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and hence 0 is pseudo-convex. Under assumption (ii) the first inequality<br />

in the above proof becomes an equality, and under assumptions (iii) or<br />

(iv) it remains an inequality.<br />

(Ill) Let assumption (i) hold, let x\ x 2 e T and let 0 < X < 1. Then<br />

0(x 2 ) < fl(x l ) «

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