3.142Clearly, if𝑓is convex, it is locally convex at everyx∈𝑆, where𝑆is the requiredneighborhood. To prove the converse, assume to the contrary that𝑓is locally convexat everyx∈𝑆but it is not globally convex. That is, there existsx1,x2∈𝑆such that𝑓(𝛼x1+ (1−𝛼)x2)> 𝛼𝑓(x1) + (1−𝛼)𝑓(x2)Letℎ(𝑡) =𝑓(𝑡x1+ (1−𝑡)x2)Local convexity implies that𝑓is continuous at everyx∈𝑆(Corollary 3.8.1), andtherefore continuous on𝑆.Therefore,ℎis continuous on [0,1].By the continuousmaximum theorem (Theorem 2.3),𝑇= argmaxx∈[x1,x2]ℎ(𝑡)is nonempty and compact. Let𝑡0= max𝑇. For every𝜖 >0,ℎ(𝑡0−𝜖)≤ℎ(𝑡0) andℎ(𝑡0+𝜖)< ℎ(𝑡0)Letx0=𝑡0x1+ (1−𝑡0)x2andx𝜖= (𝑡0+𝜖)x1+ (1−𝑡0−𝜖)x2Every neighborhood𝑉ofx0containsx−𝜖,x𝜖∈[x1,x2] with12𝑓(x−𝜖) +12𝑓(x𝜖) =12ℎ(𝑡0−𝜖) +12ℎ(𝑡0+𝜖)< ℎ(𝑡0) =𝑓(x0) =𝑓(12x−𝜖+12
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X1, Convex function, Michael Carter, Foundations of Mathematical Economics