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Inner Product Spaces and Hilbert Spaces

An inner product space is a vector space HH with an inner product ⟨⋅,⋅⟩:H×H→C\langle \cdot, \cdot \rangle : H \times H \to \mathbb{C} satisfying:

  1. ⟨x,x⟩≥0\langle x, x\rangle \geq 0 with equality iff x=0x = 0.
  2. ⟨x,y⟩=⟨y,x⟩‾\langle x, y\rangle = \overline{\langle y, x\rangle}.
  3. ⟨αx+βy,z⟩=α⟨x,z⟩+β⟨y,z⟩\langle \alpha x + \beta y, z\rangle = \alpha\langle x, z\rangle + \beta\langle y, z\rangle.

Every inner product induces a norm: ∥x∥=⟨x,x⟩\|x\| = \sqrt{\langle x, x\rangle}.

Example 1. Cn\mathbb{C}^n with ⟨x,y⟩=∑i=1nxiyi‾\langle x, y\rangle = \sum_{i=1}^n x_i \overline{y_i}.

Example 2. L2(μ)L^2(\mu) with ⟨f,g⟩=∫fg‾ dμ\langle f, g\rangle = \int f \overline{g}\, d\mu.

Example 3. ℓ2\ell^2 with ⟨x,y⟩=∑i=1∞xiyi‾\langle x, y\rangle = \sum_{i=1}^\infty x_i \overline{y_i}.

Vectors x,y∈Hx, y \in H are orthogonal (written x⊥yx \perp y) if ⟨x,y⟩=0\langle x, y\rangle = 0.

Theorem 2.1 (Pythagorean Theorem). If x⊥yx \perp y, then ∥x+y∥2=∥x∥2+∥y∥2\|x + y\|^2 = \|x\|^2 + \|y\|^2.

Theorem 2.2 (Parallelogram Law). In any inner product space:

∥x+y∥2+∥x−y∥2=2∥x∥2+2∥y∥2\|x + y\|^2 + \|x - y\|^2 = 2\|x\|^2 + 2\|y\|^2

Theorem 2.3 (Polarization Identity). In a complex inner product space:

⟨x,y⟩=14(∥x+y∥2−∥x−y∥2+i∥x+iy∥2−i∥x−iy∥2)\langle x, y\rangle = \frac{1}{4}\left(\|x + y\|^2 - \|x - y\|^2 + i\|x + iy\|^2 - i\|x - iy\|^2\right)

Theorem 2.4 (Cauchy-Schwarz Inequality). ∣⟨x,y⟩∣≤∥x∥⋅∥y∥|\langle x, y\rangle| \leq \|x\| \cdot \|y\| with equality iff xx and yy are linearly dependent.

A Hilbert space is a complete inner product space.

Theorem 2.5 (Orthogonal Projection). Let MM be a closed subspace of a Hilbert space HH. For every x∈Hx \in H, there exists a unique y∈My \in M (the orthogonal projection of xx onto MM) such that x−y⊥Mx - y \perp M. We write y=PM(x)y = P_M(x).

Theorem 2.6 (Orthogonal Decomposition). If MM is a closed subspace of HH, then H=M⊕M⊥H = M \oplus M^\perp, where M⊥={x∈H:x⊥M}M^\perp = \{x \in H : x \perp M\}.

A set {ei}i∈I⊆H\{e_i\}_{i \in I} \subseteq H is an orthonormal system if ⟨ei,ej⟩=δij\langle e_i, e_j\rangle = \delta_{ij}.

Theorem 2.7 (Bessel’s Inequality). If {ei}i=1n\{e_i\}_{i=1}^n is an orthonormal set, then ∑i=1n∣⟨x,ei⟩∣2≤∥x∥2\sum_{i=1}^n |\langle x, e_i\rangle|^2 \leq \|x\|^2.

Theorem 2.8. A Hilbert space is separable if and only if it admits a countable orthonormal basis.

Theorem 2.9 (Parseval’s Identity). If {en}\{e_n\} is an orthonormal basis for HH, then for every x∈Hx \in H:

∥x∥2=∑n=1∞∣⟨x,en⟩∣2andx=∑n=1∞⟨x,en⟩en\|x\|^2 = \sum_{n=1}^{\infty} |\langle x, e_n\rangle|^2 \quad \text{and} \quad x = \sum_{n=1}^{\infty} \langle x, e_n\rangle e_n

Theorem 2.10 (Riesz Representation). Let HH be a Hilbert space. For every bounded linear functional φ∈H∗\varphi \in H^*, there exists a unique y∈Hy \in H such that φ(x)=⟨x,y⟩\varphi(x) = \langle x, y\rangle for all x∈Hx \in H. Moreover, ∥φ∥H∗=∥y∥H\|\varphi\|_{H^*} = \|y\|_H.

Proof. If φ=0\varphi = 0, take y=0y = 0. Otherwise, ker⁡(φ)\ker(\varphi) is a closed subspace, so H=ker⁡(φ)⊕ker⁡(φ)⊥H = \ker(\varphi) \oplus \ker(\varphi)^\perp. Take z∈ker⁡(φ)⊥z \in \ker(\varphi)^\perp with ∥z∥=1\|z\| = 1. Then y=φ(z)‾⋅zy = \overline{\varphi(z)} \cdot z satisfies φ(x)=⟨x,y⟩\varphi(x) = \langle x, y\rangle for all xx. Uniqueness follows from the polarization identity. ■\blacksquare

Corollary 2.11. Every Hilbert space is isometrically isomorphic to its dual: H≅H∗H \cong H^* (anti-linearly).

StructureAxioms addedCompleteness?
Inner product spVector space + inner productNot required
Hilbert spaceInner product space + completenessYes
Banach spaceNormed vector space + completenessNorm may not come from inner product

Every Hilbert space is a Banach space, but the converse fails: LpL^p for p≠2p \neq 2 is Banach but not Hilbert. The parallelogram law characterises normed spaces whose norm comes from an inner product.

  • Assuming every Cauchy sequence converges in an inner product space. Completeness is an extra requirement. For example, C([0,1])C([0,1]) with the L2L^2 inner product is not complete.
  • Confusing orthogonality with linear independence. Orthogonal vectors are always linearly independent, but linearly independent vectors need not be orthogonal.
  • Forgetting that M⊥M^\perp is closed even when MM is not. The orthogonal complement is always a closed subspace, regardless of whether MM itself is closed.
  • Assuming ⟨x,y⟩=⟨y,x⟩\langle x, y\rangle = \langle y, x\rangle in complex spaces. Conjugate symmetry means ⟨x,y⟩=⟨y,x⟩‾\langle x, y\rangle = \overline{\langle y, x\rangle}, not equality.
  • Quantum mechanics: States are vectors in a Hilbert space; observables are self-adjoint operators; inner products give probability amplitudes.
  • Signal processing: L2(R)L^2(\mathbb{R}) is the space of finite-energy signals; orthonormal bases (Fourier, wavelet) enable efficient compression and denoising.
  • Machine learning: Kernel methods map data into a reproducing kernel Hilbert space (RKHS) where inner products correspond to kernel evaluations.
  • Numerical analysis: The Galerkin method projects PDE solutions onto finite-dimensional subspaces using orthogonal projection.

Problem 1. Let H=L2([0,1])H = L^2([0,1]) with inner product ⟨f,g⟩=∫01f(x)g(x)‾ dx\langle f,g\rangle = \int_0^1 f(x)\overline{g(x)}\,dx. Show that {e2πinx}n∈Z\{e^{2\pi i n x}\}_{n\in\mathbb{Z}} is an orthonormal system.

Solution. ⟨e2πinx,e2πimx⟩=∫01e2πi(n−m)x dx=δnm\langle e^{2\pi i n x}, e^{2\pi i m x}\rangle = \int_0^1 e^{2\pi i (n-m)x}\,dx = \delta_{nm} by orthogonality of complex exponentials. So the system is orthonormal. Completeness (that it forms a basis) is the statement of the Fourier series convergence theorem. ■\blacksquare

Problem 2. Let M={f∈L2([0,1]):∫01f(x) dx=0}M = \{f \in L^2([0,1]) : \int_0^1 f(x)\,dx = 0\}. Find M⊥M^\perp.

Solution. M⊥={g∈L2([0,1]):⟨f,g⟩=0 for all f∈M}M^\perp = \{g \in L^2([0,1]) : \langle f,g\rangle = 0 \text{ for all } f \in M\}. The constant function h(x)=ch(x) = c satisfies ⟨f,h⟩=c∫01f=0\langle f, h\rangle = c\int_0^1 f = 0 for all f∈Mf \in M. Conversely, if g∈M⊥g \in M^\perp, write g=gˉ+cg = \bar{g} + c where gˉ∈M\bar{g} \in M and c=∫01gc = \int_0^1 g. Then 0=⟨gˉ,g⟩=∥gˉ∥2+c∫gˉ=∥gˉ∥20 = \langle \bar{g}, g\rangle = \|\bar{g}\|^2 + c\int \bar{g} = \|\bar{g}\|^2, so gˉ=0\bar{g}=0 and gg is constant. Hence M⊥M^\perp is the 1-dimensional space of constant functions. ■\blacksquare