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 . Defne 2k−1 f (x)= ∑ |x − ai|, i=1 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 |x − ai|. i=1 k Figure 5.8: Subdifferential of g(x)= ∑2 1 |x − ai|. i=
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