Hilbert Spaces 17 | Riesz Representation Theorem

Q: Why does continuity of a linear functional on a Hilbert space reduce to boundedness?

For linear maps on normed spaces, continuity is equivalent to boundedness. Here boundedness means the operator norm $\|L\|=\sup_{\|x\|=1}|L(x)|$ is finite. Since the output space is one-dimensional (F), this supremum captures exactly how large L can get on unit vectors.

Q: How does the $\ell^2$ example foreshadow Riesz representation?

In $\ell^2$, a functional can be defined by combining coordinates, e.g., L(x)=x1+2x2. Using the inner product structure, coordinate extraction is realized by basis vectors: x1=⟨e1,x⟩ and x2=⟨e2,x⟩. Therefore L(x)=⟨e1+2e2, x⟩, showing how a functional becomes an inner product with a single vector.

Q: What role do kernels and orthogonal complements play in the proof?

The kernel of a continuous linear map is a closed linear subspace. Closedness allows use of orthogonal complements and the fact that for such subspaces, the double orthogonal complement returns the original subspace. If ker(L) is all of X, L is the zero map. Otherwise, ker(L) is proper, so its orthogonal complement is nontrivial, letting one choose a vector z not annihilated by L.

Q: How is the representing vector y constructed when ker(L) is not all of X?

Pick a unit vector z in (ker(L))⊥. Then z∉ker(L), so L(z)≠0. Scale z by a factor involving the complex conjugate of L(z) so that the resulting y satisfies ⟨y,x⟩=L(x) for every x. The scaling is chosen so that when x is decomposed relative to ker(L) and its orthogonal complement, the inner product reproduces the functional value.

Q: Why is the representing vector y unique?

If both y and ỹ satisfy L(x)=⟨y,x⟩=⟨ỹ,x⟩ for all x, then ⟨y−ỹ,x⟩=0 for every x. Taking x=y−ỹ gives ⟨y−ỹ,y−ỹ⟩=0, which by positive definiteness implies y−ỹ=0, so y=ỹ.

Q: How does Cauchy–Schwarz produce the norm identity \|L\|=\|y\|?

With L(x)=⟨y,x⟩, Cauchy–Schwarz gives |L(x)|≤\|y\|\|x\|. For unit vectors (\|x\|=1), this yields \|L\|≤\|y\|. The reverse inequality comes by testing x in the direction of y (normalized), where the bound becomes an equality, giving \|L\|≥\|y\|.

TL;DR

A bounded linear functional L on a Hilbert space satisfies $∥ L ∥ = sup_{∥ x ∥ = 1} ∣ L (x) ∣ < \infty$ .

Briefing Cornell Notes

Briefing

Riesz representation turns every bounded linear functional on a Hilbert space into an inner product with a single fixed vector—complete with uniqueness and an exact match of norms. That matters because it converts an abstract “function that eats vectors and spits out scalars” into a concrete geometric object inside the space, setting up later operator-theory arguments.

Start with a Hilbert space X over a field F (real or complex). A functional is a linear map L: X → F that is continuous, equivalently bounded—meaning its operator norm stays finite: $∥ L ∥ = sup_{∥ x ∥ = 1} ∣ L (x) ∣$ . In an example from $ℓ^{2}$ , a functional can take a sequence x = (x1, x2, …) and return a scalar like x1 + 2x2. Boundedness follows from the $ℓ^{2}$ norm constraint: if $∥ x ∥_{2} = 1$ , then each component cannot exceed 1 in absolute value, giving a finite bound (here, at most 3).

The key insight is that such a functional can be rewritten using the inner product. In $ℓ^{2}$ , the coordinate functionals are realized by inner products with standard basis vectors: the first component is $⟨ e_{1}, x ⟩$ , the second is $⟨ e_{2}, x ⟩$ . So the functional x1 + 2x2 becomes $⟨ e_{1} + 2 e_{2}, x ⟩$ . The Riesz representation theorem generalizes this: for any Hilbert space X and any bounded linear functional L, there exists a unique vector y in X such that $L (x) = ⟨ y, x ⟩ for all x \in X .$ Moreover, the functional’s operator norm equals the vector norm: $∥ L ∥ = ∥ y ∥$ . In other words, the representation preserves the “size” of the functional.

The proof strategy leans on orthogonality. The kernel of a continuous linear map is always a closed linear subspace, which allows use of orthogonal complements. If the kernel is all of X, then L is the zero map and y = 0 works immediately. Otherwise, the orthogonal complement of the kernel is nontrivial, so one can pick a unit vector z orthogonal to ker(L). Because z is not in the kernel, L(z) ≠ 0. Scaling z by the complex conjugate of L(z) (in the complex case) produces a vector y that forces $⟨ y, x ⟩$ to match L(x) for every x.

Uniqueness is straightforward: if two vectors y and $\tilde{y}$ both satisfy $L (x) = ⟨ y, x ⟩ = ⟨ \tilde{y}, x ⟩$ for all x, then $⟨ y - \tilde{y}, x ⟩ = 0$ for every x, forcing y = $\tilde{y}$ by positive definiteness of the inner product.

Finally, the norm identity follows from Cauchy–Schwarz. With $L (x) = ⟨ y, x ⟩$ , one gets $∣ L (x) ∣ \leq ∥ y ∥∥ x ∥$ , so $∥ L ∥ \leq ∥ y ∥$ . The reverse inequality comes by testing x in the direction of y (normalized), yielding $∥ L ∥ \geq ∥ y ∥$ . Together these give $∥ L ∥ = ∥ y ∥$ . The result turns functional analysis into geometry inside Hilbert spaces, paving the way for studying operators built from these inner-product structures.

Cornell Notes

Riesz representation theorem says that every bounded linear functional L on a Hilbert space X can be written uniquely as an inner product with a fixed vector y in X: L(x)=⟨y,x⟩ for all x. The theorem also matches sizes exactly: the operator norm of L equals the norm of y, so \|L\|=\|y\|. The construction uses orthogonal complements: if ker(L) is not all of X, pick a unit vector z orthogonal to ker(L), scale it using L(z) (with complex conjugation in the complex case), and obtain y. Uniqueness follows because if ⟨y−ỹ,x⟩=0 for all x, then y−ỹ must be the zero vector. Cauchy–Schwarz then yields the norm equality.

Why does continuity of a linear functional on a Hilbert space reduce to boundedness?

For linear maps on normed spaces, continuity is equivalent to boundedness. Here boundedness means the operator norm

∥ L ∥ = sup_{∥ x ∥ = 1} ∣ L (x) ∣

is finite. Since the output space is one-dimensional (F), this supremum captures exactly how large L can get on unit vectors.

How does the $ℓ^{2}$ example foreshadow Riesz representation?

ℓ^{2}

, a functional can be defined by combining coordinates, e.g., L(x)=x1+2x2. Using the inner product structure, coordinate extraction is realized by basis vectors: x1=⟨e1,x⟩ and x2=⟨e2,x⟩. Therefore L(x)=⟨e1+2e2, x⟩, showing how a functional becomes an inner product with a single vector.

What role do kernels and orthogonal complements play in the proof?

The kernel of a continuous linear map is a closed linear subspace. Closedness allows use of orthogonal complements and the fact that for such subspaces, the double orthogonal complement returns the original subspace. If ker(L) is all of X, L is the zero map. Otherwise, ker(L) is proper, so its orthogonal complement is nontrivial, letting one choose a vector z not annihilated by L.

How is the representing vector y constructed when ker(L) is not all of X?

Pick a unit vector z in (ker(L))⊥. Then z∉ker(L), so L(z)≠0. Scale z by a factor involving the complex conjugate of L(z) so that the resulting y satisfies ⟨y,x⟩=L(x) for every x. The scaling is chosen so that when x is decomposed relative to ker(L) and its orthogonal complement, the inner product reproduces the functional value.

Why is the representing vector y unique?

If both y and ỹ satisfy L(x)=⟨y,x⟩=⟨ỹ,x⟩ for all x, then ⟨y−ỹ,x⟩=0 for every x. Taking x=y−ỹ gives ⟨y−ỹ,y−ỹ⟩=0, which by positive definiteness implies y−ỹ=0, so y=ỹ.

How does Cauchy–Schwarz produce the norm identity \|L\|=\|y\|?

With L(x)=⟨y,x⟩, Cauchy–Schwarz gives |L(x)|≤\|y\|\|x\|. For unit vectors (\|x\|=1), this yields \|L\|≤\|y\|. The reverse inequality comes by testing x in the direction of y (normalized), where the bound becomes an equality, giving \|L\|≥\|y\|.

Review Questions

Given a bounded linear functional L on a Hilbert space X, what is the exact formula for L(x) in terms of the representing vector y?
Explain how the orthogonal complement of ker(L) guarantees the existence of a vector z with L(z)≠0.
How do Cauchy–Schwarz and the choice x=y/\|y\| combine to prove \|L\|=\|y\|?

Key Points

1
A bounded linear functional L on a Hilbert space satisfies $∥ L ∥ = sup_{∥ x ∥ = 1} ∣ L (x) ∣ < \infty$ .
2
Riesz representation guarantees a unique y in X such that $L (x) = ⟨ y, x ⟩$ for all x.
3
If ker(L)=X, then L is the zero functional and the representing vector is y=0.
4
If ker(L) is a proper closed subspace, its orthogonal complement is nontrivial, enabling a construction of y from a unit vector z in (ker(L))⊥.
5
Uniqueness follows because ⟨y−ỹ,x⟩=0 for all x forces y=ỹ by positive definiteness.
6
The operator norm matches the representing vector norm exactly: $∥ L ∥ = ∥ y ∥$ .
7
Cauchy–Schwarz gives $∥ L ∥ \leq ∥ y ∥$ , and testing x in the direction of y gives the reverse inequality.

Highlights

Every bounded linear functional on a Hilbert space is exactly an inner product with a unique vector inside the same space.

The theorem preserves norms: the functional’s operator norm equals the norm of its representing vector.

The proof hinges on orthogonal complements of ker(L), turning algebraic constraints into geometric ones.

Complex conjugation appears naturally when scaling to match inner-product conventions in the complex case.

Topics

Hilbert Spaces
Riesz Representation Theorem
Bounded Linear Functionals
Orthogonal Complements
Operator Norms