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Normed Spaces and Completeness

A norm is length for vectors, and length comes with a distance. We define normed spaces, see why completeness is the dividing line that names Banach spaces, and check completeness with the same Cauchy machinery you learned for the real line.

From length to distance

Functional analysis studies vector spaces big enough to hold functions, where a single point can be an entire signal or solution. To do analysis there we need a notion of size. A norm on a real or complex vector space V is a function ||·||: V -> [0, ∞) satisfying three axioms: ||x|| = 0 only when x = 0 (definiteness), ||λx|| = |λ| ||x|| (homogeneity), and ||x + y|| ≤ ||x|| + ||y|| (the triangle inequality). A space with such a norm is a normed vector space.

Every norm induces a metric by d(x, y) = ||x − y||. So a normed space is automatically a metric space, and every metric idea — open balls, convergence, Cauchy sequences — comes along for free. The added structure is that this metric is translation-invariant and scales linearly, which ties the geometry tightly to the linear algebra.

Completeness and Banach spaces

A sequence (x_n) is Cauchy if its terms eventually cluster: for every ε > 0 there is N with ||x_m − x_n|| < ε whenever m, n ≥ N. A normed space is complete if every Cauchy sequence converges to a point of the space. A complete normed space is exactly a Banach space — the central object of the subject. Completeness means there are no missing limits, just as the real line has no holes.

Not every normed space is complete. The classic counterexample is the polynomials, or the continuous functions on [0, 1] under the L^1 norm: a sequence can be Cauchy yet have a limit that escapes the space. Completing such a space — filling in the missing limits — is the analyst's standard move, and it is how spaces like L^p are born.

Claim: the sequence space l^1 = { x = (x_1, x_2, ...) : sum |x_k| < ∞ }
with norm ||x||_1 = sum_k |x_k| is complete.

Let (x^(n)) be Cauchy in l^1. Fix ε > 0; choose N with
    ||x^(m) - x^(n)||_1 < ε   for all m, n ≥ N.       (*)

Step 1 (find the candidate limit, coordinatewise).
For each fixed coordinate k,
    |x^(m)_k - x^(n)_k| ≤ ||x^(m) - x^(n)||_1 < ε,
so (x^(n)_k)_n is Cauchy in the scalars (which are complete).
Let x_k = lim_n x^(n)_k. Set x = (x_1, x_2, ...).

Step 2 (control finitely many coordinates, then let K grow).
From (*), for every finite K and all m, n ≥ N:
    sum_{k=1}^{K} |x^(m)_k - x^(n)_k| < ε.
Let m -> ∞ (each term converges):
    sum_{k=1}^{K} |x_k - x^(n)_k| ≤ ε   for all n ≥ N, all K.
Let K -> ∞:
    ||x - x^(n)||_1 ≤ ε   for all n ≥ N.          (**)

Step 3 (the limit lives in l^1).
x = x^(N) + (x - x^(N)); the first is in l^1, and by (**)
the second has norm ≤ ε < ∞, so x ∈ l^1.

Step 4 (conclude). (**) says x^(n) -> x in ||·||_1. So every
Cauchy sequence converges in l^1: l^1 is a Banach space.  ∎
The standard three-move proof of completeness: get a candidate limit coordinatewise, transfer the Cauchy bound to it, then confirm it lives in the space.