Consider the increasing, concave function x^(0.5) on [0,1].

Alexia Avila

Alexia Avila

Answered question

2022-11-09

Consider the increasing, concave function x 0.5 on [ 0 , 1 ].

Answer & Explanation

hitturn35

hitturn35

Beginner2022-11-10Added 20 answers

Suppose such an f exists. Then for 0 < x < 1 the condition on f implies
f
This implies f ( x ) as x 0 + .. That violates the f > 0 condition, giving a contradiction.

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