Prove it: Suppose f:R->R is a concave function and theta>1. Then theta^kf(x)>=f(theta^kx) for all k=1,2,….

Jack Ingram

Jack Ingram

Answered question

2022-10-22

Prove it:
Suppose f : R R is a concave function and θ > 1. Then
θ k f ( x ) f ( θ k x )
for all k = 1 , 2 ,

Answer & Explanation

giosgi5

giosgi5

Beginner2022-10-23Added 15 answers

What about f ( x ) = a x b for a , b > 0? Then θ k f ( x ) = θ k a x b θ k < θ k a x b = f ( θ k x )
Similarly g ( x ) = x 2 b for b > 0.
Then inequality θ k g ( x ) = θ k x 2 θ k b θ 2 k x 2 b = g ( θ k x ) is equivalent to
x 2 θ k ( θ k 1 ) θ k b b , which doesn't hold for every x.
The problem in both cases was the value at 0. Indeed, let f be concave. Then by concativity, for any λ ( 0 , 1 ), x , y R we have inequality:
f ( λ x + ( 1 λ ) y ) λ f ( x ) + ( 1 λ ) f ( y )
We want to prove a f ( x ) f ( a x ) for a > 1 , x R . Taking λ = 1 a ( 0 , 1 ), x = a t for some t R and y = 0 we get:
a f ( t ) f ( a t ) + ( a 1 ) f ( 0 )
As we see, with f ( 0 ) 0 it holds, when f ( 0 ) < 0 we have potential problems as showed in counterexamples above. I don't want to say that it does not hold for f concave such that f ( 0 ) < 0, because I do not have a proof (but I'll be glad to see one), but taking arbitrary concave f and looking at f M = f M, your inequality is equivalent to
a f M ( t ) f M ( a t )
so that
a f ( t ) a M f ( t ) M
and finally:
a M f ( t ) a 1 M f ( t ) 1
Taking arbitrary t R , a > 1, we can take M big enough so that it does not hold (since a > 1 and moreover, terms with 1 M tend to 0). In other words, we showed that if we substract enough from arbitrary concave function, then your inequality does not hold.

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