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Refers to Fourier Analysis by Stein

# Vector spaces and inner products

Definition 3.1.1 A vector space VV over the complex number C\mathbb C is a set where for all X,Y,Z∈V,∀λ,μ∈CX,Y,Z\in V,\ \forall \lambda,\mu \in\mathbb C, always X+Y,λX∈VX+Y, \lambda X\in V, and

  1. X+Y=Y+XX+Y=Y+X
  2. X+(Y+Z)=(X+Y)+ZX+(Y+Z)=(X+Y)+Z
  3. 0+X=X+0=X0+X=X+ 0=X
  4. ∃(−X)∈V,(−X)+X=X+(−X)=0\exists (-X)\in V,\ (-X)+X=X+(-X)= 0
  5. λ⋅(μ⋅X)=(λ⋅μ)⋅X\lambda \cdot(\mu \cdot X)=(\lambda \cdot \mu)\cdot X
  6. λ⋅(X+Y)=λ⋅X+λ⋅Y\lambda\cdot(X+Y)=\lambda \cdot X+\lambda \cdot Y
  7. (λ+μ)⋅X=λ⋅X+μ⋅X(\lambda +\mu)\cdot X=\lambda \cdot X+\mu \cdot X
  8. 1⋅X=X1\cdot X=X

The vector space over the real number is similar.

Definition 3.1.2 A inner product on a vector space V over the complex number (⋅,⋅)(\cdot , \cdot ) satisfies for all X,Y,Z∈V,λ,μ∈CX,Y,Z\in V, \lambda,\mu \in \mathbb C

  1. (X,Y)=\overline
  2. (λX+μY,Z)=λ(X,Y)+μ(Y,Z)(\lambda X+\mu Y,Z)=\lambda (X,Y)+\mu(Y,Z)
  3. (X,X)≥0(X,X)\geq 0, and (X,X)=0(X,X)=0 if and only if X=0X=0

These inner products are called Hermitian because of 1, and they are linear in the first variable but conjugate-linear in the second. Further more, we definite norm

∣∣X∣∣=(X,X)1/2||X||=(X,X)^{1/2}

if ∣∣X∣∣=0||X||=0 implies X=0X=0, then we say that the inner product is strictly positive-definite.

The inner product on a vector space VV over the real number is similar, where we only need to replace 1 to symmetric.

Example of inner product and norm

The inner product of two vectors Z=(z1,...,zd),W=(w1,...,wd)Z=(z_1,...,z_d),W=(w_1,...,w_d) in Cd\mathbb C^d is defined by

(Z,W)=∑k=1dzkwk‾(Z,W)=\sum^d_{k=1}z_k\overline{w_k}

while the norm of the vector ZZ is given by

∣∣Z∣∣=(Z,Z)1/2=∑k=1d∣zk∣2||Z||=(Z,Z)^{1/2}=\sqrt{\sum^d_{k=1}|z_k|^2}

Definition 3.1.3 Let VV be a vector space (over R\mathbb R or C\mathbb C) with inner product (⋅,⋅)(\cdot ,\cdot ) and associated norm ∣∣⋅∣∣||\cdot ||. We say X,Y∈VX,Y\in V are orthogonal if

(X,Y)=0(X,Y)=0

and we write X⊥YX\perp Y.

Property 3.1.4

  1. The pythagorean theorem: if X,YX,Y are orthogonal, then

∣∣X+Y∣∣2=∣∣X∣∣2+∣∣Y∣∣2||X+Y||^2=||X||^2+||Y||^2

  1. The Cauchy-Schwarz inequality: for all X,Y∈VX,Y \in V

∣(X,Y)∣≤∣∣X∣∣⋅∣∣Y∣∣|(X,Y)|\leq ||X||\cdot ||Y||

  1. The triangle inequality: for all X,Y∈VX,Y\in V

∣∣X+Y∣∣≤∣∣X∣∣+∣∣Y∣∣||X+Y||\leq ||X||+||Y||

# Hilbert space

We need to work with two infinite-dimensional vector spaces for Fourier series.

Definition 3.1.5 An inner product space with these two properties is called a Hilbert space

  1. the inne product is strictly positive-definite

∣∣X∣∣=0⟹X=0||X||=0\implies X=0

  1. the vector space VV is complete, which means for every Cauchy sequence {Xn}\{X_n\} in the norm

lim⁡n→∞Xn∈V\lim_{n\to \infty}X_n\in V

If either of the conditions above fail, the space is called a pre-Hilbert space.

Example of Hilbert space

We see that Rd,Cd\mathbb R^d,\mathbb C^d are examples of finite-dimensional Hilbert spaces, while l2(Z)l^2(\mathbb Z) is an example of an infinite-dimensional Hilbert space, which is over C\mathbb C the set of all (two-sided) infinite sequences of complex numbers

(...,a−n,...,a−1,a0,a1,...,an,...)(...,a_{-n},...,a_{-1},a_0,a_1,...,a_n,...)

such that

∑n∈Z∣an∣2<∞\sum_{n\in\mathbb Z}|a_n|^2<\infty

it is easy to tell this is a vector space. And it also fits Property 3.1.4.

Example of pre-Hilbert space

The set R={f∈R(S1)}\mathcal R=\{f\in\mathcal R(S^1)\} is a vector space over C\mathbb C with addtion

(f+g)(θ)=f(θ)+g(θ)(f+g)(\theta)=f(\theta)+g(\theta)

and multiplication by a scalar λ∈C\lambda \in\mathbb C

(λf)(θ)=λ⋅f(θ)(\lambda f)(\theta)=\lambda\cdot f(\theta)

and the inner product

(f,g)=12π∫−ππf(θ)g(θ)‾dθ(f,g)=\dfrac1{2\pi}\int^\pi_{-\pi}f(\theta)\overline{g(\theta)}\mathrm d\theta

and the norm of ff

∣∣f∣∣=(12π∫−ππ∣f(θ)∣2dθ)1/2||f||=\left(\dfrac1{2\pi}\int^\pi_{-\pi}|f(\theta)|^2\mathrm d\theta \right)^{1/2}

Incidentally, it is not a complete vector space because of the example

fn(θ)={0for∣θ∣≤1/nf(θ)for1/n<∣θ∣≤πf_n(\theta)=\left\{\begin{array}{ll}0&\text{for }|\theta|\leq 1/n\\ f(\theta)& \text{for } 1/n<|\theta|\leq \pi \end{array}\right.

where f(θ)=ln⁡(1/θ)f(\theta )=\ln (1/\theta) for 0<∣θ∣≤π0<|\theta|\leq \pi while 00 on the origin.