Average of Log of a Function: Bounded by 1 and Convex

[deathprog23]

Hello,

I am interested in the average behaviour of the log of a function.

I know the average of the function over the range of interest: [tex]F = \frac{1}{(b-a)} \int_a^b f(x) dx.[/tex]

I also know that [tex]f(x)[/tex] is convex and bounded from below by [tex]1.[/tex]

I want to know the average [tex]\frac{1}{(b-a)} \int_a^b \log( f(x) ) dx.[/tex]

In particular, under what circumstances this would be equal to the log of the average, [tex]\log(F)[/tex], up to a constant term, if [tex]F = \frac{1}{(b-a)}[/tex] and [tex](b-a)[/tex] tends to zero.

Many thanks for any help.

 

[EnumaElish]

My first observation is [itex]\lim_{b\rightarrow a}\int_a^b f(x) dx/{(b-a)} = f(a).[/itex]

Second, I am not sure how the numerator of F remains constant ([itex]\int_a^b f(x) dx[/itex] = 1) when b--->a. I expect [itex]\lim_{b\rightarrow a}\int_a^b f(x) dx[/itex] = 0. Can you explain?

 

[deathprog23]

Ah, sorry - I should have explicitly pointed out that [tex]f(x)=f(b-a,x).[/tex]

In fact, what I'm looking at is the average slope of a function [tex]g(x),[/tex] which has range [tex][0,1][/tex] and domain [tex][a,b].[/tex]

Thus [tex]f(x)=\frac{dg(x)}{dx}[/tex] and its integral over the domain must give [tex]1.[/tex]

The asymptotic properties must depend on the behaviour of [tex]f(x)[/tex] I suppose, e.g. if [tex]\limsup_{(b-a)\to 0}f(b)=K f(a)[/tex] for a constant [tex]K,[/tex] then what I ask for holds.

But what if more generally, as I ask, all I know is that [tex]f(x)\geq 1[/tex] and is convex? What about other classes of function?

Thanks for the response!