Math 2.0

Difference and functional equations Definition

Functional inequality

Builds on FunctionFunctional equation

Definition

Let DD be a set, let ℱ\mathcal{F} be a set of functions f ⁣:D→ℝf \colon D \to \R, and let n≥1n \ge 1. Let S⊆DnS \subseteq D^n be a set. Let E1E_1 and E2E_2 be rules that assign a real number to each f∈ℱf \in \mathcal{F} and each (x1,…,xn)∈S(x_1, \ldots, x_n) \in S. The rules are built from ff, the variables and fixed given functions by composition and arithmetic operations. The functional inequality given by E1E_1 and E2E_2 is the condition E1(f;x1,…,xn)≤E2(f;x1,…,xn)for all (x1,…,xn)∈S,E_1(f; x_1, \ldots, x_n) \le E_2(f; x_1, \ldots, x_n) \quad \text{for all } (x_1, \ldots, x_n) \in S, and a solution is a function f∈ℱf \in \mathcal{F} that satisfies it.

Remarks

Example: the subadditivity inequality f(x+y)≤f(x)+f(y)f(x + y) \le f(x) + f(y) on D=[0,∞)D = [0, \infty), with S=D2S = D^2. The function f(x)=xf(x) = \sqrt{x} is a solution: both sides are nonnegative, and squaring gives x+y≤x+y+2xyx + y \le x + y + 2\sqrt{xy}.

Non-example: f(x)=x2f(x) = x^2 is not a solution of the same inequality, since f(1+1)=4>2=f(1)+f(1)f(1 + 1) = 4 > 2 = f(1) + f(1).

As with a functional equation, the class ℱ\mathcal{F} matters: one may ask for all solutions, or only for continuous, measurable or bounded ones. A system that mixes equations and inequalities is also common, for example a function that is additive and bounded above.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Functional_equation
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