Semi-Hilbert space

In functional analysis, a semi-Hilbert space is a complex or real vector space equipped with a positive semidefinite sesquilinear form whose associated quotient space is a Hilbert space. Semi-Hilbert spaces generalize ordinary Hilbert spaces by allowing nonzero vectors to have zero length.

Unlike the inner product of a Hilbert space, the form defining a semi-Hilbert space need not be positive definite. Consequently, it induces a seminorm rather than necessarily a norm. The degeneracy of this seminorm is removed by taking the quotient with respect to its kernel.

Formal definition

Let V be a vector space over 𝕂, where 𝕂=ℝ or ℂ.

Suppose that

⟨⋅,⋅⟩:V×V⟶𝕂

is a positive semidefinite symmetric bilinear form in the real case, or a positive semidefinite Hermitian sesquilinear form in the complex case.

Using the convention that the form is linear in its first argument, it satisfies the following properties.

Sesquilinearity

For all x,y,z∈V and a,b∈𝕂,

⟨ax+by,z⟩=a⟨x,z⟩+b⟨y,z⟩.

In the complex case, conjugate linearity in the second argument gives

⟨x,ay+bz⟩=a‾⟨x,y⟩+b‾⟨x,z⟩.

Hermitian symmetry

For every x,y∈V,

⟨x,y⟩=⟨y,x⟩‾.

Positive semidefiniteness

For every x∈V,

⟨x,x⟩≥0.

Unlike a Hilbert space, the condition

⟨x,x⟩=0⟹x=0

is not required.

Define the associated seminorm by

p(x)=⟨x,x⟩.

Its kernel is the vector subspace

N=ker⁡p={x∈V:⟨x,x⟩=0}.

The quotient vector space

V/N

inherits a positive definite inner product given by

⟨x+N,y+N⟩V/N=⟨x,y⟩V.

A semi-Hilbert space is obtained when this quotient is complete with respect to its induced norm.

Equivalently,

(V,⟨⋅,⋅⟩) is semi-Hilbert⟺V/N is Hilbert.

Fundamental properties

Cauchy–Schwarz inequality

Every positive semidefinite Hermitian form satisfies the Cauchy–Schwarz inequality

|⟨x,y⟩|2≤⟨x,x⟩⟨y,y⟩.

Proof.

Suppose first that ⟨y,y⟩>0.

For every scalar λ∈𝕂, positive semidefiniteness implies

0≤⟨x−λy,x−λy⟩.

Expanding,

0≤⟨x,x⟩−λ‾⟨x,y⟩−λ⟨y,x⟩+|λ|2⟨y,y⟩.

Choose

λ=⟨x,y⟩⟨y,y⟩.

Substitution yields

0≤⟨x,x⟩−|⟨x,y⟩|2⟨y,y⟩.

Therefore,

|⟨x,y⟩|2≤⟨x,x⟩⟨y,y⟩.

If ⟨y,y⟩=0, positivity of

⟨x+ty,x+ty⟩

for arbitrary real or complex scalars t implies that ⟨x,y⟩=0. The inequality follows in this case as well.

Characterization of the null space

Proposition. The kernel of the induced seminorm coincides with the radical of the positive semidefinite form:

N={x∈V:⟨x,y⟩=0 for every y∈V}.

Proof.

If x∈N, then

⟨x,x⟩=0.

By the Cauchy–Schwarz inequality,

|⟨x,y⟩|2≤⟨x,x⟩⟨y,y⟩=0.

Hence

⟨x,y⟩=0

for every y∈V.

Conversely, if ⟨x,y⟩=0 for every y∈V, choosing y=x gives

⟨x,x⟩=0.

Thus x∈N, proving the equality.

Quotient Hilbert space

Theorem. Let V be a vector space equipped with a positive semidefinite Hermitian form, and let N be its null space. Then V/N is an inner product space. It is a Hilbert space if and only if it is complete in the induced norm.

Proof.

Define

⟨[x],[y]⟩=⟨x,y⟩,

where [x]=x+N denotes the equivalence class of x.

To establish that this definition is independent of the representatives, suppose

x′=x+n1,y′=y+n2,

with n1,n2∈N.

Then

⟨x′,y′⟩=⟨x+n1,y+n2⟩=⟨x,y⟩+⟨n1,y⟩+⟨x,n2⟩+⟨n1,n2⟩=⟨x,y⟩.

The final equality follows from the characterization of the null space.

Therefore the quotient form is well-defined.

Furthermore,

⟨[x],[x]⟩=0⟺x∈N⟺[x]=0.

Thus the induced form is positive definite.

Consequently, V/N is an inner product space, and it is a Hilbert space precisely when it is complete.

Completeness

Completeness of the quotient is an essential part of the definition.

A positive semidefinite form alone does not guarantee that the quotient is complete.

For example, let

V=Cc∞(ℝ),

the vector space of smooth compactly supported functions on the real line.

Equip it with the positive definite form

⟨f,g⟩=∫ℝf(x)g(x)‾dx.

Its null space is trivial:

N={0}.

However, V is not complete in the induced L2 norm.

Its completion is

L2(ℝ).

Thus V, with this form, is a pre-Hilbert space but not a semi-Hilbert space under the quotient-completeness definition.

More generally, every positive semidefinite Hermitian form determines a Hilbert space after quotienting by its null space and subsequently taking the metric completion.

Examples

Degenerate Euclidean space

Consider the vector space

V=ℝ3

with the bilinear form

⟨x,y⟩=x1y1+x2y2.

The associated seminorm is

p(x)=x12+x22.

The null space is

N={(0,0,t):t∈ℝ}.

Hence

V/N≅ℝ2.

Since finite-dimensional inner product spaces are complete, the quotient is a Hilbert space.

Therefore V is a semi-Hilbert space.

Sobolev-type seminorm

Let

V=H1(0,1),

the first-order Sobolev space on the interval (0,1).

Define

⟨f,g⟩D=∫01f′(x)g′(x)‾dx.

The induced seminorm is

p(f)=(∫01|f′(x)|2dx)1/2.

Its null space consists precisely of the almost-everywhere constant functions:

N={f∈H1(0,1):f′=0}.

Consequently,

H1(0,1)/N

is an inner product space with norm determined by the weak derivative.

By the Poincaré inequality, this quotient is complete.

It is therefore a Hilbert space, making H1(0,1) a semi-Hilbert space under the derivative form.

Relationship to Hilbert spaces

Every Hilbert space is a semi-Hilbert space with trivial null space.

Conversely, a semi-Hilbert space is a Hilbert space under its defining form if and only if

N={0}.

The canonical quotient mapping

q:V⟶V/N

satisfies

‖q(x)‖V/N=p(x).

Thus the quotient identifies vectors whose difference has zero seminorm.

In particular,

q(x)=q(y)⟺p(x−y)=0.

This construction allows positive semidefinite forms to be studied using the ordinary theory of Hilbert spaces.

norm.