Math Labs

Analysis

Functional analysis

Studies infinite-dimensional vector spaces of functions and the linear operators acting on them.

IntuitionWhen can we do calculus with infinitely many coordinates?

A vector in Rn\mathbb{R}^n is just a list of nn numbers, and calculus on Rn\mathbb{R}^n works because we can measure length (∥x∥\|x\|) and angle (inner product) between such lists. A function like a sound wave, an image, or a quantum wavefunction is really an infinite list of numbers — its value at every point, or equivalently its infinitely many Fourier coefficients. Functional analysis asks: can we still measure "length" and "angle" between functions, and still do calculus (limits, derivatives, optimization) on the resulting infinite-dimensional space? The answer is yes, provided the space is complete — infinite sums and limits do not "leak out" of the space — which is exactly what turns a plain vector space of functions into a Banach space or, when an inner product is available, a Hilbert space.

Directed diagram of the continuous embeddings H¹ into L² and L² into distributions.
The diagram isolates the inclusion chain H1↪L2↪D′H^1\hookrightarrow L^2\hookrightarrow\mathcal D': the two arrows are continuous embeddings, and the focus control highlights one inclusion.

UndergraduateNorms, completeness, and inner products

Definition: Normed vector space and Banach space

A norm on a vector space XX over R\mathbb{R} or C\mathbb{C} is a map ∥x∥\|x\| satisfying ∥x∥\|x\|=0=0 iff x=0x=0, ∥λx∥=∣λ∣∥x∥\|\lambda x\|=|\lambda|\|x\|, and the triangle inequality ∥x+y∥≤∥x∥+∥y∥\|x+y\|\le\|x\|+\|y\|. A sequence (xn)(x_n) is a Cauchy sequence if its terms eventually get arbitrarily close together; XX is complete, and called a Banach space, if every Cauchy sequence in XX converges to some point of XX (not just of a larger space containing it). Completeness is what lets us build a limit function out of an infinite process — an infinite series of functions, an iterative approximation scheme — and be sure the limit is still a bona fide element of the space.

Written out, the Cauchy condition says the tail of the sequence gets tightly bunched together in norm:

∀ε>0 ∃N: n,m≥N⟹∥xn−xm∥<ε\forall\varepsilon>0\ \exists N:\ n,m\ge N \Longrightarrow \|x_n-x_m\|<\varepsilon

Here ε\varepsilon is the tolerance, NN is the point beyond which every pair of terms is within that tolerance, and ∥x∥\|x\| is the norm on XX. When XX additionally carries an inner product ⟨x,x⟩\langle x,x\rangle — a bilinear (or sesquilinear) pairing generalizing the dot product, from which the norm is recovered as ∥x∥=⟨x,x⟩\|x\|=\sqrt{\langle x,x\rangle} — and is complete in that induced norm, XX is called a Hilbert space, usually written HH. The inner product is what lets us speak of orthogonality and angle between functions, not just their length.

∣⟨x,y⟩∣≤∥x∥ ∥y∥|\langle x,y\rangle|\le\|x\|\,\|y\|

This is the Cauchy–Schwarz inequality: the inner product of two vectors can never exceed the product of their lengths, exactly like u⃗⋅v⃗=∥u⃗∥∥v⃗∥cos⁡θ\vec u\cdot\vec v=\|\vec u\|\|\vec v\|\cos\theta in ordinary 3D space, where ∣cos⁡θ∣≤1|\cos\theta|\le1. Hilbert spaces are further characterized among Banach spaces by the parallelogram law, ∥x+y∥2+∥x−y∥2=2∥x∥2+2∥y∥2\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2 — a purely norm-based identity (no inner product needed to state it) that a norm must satisfy if and only if it actually comes from an inner product. It says the sum of the squared diagonals of a parallelogram equals the sum of the squared sides, generalizing the Pythagorean-flavored geometry of Euclidean space to infinite dimensions.

Common Banach and Hilbert spaces
SpaceNormComplete?Hilbert space?
ℓ2\ell^2(∑i∣xi∣2)1/2\left(\sum_i|x_i|^2\right)^{1/2}YesYes
ℓp\ell^p, p≠2p\ne2(∑i∣xi∣p)1/p\left(\sum_i|x_i|^p\right)^{1/p}YesNo
C[0,1]C[0,1]sup⁡t∣f(t)∣\sup_{t}|f(t)|YesNo
L2[0,1]L^2[0,1](∫01∣f(t)∣2dt)1/2\left(\int_0^1|f(t)|^2dt\right)^{1/2}YesYes

UndergraduateTwo pillars: the Riesz representation theorem and the Hahn–Banach theorem

Let HH be a Hilbert space and φ\varphi a bounded (continuous) linear functional on HH. Then there is a unique y∈Hy\in H such that φ(x)=⟨x,y⟩\varphi(x)=\langle x,y\rangle for every x∈Hx\in H, and moreover ∥φ∥=∥y∥\|\varphi\|=\|y\|.

Why is it true?

It says every way of assigning a number to each vector linearly and continuously is secretly just "taking the inner product with some fixed vector" — abstract functionals are no more general than the most concrete ones you already know.

Proof

If φ=0\varphi=0, take y=0y=0 and we are done. Otherwise let N=ker⁡φ={x∈H:φ(x)=0}N=\ker\varphi=\{x\in H:\varphi(x)=0\}; since φ\varphi is continuous and linear, NN is a closed proper subspace of HH. Because HH is a Hilbert space, the projection theorem gives an orthogonal decomposition H=N⊕N⊥H=N\oplus N^{\perp}, and since NN is a proper subspace, N⊥N^{\perp} contains some z≠0z\ne0.

For any x∈Hx\in H, consider the vector u=φ(x)z−φ(z)xu=\varphi(x)z-\varphi(z)x. Applying φ\varphi gives φ(u)=φ(x)φ(z)−φ(z)φ(x)=0\varphi(u)=\varphi(x)\varphi(z)-\varphi(z)\varphi(x)=0, so u∈Nu\in N. Since z∈N⊥z\in N^{\perp}, we get ⟨u,z⟩=0\langle u,z\rangle=0, i.e. φ(x)⟨z,z⟩−φ(z)⟨x,z⟩=0\varphi(x)\langle z,z\rangle-\varphi(z)\langle x,z\rangle=0. Solving for φ(x)\varphi(x) gives φ(x)=φ(z)∥z∥2⟨x,z⟩=⟨x,φ(z)‾∥z∥2z⟩\varphi(x)=\dfrac{\varphi(z)}{\|z\|^2}\langle x,z\rangle=\left\langle x,\dfrac{\overline{\varphi(z)}}{\|z\|^2}z\right\rangle, so y=φ(z)‾∥z∥2zy=\dfrac{\overline{\varphi(z)}}{\|z\|^2}z represents φ\varphi.

For uniqueness, if ⟨x,y1⟩=⟨x,y2⟩\langle x,y_1\rangle=\langle x,y_2\rangle for all xx, take x=y1−y2x=y_1-y_2 to get ∥y1−y2∥2=0\|y_1-y_2\|^2=0, so y1=y2y_1=y_2. For the norm identity, Cauchy–Schwarz gives ∣φ(x)∣=∣⟨x,y⟩∣≤∥y∥∥x∥|\varphi(x)|=|\langle x,y\rangle|\le\|y\|\|x\|, so ∥φ∥≤∥y∥\|\varphi\|\le\|y\|; and plugging in x=yx=y gives φ(y)=∥y∥2\varphi(y)=\|y\|^2, so ∥φ∥≥∣φ(y)∣/∥y∥=∥y∥\|\varphi\|\ge|\varphi(y)|/\|y\|=\|y\|. Together, ∥φ∥=∥y∥\|\varphi\|=\|y\|.

Let XX be a real normed vector space, YY a linear subspace of XX, and φ\varphi a bounded linear functional on YY with ∣φ(y)∣≤M∥y∥|\varphi(y)|\le M\|y\| for all y∈Yy\in Y. Then there exists a bounded linear functional Φ\Phi on all of XX such that Φ(y)=φ(y)\Phi(y)=\varphi(y) for every y∈Yy\in Y, and ∣Φ(x)∣≤M∥x∥|\Phi(x)|\le M\|x\| for every x∈Xx\in X.

Why is it true?

It guarantees that a linear functional defined only on a small subspace — for instance, only knowing "the value of a signal at a few sample points" — can always be extended to the whole space without growing its bound; nothing ever forces us to be stuck working on a subspace.

Proof

First extend by one dimension. Pick x0∉Yx_0\notin Y and let Y1=Y⊕Rx0Y_1=Y\oplus\mathbb{R}x_0. We must choose the value c=Φ(x0)c=\Phi(x_0) so that ∣φ(y)+tc∣≤M∥y+tx0∥|\varphi(y)+tc|\le M\|y+tx_0\| for all y∈Y,t∈Ry\in Y,t\in\mathbb{R}; dividing by t≠0t\ne0 and substituting y/t→yy/t\to y, this reduces to needing cc with φ(y)−M∥y−x0∥≤c≤M∥y+x0∥−φ(y)\varphi(y)-M\|y-x_0\|\le c\le M\|y+x_0\|-\varphi(y) for all y∈Yy\in Y. Using φ(y1)−φ(y2)=φ(y1−y2)≤M∥y1−y2∥≤M∥y1+x0∥+M∥y2−x0∥\varphi(y_1)-\varphi(y_2)=\varphi(y_1-y_2)\le M\|y_1-y_2\|\le M\|y_1+x_0\|+M\|y_2-x_0\|, one checks the supremum of the left expressions over y1y_1 never exceeds the infimum of the right expressions over y2y_2, so a valid cc in that interval exists; this defines Φ\Phi on Y1Y_1 with the same bound MM.

Next, extend to all of XX using Zorn's Lemma. Consider the set of all pairs (Z,Ψ)(Z,\Psi) where ZZ is a subspace with Y⊆Z⊆XY\subseteq Z\subseteq X and Ψ\Psi extends φ\varphi to ZZ with bound MM, partially ordered by (Z1,Ψ1)≤(Z2,Ψ2)(Z_1,\Psi_1)\le(Z_2,\Psi_2) iff Z1⊆Z2Z_1\subseteq Z_2 and Ψ2∣Z1=Ψ1\Psi_2|_{Z_1}=\Psi_1. Every chain has an upper bound (take the union of the subspaces and the functional agreeing with each on its domain), so Zorn's Lemma gives a maximal element (Z∗,Φ)(Z^*,\Phi).

Finally, if Z∗≠XZ^*\ne X, the one-dimension extension step above applied to Z∗Z^* and any x0∈X∖Z∗x_0\in X\setminus Z^* would produce a strictly larger admissible pair, contradicting maximality of (Z∗,Φ)(Z^*,\Phi). Hence Z∗=XZ^*=X, and Φ\Phi is the desired bound-preserving extension to all of XX.

UndergraduateReal-World Applications and Worked Examples

Functional analysis is the mathematical backbone of Fourier analysis and signal processing (a signal lives in L2[0,2π]L^2[0,2\pi] or L2[0,1]L^2[0,1]), of quantum mechanics (states live in a Hilbert space HH), and of statistics, engineering, and machine learning problems that boil down to finding the closest point in a subspace — from least-squares regression to Tikhonov-regularized inverse problems.

Example: Signal processing: does sin⁡(nt)\sin(nt) settle down as n→∞n\to\infty?

Fix f∈L2[0,2π]f\in L^2[0,2\pi]. Show that ∫02πf(t)sin⁡(nt) dt→0\int_0^{2\pi} f(t)\sin(nt)\,dt\to 0 as n→∞n\to\infty (i.e. sin⁡(nt)⇀0\sin(nt)\rightharpoonup 0 weakly), while the norm ∥sin⁡(nt)∥2=π\|\sin(nt)\|_2=\sqrt{\pi} stays constant — so sin⁡(nt)\sin(nt) itself never settles down in the strong (norm) sense.

Solution

Consider the orthonormal system en(t)=sin⁡(nt)/πe_n(t)=\sin(nt)/\sqrt\pi in L2[0,2π]L^2[0,2\pi] (orthogonality follows from ∫02πsin⁡(nt)sin⁡(mt) dt=0\int_0^{2\pi}\sin(nt)\sin(mt)\,dt=0 for n≠mn\ne m, and ∫02πsin⁡2(nt) dt=π\int_0^{2\pi}\sin^2(nt)\,dt=\pi).

By Bessel's inequality, ∑n=1∞∣⟨f,en⟩∣2≤∥f∥2\sum_{n=1}^{\infty}|\langle f,e_n\rangle|^2\le\|f\|^2, the series of squared Fourier coefficients of ff against this orthonormal system converges, since ∥f∥2<∞\|f\|^2<\infty. A convergent series has terms tending to zero, so ∣⟨f,en⟩∣2→0|\langle f,e_n\rangle|^2\to0, i.e. ⟨f,en⟩→0\langle f,e_n\rangle\to0.

Since ⟨f,en⟩=1π∫02πf(t)sin⁡(nt) dt\langle f,e_n\rangle=\frac{1}{\sqrt\pi}\int_0^{2\pi}f(t)\sin(nt)\,dt, this is exactly ∫02πf(t)sin⁡(nt) dt→0\int_0^{2\pi} f(t)\sin(nt)\,dt\to 0 — this is the Riemann–Lebesgue lemma in disguise, and precisely the statement sin⁡(nt)⇀0\sin(nt)\rightharpoonup 0.

On the other hand, direct computation gives ∥sin⁡(nt)∥22=∫02πsin⁡2(nt) dt=π\|\sin(nt)\|_2^2=\int_0^{2\pi}\sin^2(nt)\,dt=\pi for every nn, so ∥sin⁡(nt)∥2=π\|\sin(nt)\|_2=\sqrt{\pi} for all nn — the norm never shrinks. So sin⁡(nt)\sin(nt) converges to 00 when tested against any fixed ff (weak convergence), but never converges to 00 in norm (no strong convergence): the oscillations get faster and "average out" against every fixed probe, without the signal itself losing energy.

Example: Statistics and engineering: least-squares regression as an orthogonal projection

Given data matrix AA (columns = predictors) and observations bb, the least-squares fit minimizes ∥Ax−b∥2\|Ax-b\|^2 over xx. Use the Hilbert-space projection idea (the finite-dimensional case of the Riesz/orthogonality argument above, with H=RnH=\mathbb{R}^n and the dot product) to derive the normal equations ATAx^=ATbA^{\mathsf T}A\hat x=A^{\mathsf T}b that the minimizer x^\hat x must satisfy.

Solution

The set of achievable outputs {Ax:x∈Rn}\{Ax:x\in\mathbb{R}^n\} is a subspace ran(A)⊆Rm\mathrm{ran}(A)\subseteq\mathbb{R}^m (a finite-dimensional, automatically complete, hence closed subspace of the Hilbert space Rm\mathbb{R}^m). Minimizing ∥Ax−b∥2\|Ax-b\|^2 is exactly the problem of finding the point of ran(A)\mathrm{ran}(A) closest to bb.

By the Hilbert-space projection theorem (the same orthogonal-decomposition idea used to prove the Riesz representation theorem above), the closest point Ax^A\hat x in a closed subspace to bb is characterized by the residual b−Ax^b-A\hat x being orthogonal to the entire subspace, i.e. ⟨b−Ax^, Av⟩=0  ∀v\langle b-A\hat x,\,Av\rangle=0\ \ \forall v.

Writing Av=A(v1,…,vn)Av=A(v_1,\dots,v_n) ranges over all columns of AA as vv ranges over the standard basis vectors, the condition ⟨b−Ax^, Av⟩=0  ∀v\langle b-A\hat x,\,Av\rangle=0\ \ \forall v applied to each column aja_j of AA says ⟨b−Ax^,aj⟩=0\langle b-A\hat x,a_j\rangle=0 for every jj, which packaged as a single matrix equation is exactly AT(b−Ax^)=0A^{\mathsf T}(b-A\hat x)=0.

Expanding gives ATb−ATAx^=0A^{\mathsf T}b-A^{\mathsf T}A\hat x=0, i.e. the normal equations ATAx^=ATbA^{\mathsf T}A\hat x=A^{\mathsf T}b. So the abstract Hilbert-space fact "the residual of a best approximation is orthogonal to the approximating subspace" is, in disguise, exactly the linear-algebra formula used every day to fit a regression line or a financial factor model.

ResearchOpen directions: the geometry of infinite-dimensional spaces

Which of the following Banach spaces is not a Hilbert space (its norm does not come from an inner product)?

In a Hilbert space, if ∥x∥=3\|x\|=3 and ∥y∥=4\|y\|=4, what is ∥x+y∥2+∥x−y∥2\|x+y\|^2+\|x-y\|^2?

What does the Riesz representation theorem guarantee about a bounded linear functional φ\varphi on a Hilbert space HH?

In the sin⁡(nt)\sin(nt) example, why does the signal sin⁡(nt)\sin(nt) "average out to zero" against every fixed probe ff as n→∞n\to\infty, even though it never loses energy?

References

  1. Walter Rudin (1991). Functional Analysis
  2. John B. Conway (2007). A Course in Functional Analysis
  3. Assaf Naor (2012). An introduction to the Ribe program · arXiv:1205.5993