147 Proof: Suppose f has an absolute minimum at x0. Then f (x0) ≤f (x) for all x ∈R. This implies 0· (x−x0) =0≤f (x)−f (x0) for all x ∈R. It follows from (5.8) that 0∈∂ f (x0). Conversely, if 0∈∂ f (x0), again, by (5.8), 0· (x−x0) =0≤f (x)−f (x0) for all x ∈R. Thus, f has an absolute minimum at x0. □ ■ Example 5.6.4 Let k be a positive integer and a1 <a2 <· · · <a2k−1. Define f (x) = 2k−1 ∑ i=1 | x−ai|, for x ∈R. It follows from the subdifferential formula in Example 5.6.3 that 0∈∂ f (x0) if and only if x0 =ak. Thus, f has a unique absolute minimum at ak. Figure 5.7: Subdifferential of f (x) =∑2k−1 i=1 |x−ai|. Figure 5.8: Subdifferential of g(x) =∑2k i=1|x−ai|.
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