64 Classical Optimization Techniques
Figure 2.1 Relative and global minima.exists as a definite number, which we want to prove to be zero. Sincex∗is a relative
minimum, we have
f (x∗) ≤f(x∗+ h)for all values ofhsufficiently close to zero. Hence
f (x∗+ h)−f (x∗)
h≥0 ifh> 0f (x∗+ h)−f (x∗)
h≤0 ifh < 0Thus Eq. (2.1) gives the limit ashtends to zero through positive values asf′(x∗)≥ 0 (2.2)while it gives the limit ashtends to zero through negative values asf′(x∗)≤ 0 (2.3)The only way to satisfy both Eqs. (2.2) and (2.3) is to havef′(x∗)= 0 (2.4)This proves the theorem.
Notes:
1.This theorem can be proved even ifx∗is a relative maximum.
2 .The theorem does not say what happens if a minimum or maximum occurs at
a pointx∗where the derivative fails to exist. For example, in Fig. 2.2,lim
h→ 0f (x∗+ h)−f (x∗)
h=m+( ositivep )orm−( egativen )depending on whetherhapproaches zero through positive or negative values,
respectively. Unless the numbersm+andm−are equal, the derivativef′(x∗)
does not exist. Iff′(x∗) does not exist, the theorem is not applicable.