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: F = \frac{1}{(b-a)} \int_a^b f(x) dx.

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

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

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

Many thanks for any help.

 

[EnumaElish]

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

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

 

[deathprog23]

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

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

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

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

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

Thanks for the response!