Consider g in C^1[0,1]. We say that g is nowhere convex (concave, resp.) on [0,1] if there is no open interval I sube[0,1] on which g is convex (concave, resp.) Is it possible to find a function g which satisfies: (strong verison) g is C^2[0,1] and g is nowhere convex and nowhere concave on [0,1]?

rhenan5v

rhenan5v

Answered question

2022-10-30

Consider g C 1 [ 0 , 1 ]. We say that g is nowhere convex (concave, resp.) on [ 0 , 1 ] if there is no open interval I [ 0 , 1 ] on which g is convex (concave, resp.) Is it possible to find a function g which satisfies:
g is C 2 [ 0 , 1 ] and g is nowhere convex and nowhere concave on [ 0 , 1 ]?

Answer & Explanation

n8ar1val

n8ar1val

Beginner2022-10-31Added 12 answers

The answer is no. If g ( a ) > 0 for some a [ 0 , 1 ] ,, then g > 0 in a neighborhood of a by the continuity of g .. Hence g is strictly convex in that neighborhood. Similarly, if g ( a ) < 0 for some a [ 0 , 1 ] ,, then g is strictly concave in that neighborhood. We're left with the case f 0.. But this implies f ( x ) = a x + b on [ 0 , 1 ] ,, hence f is both convex and concave everywhere on [ 0 , 1 ] ..

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