Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
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The quotient seminorm satisfies the triangle inequality

Statement

Let X be a normed space and let MX. Then for all x,yX,

(x+y)+MX/Mx+MX/M+y+MX/M.

Facts & Assumptions

Given: A normed space X, a linear subspace MX, vectors x,yX, and a real ε>0.

[L1]

The quotient seminorm is x+MX/M=infmMx+m (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L2]

The quotient seminorm is representative-independent, so (x+y)+(m1+m2)+M=(x+y)+M may be read with any m1,m2M (The quotient seminorm is independent of the chosen coset representative).

Proof

technique · direct
1.1

By [L1], choose m1,m2M such that x+m1<x+MX/M+ε/2 and y+m2<y+MX/M+ε/2.

L1choose
2.1

Since m1+m2M, [L2] lets us evaluate the quotient seminorm of (x+y)+M at the representative x+y+m1+m2. Hence (x+y)+MX/Mx+y+m1+m2x+m1+y+m2<x+MX/M+y+MX/M+ε by step 1.1.

step 1.1L1L2algebra
3.1

Since ε>0 was arbitrary, the displayed strict inequality of step 2.1 implies the stated triangle inequality.

step 2.1given

Depends on

Used by

Dependency tree · two levels

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