The Riesz representation theorem

Theorem

The Riesz representation theorem

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Statement

Let \( H \) be a Hilbert space over \( \mathbb{K} \) (where \( \mathbb{K} = \mathbb{R} \) or \( \mathbb{K} = \mathbb{C} \)), with inner product \( \langle \cdot,\cdot\rangle \) linear in the first argument and conjugate-linear in the second, and induced norm \( \|x\| = \langle x,x\rangle^{1/2} \). Let \( f : H \to \mathbb{K} \) be a bounded (equivalently, continuous) linear functional, so that \( \|f\| := \sup_{\|x\|\le 1} |f(x)| \lt \infty \). Then there exists a unique \( y \in H \) such that \[ f(x) = \langle x, y\rangle \qquad \text{for all } x \in H, \] and moreover \( \|f\| = \|y\| \). Conversely every \( y \in H \) defines a bounded linear functional \( x \mapsto \langle x,y\rangle \) of norm \( \|y\| \), so the correspondence \( y \mapsto \langle \cdot, y\rangle \) is a bijective, conjugate-linear, norm-preserving map from \( H \) onto its continuous dual \( H^{*} \).

Why it matters

The theorem says that a Hilbert space is self-dual: every continuous linear functional, an object defined purely in terms of the topology and linear structure, is secretly geometric, given by "measuring the angle against a fixed vector". This identification \( H \cong H^{*} \) is the single fact that makes weak topologies on Hilbert spaces so tractable, and it underlies the definition of the Hilbert-space adjoint of a bounded operator, the variational (weak) formulation of PDEs via the Lax–Milgram theorem, and the existence of orthogonal projections and conditional expectations.

It is also the prototype for a whole family of "representation theorems" in analysis (Riesz–Markov, Radon–Nikodym) in which an abstract linear functional on a space of functions is shown to be integration against a fixed object — here the object is simply a vector.

Hypotheses

\( H \) is a Hilbert space (complete inner product space). Drop completeness and the theorem fails: let \( H_0 \subset \ell^2(\mathbb{N}) \) be the (incomplete) inner-product space of finitely supported sequences, with the \( \ell^2 \) inner product. Define \( f(x) = \sum_n x_n / n \). By Cauchy–Schwarz applied in \( \ell^2 \), \( f \) is bounded on \( H_0 \) with \( \|f\| = \|(1/n)_n\|_{\ell^2} \), since \( (1/n)_n \in \ell^2 \). But the only candidate representing vector is \( (1/n)_n \), which does not lie in \( H_0 \) (it has infinite support) — no \( y \in H_0 \) represents \( f \).

\( f \) is linear. A nonlinear functional cannot equal \( x \mapsto \langle x,y\rangle \), which is linear by definition of the inner product; e.g. \( f(x) = \|x\|^2 \) on any nonzero Hilbert space satisfies \( f(2x) = 4f(x) \ne 2f(x) \) unless \( f(x)=0 \), so it is not of Riesz form.

\( f \) is bounded (continuous). On an infinite-dimensional Hilbert space, unbounded linear functionals exist: extend a Hamel basis of a dense proper subspace using the Axiom of Choice (via a basis of \( H \) as a \( \mathbb{K} \)-vector space) and define \( f \) to blow up along the basis vectors. Such \( f \) cannot equal \( \langle \cdot, y\rangle \) for any \( y \in H \), since \( |\langle x,y\rangle| \le \|y\|\,\|x\| \) by Cauchy–Schwarz is automatically bounded. So boundedness is not merely convenient but strictly necessary, and the counterexample is inherently non-constructive.

Proof

\text{If } f = 0, \text{ take } y = 0. \text{ Assume henceforth } f \ne 0.

Trivial case disposed of first so the construction below may assume a nonzero kernel-complement exists. A

N := \ker f = \{x \in H : f(x) = 0\} \text{ is a closed linear subspace of } H.

\( N \) is a subspace since \( f \) is linear; it is closed because \( f \) is continuous (boundedness of a linear map is equivalent to continuity) and \( \{0\} \) is closed in \( \mathbb{K} \), so \( N = f^{-1}(\{0\}) \) is the preimage of a closed set under a continuous map. A

H = N \oplus N^{\perp}, \qquad N^{\perp} \ne \{0\}.

Because \( H \) is complete and \( N \) is closed, the Hilbert Projection Theorem applies: every \( x \in H \) decomposes uniquely as \( x = P_N x + P_{N^\perp} x \) with \( P_N x \in N \), \( P_{N^\perp}x \in N^\perp \), \( \langle P_N x, P_{N^\perp}x\rangle = 0 \). Since \( f \ne 0 \), \( N \ne H \), so this decomposition is nontrivial and \( N^\perp \ne \{0\} \) (otherwise \( H = N \)). B

\text{Fix } z \in N^{\perp}, \; z \ne 0. \text{ Then } f(z) \ne 0.

If \( f(z) = 0 \) then \( z \in N \); but also \( z \in N^\perp \), so \( z \in N \cap N^\perp \), forcing \( \langle z,z\rangle = 0 \), i.e. \( z = 0 \), contradicting the choice of \( z \). A

\text{For every } x \in H: \quad w := f(x)\,z - f(z)\,x \in N.

Linearity of \( f \) gives \( f(w) = f(x)f(z) - f(z)f(x) = 0 \), so \( w \in \ker f = N \) by definition. B

\langle w, z\rangle = 0 \quad\Longrightarrow\quad f(x)\,\|z\|^2 = f(z)\,\langle x, z\rangle.

\( w \in N \) and \( z \in N^\perp \), so \( \langle w,z\rangle = 0 \) by definition of orthogonal complement. Expanding \( w \) by linearity of the inner product in its first slot: \( \langle f(x)z - f(z)x,\, z\rangle = f(x)\langle z,z\rangle - f(z)\langle x,z\rangle = 0 \). A

f(x) = \left\langle x,\; \frac{\overline{f(z)}}{\|z\|^2}\, z \right\rangle \quad \text{for all } x \in H. \text{ Set } y := \frac{\overline{f(z)}}{\|z\|^2}\, z.

Dividing step 6 by \( \|z\|^2 \ne 0 \) gives \( f(x) = \dfrac{f(z)}{\|z\|^2}\langle x,z\rangle \). Since the inner product is conjugate-linear in its second slot, \( \left\langle x, \dfrac{\overline{f(z)}}{\|z\|^2} z\right\rangle = \overline{\left(\dfrac{\overline{f(z)}}{\|z\|^2}\right)}\langle x,z\rangle = \dfrac{f(z)}{\|z\|^2}\langle x,z\rangle \), which matches. This defines the representing vector \( y \) explicitly. C

\text{Uniqueness: if } \langle x,y_1\rangle = \langle x,y_2\rangle \; \forall x \in H, \text{ then } y_1 = y_2.

Take \( x = y_1 - y_2 \): then \( \langle y_1-y_2,\, y_1-y_2\rangle = \langle y_1-y_2,y_1\rangle - \langle y_1-y_2,y_2\rangle = \overline{f(y_1-y_2)} - \overline{f(y_1-y_2)}=0\) using the hypothesis with the roles matched via conjugate symmetry; directly, \( \langle x,y_1-y_2\rangle = 0\) for all \(x\), so with \(x=y_1-y_2\) we get \( \|y_1-y_2\|^2 = 0 \), hence \( y_1 = y_2 \) by positive-definiteness of the inner product. A

\|f\| = \|y\|.

(\(\le\)) By Cauchy–Schwarz, \( |f(x)| = |\langle x,y\rangle| \le \|x\|\,\|y\| \) for all \( x \), so \( \|f\| \le \|y\| \). (\(\ge\)) If \( y = 0 \) this is trivial. Otherwise \( f(y) = \langle y,y\rangle = \|y\|^2 \), and \( |f(y)| \le \|f\|\,\|y\| \) by definition of the operator norm, giving \( \|y\|^2 \le \|f\|\,\|y\| \), so \( \|y\| \le \|f\| \). Combining both inequalities, \( \|f\|=\|y\| \). B

Result

\forall f \in H^{*} \;\; \exists! \, y \in H \text{ such that } f(x) = \langle x,y\rangle \;\; \forall x \in H, \quad \|f\| = \|y\|

Reading. Every way of continuously and linearly extracting a scalar from a vector in a Hilbert space is really just "take the inner product with some fixed vector \( y \)", and that \( y \) is unique and has exactly the same size (operator norm equals vector norm) as the functional it represents.

Scope. Applies to any Hilbert space over \( \mathbb{R} \) or \( \mathbb{C} \), of any dimension (finite or infinite, separable or not). It applies specifically to the continuous dual \( H^{*} \); it is false, in the stated form, for merely complete normed spaces (Banach spaces) that are not Hilbert spaces, and false for incomplete inner product spaces.

Corollaries & converses

The map \( R : H \to H^{*} \), \( R(y) = \langle \cdot, y\rangle \), is a bijective isometry that is conjugate-linear (linear if \( \mathbb{K}=\mathbb{R}\)); it identifies \( H \) with its own dual, so \( H \) is automatically reflexive.
Every Hilbert space is weakly sequentially complete and has weakly compact closed bounded sets (Banach–Alaoglu plus reflexivity), a fact used throughout to extract weakly convergent minimising sequences.
The Riesz map lets one define the Hilbert-adjoint \( T^{*} \) of a bounded operator \( T : H \to H \) via \( \langle Tx,y\rangle = \langle x, T^{*}y\rangle \), by applying the theorem to the functional \( x \mapsto \langle Tx,y\rangle \) for each fixed \( y \).
Converse holds, trivially, and is half of the statement. Every \( y \in H \) does define a bounded functional \( \langle\cdot,y\rangle \) with norm \( \|y\| \) — this is immediate from Cauchy–Schwarz and is the "easy direction" already packaged into the theorem above; there is no separate converse to fail.

Fails without

Completeness dropped: in \( H_0 \subset \ell^2 \), the finitely-supported sequences with the \( \ell^2 \) inner product, the bounded functional \( f(x)=\sum_n x_n/n \) has no representing vector in \( H_0 \) (see Hypotheses); the Projection Theorem (step 3) fails because \( \ker f \) need not have a complement inside an incomplete space.
Boundedness dropped: a discontinuous linear functional built from a Hamel basis via the Axiom of Choice cannot be written as \( \langle \cdot, y\rangle \) for any \( y \in H \), since every such expression is automatically continuous by Cauchy–Schwarz; existence fails outright.
Genuine Banach space instead of Hilbert space: in \( C[0,1] \) with the sup norm, the evaluation functional \( f(x)=x(0) \) is bounded and linear, but \( C[0,1] \) carries no inner product inducing the sup norm (it fails the parallelogram law), so the very statement "\(f(x)=\langle x,y\rangle\)" is meaningless; more strikingly, \( (C[0,1])^{*} \) is the space of signed Borel measures (Riesz–Markov), not isomorphic to \( C[0,1] \) itself.

Common errors

Forgetting conjugate-linearity in the complex case and writing \( y = f(z)z/\|z\|^2 \) instead of \( \overline{f(z)}z/\|z\|^2 \), then failing to reproduce \( f(z) \) when checking the answer.
Confusing the two possible conventions for which slot of \( \langle\cdot,\cdot\rangle \) is conjugate-linear, and so getting \( \bar y \) instead of \( y \) (or vice versa) when comparing to a textbook.
Trying to apply the theorem to a functional on a dense but non-closed subspace, forgetting that \( \ker f \) must be closed in the whole space for the Projection Theorem to supply a nonzero complement.
Believing the theorem gives an algorithm for computing \( y \) in general — in practice one still has to solve \( f(x) = \langle x,y\rangle \) for all \( x \) directly (e.g. by testing against a basis or an explicit integral kernel); the existence proof (via an arbitrary \( z \in N^\perp \)) is usually not the fastest way to find \( y \) by hand.
Applying the real-Hilbert-space formula (\(y=f(z)z/\|z\|^2\), no conjugate) unthinkingly in a complex Hilbert space problem.

Discussion

The theorem was proved independently by Frigyes Riesz and Maurice Fréchet in 1907, in the context of what we would now call \( L^2 \) and \( \ell^2 \); it is one of the results (alongside the Riesz–Fischer theorem) that cemented the modern, coordinate-free notion of Hilbert space as the right abstract setting for these ideas. Its significance is less about a difficult computation and more about collapsing an entire category of objects, bounded linear functionals, onto a category we already understand concretely, vectors.

A far-reaching generalisation is the Lax–Milgram theorem, which replaces the inner product \( \langle x,y\rangle \) with a bounded, coercive (not necessarily symmetric) bilinear (or sesquilinear) form \( a(x,y) \), and shows that every bounded functional \( f \) can still be written \( f(x) = a(x,y) \) for a unique \( y \). This is the engine behind the weak/variational formulation of elliptic PDEs (e.g. the Poisson equation), where \( a \) comes from integrating derivatives against test functions.

A subtlety worth flagging: the Riesz map \( R : H \to H^{*} \) is an isometry of the underlying real normed spaces, but over \( \mathbb{C} \) it is only real-linear, not complex-linear — it is conjugate-linear. Consequently \( H \) and \( H^{*} \) are isomorphic as real Banach spaces (and even as complex Banach spaces via \( y \mapsto \overline{R(y)} \) composed appropriately), but the canonical identification itself is not a \( \mathbb{C} \)-linear map. This is precisely why the adjoint \( T^{*} \) of a bounded operator, built via Riesz, is conjugate-linear in the scalar it "absorbs" from \( T \) in some bookkeeping conventions, and why care is needed when moving between operator-theoretic and functional-analytic conventions.

Common misconception: students often think the theorem is about existence of an inner product recovering \( f \) — but the inner product on \( H \) is given data, fixed in advance; the theorem finds the representing vector \( y \) for that fixed inner product, not a new inner product.

Worked examples

Ex1

H = L^2[0,1] \text{ (real-valued)}, \qquad f(x) = \int_0^1 x(t)\, t \, dt.

Set-up: identify the representing vector for a concrete integral functional. A

Ex1.1

|f(x)| = \left|\int_0^1 x(t)\,t\,dt\right| = |\langle x, \mathrm{id}\rangle_{L^2}| \le \|x\|_{L^2}\, \|\mathrm{id}\|_{L^2}, \quad \mathrm{id}(t):=t,

Cauchy–Schwarz in \( L^2[0,1] \), where \( \langle x,g\rangle_{L^2} = \int_0^1 x(t)g(t)\,dt \) (real case, no conjugation needed) and \( \|\mathrm{id}\|_{L^2}^2 = \int_0^1 t^2\,dt = 1/3 \). Hence \( f \) is bounded with \( \|f\| \le 1/\sqrt3 \). A

Ex1.2

f(x) = \int_0^1 x(t)\cdot t\, dt = \langle x, \mathrm{id}\rangle_{L^2} \quad\Rightarrow\quad y(t) = t.

The functional is already, by inspection, exactly \( \langle x, y\rangle \) with \( y(t)=t \); by the uniqueness clause of the theorem (step 8) this is the representing vector — no other candidate can work. B

y(t) = t, \qquad \|f\| = \|y\|_{L^2} = \left(\int_0^1 t^2\,dt\right)^{1/2} = \frac{1}{\sqrt3}

Reading. The functional "integrate against \(t\)" is represented by the function \(t\) itself, and its operator norm equals the \(L^2\) norm of that function, matching the Cauchy–Schwarz bound in step Ex1.1 exactly.

Ex2

H = \ell^2(\mathbb{N}), \qquad f(x) = \sum_{n=1}^{\infty} \frac{x_n}{2^n}.

Set-up: a sequence-space functional; must first confirm boundedness before invoking the theorem. A

Ex2.1

\left(\sum_{n=1}^\infty \frac{1}{4^n}\right) = \frac{1/4}{1-1/4} = \frac13 \lt \infty \quad\Rightarrow\quad c:=\left(\tfrac{1}{2^n}\right)_n \in \ell^2.

Direct computation of the geometric series confirms \( c \in \ell^2 \), so by Cauchy–Schwarz \( |f(x)| = |\langle x,c\rangle_{\ell^2}| \le \|x\|_{\ell^2}\|c\|_{\ell^2} \), i.e. \( f \) is bounded, with \( \|f\|\le \|c\|_{\ell^2} \). A

Ex2.2

f(x) = \sum_n x_n \cdot \frac{1}{2^n} = \langle x, c\rangle_{\ell^2} \quad\Rightarrow\quad y = c = \left(\tfrac{1}{2^n}\right)_n,

Again the functional is visibly already of inner-product form (real scalars, so no conjugation is needed); uniqueness (step 8) identifies \( y=c \) as the only possible representer. B

y = \left(\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots\right), \qquad \|f\| = \|y\|_{\ell^2} = \sqrt{\frac13} = \frac{1}{\sqrt3}

Reading. The "weighted sum" functional on \( \ell^2 \) is represented by the geometric sequence of its own weights, and the operator norm is again the \( \ell^2 \)-norm of that sequence, computed here as \( 1/\sqrt3 \) via the geometric series.

Problems

Let \( H = \mathbb{R}^3 \) with the standard inner product, and let \( f(x_1,x_2,x_3) = 2x_1 - x_2 + 3x_3 \). Find the vector \( y \) representing \( f \), and verify \( \|f\| = \|y\| \) by computing \( \sup_{\|x\|=1} |f(x)| \) directly using Cauchy–Schwarz.
Solution

By inspection \( f(x) = \langle x, y\rangle \) with \( y = (2,-1,3) \), since \( \langle x,y\rangle = 2x_1 - x_2 + 3x_3 \) matches \( f \) term by term; uniqueness guarantees this is the only such \( y \). Directly, Cauchy–Schwarz gives \( |f(x)| = |\langle x,y\rangle| \le \|x\|\,\|y\| \), with equality when \( x = y/\|y\| \), so \( \|f\| = \|y\| = \sqrt{4+1+9} = \sqrt{14} \).
Let \( H = \mathbb{C}^2 \) with \( \langle x,w\rangle = x_1\overline{w_1} + x_2\overline{w_2} \), and \( f(x_1,x_2) = ix_1 + (1-i)x_2 \). Find \( y \) with \( f(x) = \langle x,y\rangle \) for all \( x \), being careful with conjugation.
Solution

We need \( \langle x,y\rangle = x_1\overline{y_1}+x_2\overline{y_2} = ix_1 + (1-i)x_2 \), so \( \overline{y_1} = i \Rightarrow y_1 = \overline{i} = -i \), and \( \overline{y_2} = 1-i \Rightarrow y_2 = \overline{1-i} = 1+i \). So \( y = (-i,\, 1+i) \). Check: \( \langle x,y\rangle = x_1\overline{(-i)} + x_2\overline{(1+i)} = x_1 i + x_2(1-i) \), which matches \( f \). This is the step where forgetting to conjugate (a common error) would give the wrong sign/factor on the imaginary parts.
Let \( H \) be a Hilbert space and \( T : H \to H \) a bounded linear operator. Using the Riesz representation theorem, show that there is a unique bounded linear operator \( T^{*} : H \to H \) (the adjoint) with \( \langle Tx,y\rangle = \langle x,T^{*}y\rangle \) for all \( x,y \in H \). (You may assume \( \|T^*\|=\|T\|\) without proving it.)
Solution

Fix \( y \in H \). The map \( g_y : H \to \mathbb{K} \), \( g_y(x) = \langle Tx, y\rangle \), is linear in \( x \) (since \( T \) and \( \langle\cdot,y\rangle \) are linear) and bounded: \( |g_y(x)| = |\langle Tx,y\rangle| \le \|Tx\|\,\|y\| \le \|T\|\,\|y\|\,\|x\| \) by Cauchy–Schwarz and boundedness of \( T \). By the Riesz Representation Theorem, there is a unique vector, call it \( T^{*}y \in H \), with \( g_y(x) = \langle x, T^{*}y\rangle \) for all \( x \), i.e. \( \langle Tx,y\rangle = \langle x,T^{*}y\rangle \). Linearity of \( y \mapsto T^{*}y \) follows from uniqueness applied to \( g_{\alpha y_1+\beta y_2} = \bar\alpha^{-1}\)-type bookkeeping — concretely, for scalars \( \alpha,\beta \) and vectors \( y_1,y_2 \), both \( T^*(\alpha y_1+\beta y_2) \) and \( \alpha T^*y_1 + \beta T^* y_2 \) represent the functional \( x\mapsto \alpha\langle Tx,y_1\rangle+\beta\langle Tx,y_2\rangle \), so they are equal by uniqueness. Existence and uniqueness of \( T^* \) as a well-defined map on all of \( H \) is thus established; boundedness with \( \|T^*\|=\|T\| \) is a standard further computation, assumed here as stated.
Show, by direct example, that the Riesz Representation Theorem can fail for an incomplete inner product space: let \( H_0 \) be the space of polynomials on \( [0,1] \) with the \( L^2[0,1] \) inner product (an incomplete inner product space), and let \( f(p) = p(1) \) (evaluation at \( 1 \)). Show \( f \) is unbounded on \( H_0 \) (so this particular \( f \) does not even test the completeness hypothesis in isolation) — then explain what feature of \( H_0 \) you would need to fix, and what functional would work instead as a genuine test of completeness.
Solution

Take \( p_n(t) = t^n \). Then \( \|p_n\|_{L^2}^2 = \int_0^1 t^{2n}dt = \frac{1}{2n+1} \to 0 \), so \( \|p_n\|_{L^2}\to 0 \), while \( f(p_n) = p_n(1) = 1 \) for every \( n \). Hence \( f(p_n)/\|p_n\|_{L^2} \to \infty \), so \( f \) is unbounded on \( H_0 \); it fails the boundedness hypothesis, not completeness specifically, so it is not the right example to isolate completeness. A genuine completeness-only counterexample instead needs a functional that is bounded on the incomplete space but whose representing vector lies only in the completion — exactly the construction in the Hypotheses section: \( H_0=\) finitely supported sequences in \( \ell^2 \), \( f(x)=\sum x_n/n \), bounded by Cauchy–Schwarz via \( (1/n)_n \in \ell^2\), but \( (1/n)_n \notin H_0 \).
Let \( H = L^2[0,1] \) (real) and let \( K(s,t) \) be a continuous kernel on \( [0,1]^2 \). Define \( T:H\to H \) by \( (Tx)(s) = \int_0^1 K(s,t)x(t)\,dt \), and fix \( s_0 \in [0,1] \). Consider the functional \( f(x) = (Tx)(s_0) \). Show \( f \) is bounded on \( H\) and find its Riesz representer.
Solution

\( f(x) = \int_0^1 K(s_0,t)x(t)\,dt = \langle x, K(s_0,\cdot)\rangle_{L^2} \), where \( K(s_0,\cdot) \in L^2[0,1] \) because \( K \) is continuous on the compact square \( [0,1]^2\) hence bounded, so \( K(s_0,\cdot) \) is a bounded (hence square-integrable) function on \( [0,1] \). Boundedness of \( f \) then follows from Cauchy–Schwarz exactly as in step 9 of the proof, with \( \|f\|\le \|K(s_0,\cdot)\|_{L^2}\). By inspection \( f \) is already in inner-product form, so by uniqueness the Riesz representer is \( y(t) = K(s_0,t) \), i.e. evaluating an integral operator at a point is represented by the corresponding slice of its kernel.