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FUNCAN · 1

Normed and Banach spaces

2 concepts · 2 questions

Norms, Banach spaces, the standard examples (ℓᵖ, C[a,b] with the sup norm, Lᵖ), and why infinite dimensions differ: closed balls are not compact (Riesz's lemma).

Definition 1.1Normed and Banach spaces

A normed space (X,∥⋅∥)(X, \|\cdot\|) is a vector space with a norm; it is a metric space via d(x,y)=∥x−y∥d(x, y) = \|x - y\|. A Banach space is a complete normed space. Examples: Kn\mathbb{K}^n (any norm); ℓp={(xn):∑∣xn∣p<∞}\ell^p = \{(x_n) : \sum|x_n|^p < \infty\}; ℓ∞\ell^\infty; C[a,b]C[a, b] with ∥⋅∥∞\|\cdot\|_\infty (complete); LpL^p (Riesz–Fischer). Non-example: C[0,1]C[0, 1] with ∥⋅∥1\|\cdot\|_1. Absolutely convergent series test: XX is Banach iff every series with ∑∥xn∥<∞\sum\|x_n\| < \infty converges.

Banach space: a complete normed vector space, the setting for analysis in infinite dimensions.

Analysis II already proved (C[a,b],∥⋅∥∞)(C[a,b], \|\cdot\|_\infty) is Banach and (C[0,1],d1)(C[0,1], d_1) isn't.

Theorem 1.2Finite vs infinite dimensions (Riesz's lemma)

(i) All norms on a finite-dimensional space are equivalent, and finite-dimensional subspaces are closed and complete. (ii) Riesz's lemma: if Y⊊XY\subsetneq X is a closed subspace, then for every θ<1\theta < 1 there is a unit vector xx with d(x,Y)≥θd(x, Y)\ge\theta. (iii) Hence the closed unit ball of XX is compact iff dim⁡X<∞\dim X < \infty.

Infinite dimensions are different: the closed unit ball is never compact. (In ℓ2\ell^2, the basis vectors ene_n are all 2\sqrt2 apart.)

So compactness arguments need replacing, e.g. with compact operators or weak topologies.

Proof