Mathematics.

Functional Analysis

Banach Spaces

Real Analysis90 minDifficulty8 out of 10

You should know: metric spaces

Overview

A Banach space is a normed vector space that is complete with respect to the metric induced by its norm. Completeness means every Cauchy sequence of vectors converges to a limit within the space. Banach spaces are the natural setting for most of functional analysis: they include all finite-dimensional normed spaces, the sequence spaces \ell^p (1 \leq p \leq \infty), the function spaces L^p(\mu) for a measure \mu, and the space C([a,b]) of continuous functions with the sup-norm. The structure of Banach spaces underlies the three pillars of functional analysis: the Hahn-Banach theorem, the Open Mapping theorem, and the Uniform Boundedness Principle.

Intuition

A Banach space is a vector space equipped with a notion of length (a norm) that is 'complete' — you can take limits freely, and they always land back inside the space. This is the infinite-dimensional analogue of the completeness of \mathbb{R}: just as every Cauchy sequence of real numbers converges, in a Banach space every Cauchy sequence of vectors converges. Without completeness, limit operations in analysis are unreliable.

Formal Definition

Definition

A Banach space is a normed vector space (X, \|\cdot\|) over \mathbb{R} or \mathbb{C} that is complete:

(X,∥⋅∥) is a Banach space  ⟺  every Cauchy sequence in X converges in X.(X, \|\cdot\|) \text{ is a Banach space} \iff \text{every Cauchy sequence in } X \text{ converges in } X.
Definition
{xn} Cauchy:∀ε>0, ∃N, m,n≥N  ⟹  ∥xm−xn∥<ε.\{x_n\} \text{ Cauchy}: \quad \forall \varepsilon > 0,\ \exists N,\ m,n \geq N \implies \|x_m - x_n\| < \varepsilon.
Cauchy condition
ℓp={(xn):∑n=1∞∣xn∣p<∞},∥(xn)∥p=(∑∣xn∣p)1/p,1≤p<∞.\ell^p = \left\{ (x_n) : \sum_{n=1}^{\infty} |x_n|^p < \infty \right\}, \quad \|(x_n)\|_p = \left(\sum |x_n|^p\right)^{1/p}, \quad 1 \leq p < \infty.
Example: sequence space ell^p
Lp(μ)={f:∫∣f∣p dμ<∞},∥f∥p=(∫∣f∣p dμ)1/p.L^p(\mu) = \left\{ f : \int |f|^p\, d\mu < \infty \right\}, \quad \|f\|_p = \left(\int |f|^p\, d\mu\right)^{1/p}.
Example: L^p space

Properties

Bounded linear operators

AlinearmapT:X→Ybetweennormedspacesisboundediff∥T∥:=sup⁡∥x∥=1∥Tx∥<∞.A linear map T: X \to Y between normed spaces is bounded iff \|T\| := \sup_{\|x\|=1} \|Tx\| < \infty.

Dual space

ThedualX∗=B(X,F)ofallboundedlinearfunctionalsonXisitselfaBanachspace.The dual X^* = B(X, \mathbb{F}) of all bounded linear functionals on X is itself a Banach space.

Reflexivity

XisreflexiveifthecanonicalembeddingJ:X→X∗∗issurjective.EveryHilbertspaceandeveryLp(1<p<∞)isreflexive.X is reflexive if the canonical embedding J: X \to X^{**} is surjective. Every Hilbert space and every L^p (1 < p < \infty) is reflexive.

Worked Examples

  1. Let (f_n) be a Cauchy sequence in C([0,1]) with the sup-norm \|f\|_\infty = \sup_{x \in [0,1]} |f(x)|.

    ∥fn−fm∥∞→0 as m,n→∞.\|f_n - f_m\|_\infty \to 0 \text{ as } m,n \to \infty.

  2. For each fixed x, |f_n(x) - f_m(x)| \leq \|f_n - f_m\|_\infty \to 0, so (f_n(x)) is Cauchy in \mathbb{R} and converges to some f(x).

  3. The convergence f_n \to f is uniform: given \varepsilon > 0, choose N so \|f_n - f_m\|_\infty < \varepsilon for m,n \geq N. Taking m \to \infty gives |f_n(x) - f(x)| \leq \varepsilon for all x.

  4. A uniform limit of continuous functions is continuous, so f \in C([0,1]). Thus (C([0,1]), \|\cdot\|_\infty) is complete.

Answer: C([0,1]) with the sup-norm is a Banach space because uniform limits of continuous functions are continuous.

Practice Problems

Difficulty 6/10

Is (\mathbb{Q}, |\cdot|) a Banach space over \mathbb{Q}? Why or why not?

Difficulty 7/10

Show that \ell^\infty (bounded sequences with sup-norm) is a Banach space.

Difficulty 8/10

State the Uniform Boundedness Principle (Banach-Steinhaus theorem) and describe a key application.

Common Mistakes

Common Mistake

Every normed space is a Banach space.

Completeness is an additional requirement. The rationals \mathbb{Q} and C([0,1]) with the L^1-norm are normed but not Banach.

Common Mistake

A closed subspace of a Banach space need not be a Banach space.

Closed subspaces of a Banach space are always Banach (closed subsets of complete metric spaces are complete).

Quiz

A Banach space is a normed vector space that is:
Which of the following is NOT a Banach space?

Summary

  • A Banach space is a normed vector space that is complete: every Cauchy sequence converges.
  • Examples include C([a,b]) with sup-norm, \ell^p and L^p spaces (1 \leq p \leq \infty).
  • The dual space X^* of bounded linear functionals on a Banach space is itself a Banach space.
  • Key theorems — Hahn-Banach, Open Mapping, Closed Graph, Uniform Boundedness — all rely on completeness.
  • Hilbert spaces are the special Banach spaces where the norm comes from an inner product.

References

  1. BookRudin, W. — Functional Analysis (2nd ed.), McGraw-Hill, 1991.
  2. WebsiteWikipedia — Banach space