Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Vector addition and scalar multiplication are continuous in a normed space

Statement

Let V be a normed space over R or C. Then

+:V×V→V,(x,y)↦x+y,

and

m:K×V→V,(λ,x)↦λx,

are continuous for the product topology on the domain and the norm topology on the target.

Facts & Assumptions

Given: Scalars λ,α∈K and vectors x,y,h,k∈V.

[L1]

The scalar field on this page is either R with its usual absolute value or C with its modulus, and the same norm estimates are valid in both cases (Real and complex scalar conventions for normed spaces).

[L2]

The norm is 1-Lipschitz for its metric, so in particular ∥u+v∥≤∥u∥+∥v∥ for all vectors and ∣∥u∥−∥v∥∣≤∥u−v∥ (The reverse triangle inequality in a normed space).

Proof

technique · direct
1.1L2

For addition, ∥(x+h)+(y+k)−(x+y)∥=∥h+k∥≤∥h∥+∥k∥ by [L2].

1.2L1L2algebra

For scalar multiplication, (λ+α)(x+h)−λx=λh+αx+αh, so ∥(λ+α)(x+h)−λx∥≤∣λ∣ ∥h∥+∣α∣ ∥x∥+∣α∣ ∥h∥ by the triangle inequality and absolute homogeneity from [L1] and [L2].

2.1step 1.1given

Given ε>0, step 1.1 shows that if ∥h∥<ε/2 and ∥k∥<ε/2, then (x+h)+(y+k) lies in the ε-ball around x+y; this is continuity of addition at (x,y).

2.2step 1.2givenalgebra

For scalar multiplication at (λ,x), first require ∣α∣<1 and then choose δ>0 so that ∣λ∣ δ+δ(∥x∥+δ)<ε; step 1.2 then gives ∥(λ+α)(x+h)−λx∥<ε whenever ∣α∣<δ and ∥h∥<δ.

3.1step 2.1step 2.2∎

Since the point (x,y) was arbitrary, step 2.1 proves continuity of addition everywhere, and step 2.2 proves continuity of scalar multiplication everywhere.

Depends on

Used by

Dependency tree · two levels

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Sources