*2.1. Deducing Monotonicity*

The monotonicity significantly helps in global optimization. If a function *f*(*x*) is monotonically nondecreasing on a segment [*a*, *<sup>b</sup>*], then min*x*∈[*a*,*b*] *<sup>f</sup>*(*x*) = *<sup>f</sup>*(*a*), max*x*∈[*a*,*b*] *<sup>f</sup>*(*x*) = *<sup>f</sup>*(*b*) and the segment [*a*, *b*] can be eliminated from further consideration after updating the record (best known solution so far). A similar statement is valid for a nonincreasing function. This techniques is known as the monotonicity test [26,41,42]. Moreover, as it is shown below, the monotonicity is crucial for evaluating the convexity/concavity of a composite function.

The usual way to ensure the monotonicity of a differentiable univariate function *f*(*x*) on an interval [*a*, *b*] is to compute an interval extension for its derivative [*c*, *d*] = **f** ([*a*, *b*]). If *c* ≥ 0, then the function is nondecreasing monotonic on [*a*, *b*]. Similarly, if *d* ≤ 0, then the function is nonincreasing monotonic on [*a*, *b*].

If a function is not differentiable, its monotonicity can still be evaluated using the rules described below. The Proposition 1 lists rules for evaluating an expression's monotonicity composed with the simple arithmetic operations.

**Proposition 1.** *The following rules hold:*


The proof of Proposition 1 is obvious. The rules for evaluating the monotonicity of the composition of functions are summarized in Proposition 2. The proof is intuitive and not presented here.

**Proposition 2.** *Let f*(*x*) *be a composition of univariate functions g*(*x*) *and h*(*x*)*: f*(*x*) = *g*(*h*(*x*))*. Then, the following four statements hold.*

*1. If h*(*x*) *on* [*a*, *b*]*, g*(*x*) *on* [*c*, *d*] *and* R*h*([*a*, *b*]) ⊆ [*c*, *d*] *then f*(*x*) *on* [*a*, *b*]*.*


The monotonicity of elementary univariate functions on a given interval can easily be established as these functions' behavior is well-known (Table 2).

**Table 2.** The monotonicity of elementary functions.


The monotonicity of a composite function defined by an arbitrary complex algebraic expression can be evaluated automatically using Propositions 1 and 2 and the data from the Tables 2. Let us consider an example.

**Example 1.** *Evaluate the monotonicity of the function f*(*x*) = max(3 − *x*, 1/ ln(*x* + 1))*, where x* ∈ [0.1, 3]*. This function is nonsmooth: it can be easily shown that f*(*x*) *does not have derivatives in two points on* [0.1, 3]*. Apply rules from Propositions 1 and 2:*

$$\begin{aligned} &\ge \nearrow \text{ on } [0.1, 3], 1 \nearrow \text{ on } [0.1, 3] \Rightarrow \ge +1 \nearrow \text{ on } [0.1, 3], \\ &\ge \ge +1 \nearrow \text{ on } [0.1, 3], \ln(\mathbf{x}) \nearrow \text{ on } [1.1, 4] \Rightarrow \ln(\mathbf{x} + 1) \nearrow \text{ on } [0.1, 3], \\ &\ln(\mathbf{x} + 1) \nearrow \text{ on } [0.1, 3], \ln(\mathbf{x} + 1) > 0 \newline \text{ on } [0.1, 3] \Rightarrow 1/\ln(\mathbf{x} + 1) \searrow \text{ on } [0.1, 3]. \end{aligned}$$

*Thus,* 1/ ln(*x* + 1) *is nonincreasing monotonic on* [0.1, 3]*. In the same way, it can be established that* 3 − *x is nonincreasing monotonic on* [0.1, 3]*. From the Proposition 1, it follows that f*(*x*) = max(3 − *x*, 1/ ln(*x* + 1)) *is also nonincreasing monotonic on* [0.1, 3]*.*

It is worth noting that the rules outlined above help to prove the monotonicity of nondifferentiable functions. However, for differentiable functions, the analysis of the the range of the first derivative is a better way to establish monotonicity. For example, a function *f*(*x*) = *e<sup>x</sup>* + sin(*x*) is monotonic on an interval [0, 2*π*]. Indeed, the range [0,*e*2*<sup>π</sup>* + 1] of its first derivative *f* (*x*) = *e<sup>x</sup>* + cos(*x*) computed by the natural interval expansion is non-negative. However, its monotonicty cannot be established by the outlined rules since sin(*x*) is not monotonic on [0, 2*π*]. The general recommendation is to compute the first derivative's range when the function is smooth and use Propositions 1 and 2 otherwise.

Monotonicity itself plays a vital role in optimization. The following obviously valid Proposition shows how the interval bounds can be computed for a monotonic function.

**Proposition 3.** *Let f*(*x*) *be a monotonic function on an interval* [*a*, *b*]*. Then*

$$\min\_{\mathbf{x}\in[a,b]} f(\mathbf{x}) = \min(f(a), f(b)),$$

$$\max\_{\mathbf{x}\in[a,b]} f(\mathbf{x}) = \max(f(a), f(b)).$$
