---
tag: TTC40N49M6
kind: statement
role: theorem
name: Riesz representation theorem
msc: [46C05]
uses: [HM2HHE3J51, 4A82S5FVRT, 07YQ1C78G0, 0XJXTBCB2T, J5YNTHQMFX, 04GYPXDXNH]
about: [HM2HHE3J51, 4A82S5FVRT]
sources:
  - url: https://en.wikipedia.org/wiki/Riesz_representation_theorem
same_as:
  wikidata: Q1357684
  wikipedia: Riesz representation theorem
---

## Statement

Let $\mathbb{K}$ be $\R$ or $\C$, and let $H$ be a [[Hilbert space|HM2HHE3J51]] over $\mathbb{K}$ whose inner product $\ip{\cdot}{\cdot}$ is linear in the first argument.
Let $\varphi \colon H \to \mathbb{K}$ be a continuous [[linear form|07YQ1C78G0]].
Then there is exactly one vector $y \in H$ such that $\varphi(x) = \ip{x}{y}$ for all $x \in H$.
Moreover $\norm{\varphi} = \norm{y}$, where $\norm{\varphi}$ is the operator norm of $\varphi$.
Hence the map that sends $\varphi$ to $y$ is a bijective, conjugate-linear isometry from the [[continuous dual|4A82S5FVRT]] $H'$ onto $H$.

## 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|J5YNTHQMFX]], its [[orthogonal complement|0XJXTBCB2T]] $N^\perp$ 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|04GYPXDXNH]].

## Explanations

### L1 human

On a Hilbert space, every reasonable way to turn vectors into numbers linearly is "take the inner product with a fixed vector".
So the space of such linear measurements looks exactly like the space itself.
For example, taking the average of a function on $[0,1]$ is the inner product with the constant function $1$.

### L2 human

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.