$ has dimension $1$. Then $y = \overline{\varphi(z)}\, z / \norm{z}^2$ works, and uniqueness follows because $\ip{x}{y - y'} = 0$ for all $x$ forces $y = y'$.

Example: on $L^2([0,1])$ the form $\varphi(f) = \int_0^1 f(t)\, dt$ is represented by the constant function $y = 1$, and $\norm{\varphi} = 1$.

Completeness is necessary. Let $c_{00}$ be the space of finitely supported sequences with the inner product of $\ell^2$. The form $\varphi(x) = \sum_n x_n / n$ is continuous on $c_{00}$ by the Cauchy–Schwarz inequality. The only candidate is $y = (1/n)_n$, which is not finitely supported, so no $y \in c_{00}$ represents $\varphi$.

In finite dimension the result reduces to the Riesz representation theorem in finite dimensions.

Going deeper

For a student

The theorem identifies a Hilbert space with its dual, up to complex conjugation. It gives the adjoint of a bounded operator: $\ip{Tx}{y} = \ip{x}{T^*y}$ defines $T^*y$ because $x \mapsto \ip{Tx}{y}$ is a continuous linear form. It also shows that every Hilbert space is reflexive. The Lax–Milgram theorem, used for weak solutions of elliptic equations, extends it to bounded coercive bilinear forms.