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!