Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Composition satisfies |ST|\le|S|,|T|

Statement

Let X, Y, and Z be normed spaces over the same scalar field. If T∈B(X,Y) and S∈B(Y,Z), then

∥ST∥≤∥S∥ ∥T∥.

Facts & Assumptions

Given: Bounded linear operators T:X→Y and S:Y→Z.

[L1]

The operator norm is the unit-ball supremum and satisfies ∥Tu∥≤∥T∥ ∥u∥ for every vector u (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L2]

Bounded linear operators compose to a linear map, and B(X,Y) denotes the bounded ones (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

Proof

technique · direct
1.1

Let x∈X satisfy ∥x∥≤1. Then [L1] gives ∥Tx∥≤∥T∥, and applying [L1] again to S yields ∥STx∥≤∥S∥ ∥Tx∥≤∥S∥ ∥T∥.

L1L2
2.1

Step 1.1 holds for every x in the unit ball of X, so taking the supremum over that ball gives ∥ST∥≤∥S∥ ∥T∥.

step 1.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources