Math 2.0

Functional analysis Definition

Banach lattice

Builds on Banach spacePartial order

Definition

Let EE be a real Banach space with norm ∥⋅∥\norm{\cdot}, and let ≤\le be a partial order on EE. The pair is a Banach lattice if the following hold.

  1. If x≤yx \le y, then x+z≤y+zx + z \le y + z for all z∈Ez \in E.
  2. If x≤yx \le y and t≥0t \ge 0 is real, then tx≤tyt x \le t y.
  3. Every two elements x,yx, y have a least upper bound x∨yx \vee y in EE.
  4. If ∣x∣≤∣y∣\abs{x} \le \abs{y}, then ∥x∥≤∥y∥\norm{x} \le \norm{y}, where ∣x∣=x∨(−x)\abs{x} = x \vee (-x).

Remarks

Clauses 1 to 3 say that EE is a Riesz space, and clause 4 says that the norm is a lattice norm.

Example: C(K)C(K) for a compact Hausdorff space KK, with f≤gf \le g when f(t)≤g(t)f(t) \le g(t) for all t∈Kt \in K. Clauses 1 and 2 are clear, and f∨gf \vee g is the pointwise maximum, which is continuous. Then ∣f∣\abs{f} is the pointwise absolute value, and ∣f∣≤∣g∣\abs{f} \le \abs{g} gives sup⁡∣f∣≤sup⁡∣g∣\sup \abs{f} \le \sup \abs{g}.

Non-example: ℝ2\R^2 with the coordinatewise order and the norm ∥(a,b)∥=∣a∣+∣a−b∣\norm{(a, b)} = \abs{a} + \abs{a - b}. Clauses 1 to 3 hold, and this is a norm. For x=(1,0)x = (1, 0) and y=(1,1)y = (1, 1), one has ∣x∣≤∣y∣\abs{x} \le \abs{y}, but ∥x∥=2>1=∥y∥\norm{x} = 2 > 1 = \norm{y}, so clause 4 fails.

Reading path

To understand this page from the ground up, read these 3 pages in order. The foundations come first.

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  1. Banach space concept
  2. Partial order concept
  3. Banach lattice definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Banach_lattice
About this page
TagG4CA22W2XX
URIm2://G4CA22W2XX
Kinddefinition
WritingW0 stub
MSC46B42, 46A40
DefinesBanach lattice
UsesBanach space, Partial order
Changed2 November 2026

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