Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a compact topological group (feel free to add hypotheses if necessary). Is there any mean value theorem for its (normalized to 1) Haar integral?

In general, are there mean value theorems for abstract spaces with measures? (Or at least for Borel measures?)

Later edit: After reading the first two comments, let me be more precise; I'm looking for a theorem giving something like: for any continuous $f$ on $G$, there exist $x \in G$ such that $\int_G f(g) \mathrm{d}g = f(x)$. Does such an $x$ really exist? Can anything else be said about it (the integral being so special, maybe this $x$ can be made more precise)?

share|improve this question
    
What is your substitute for the notion of an interval when you move outside the setting of R? Already for domains in R^2 one has to be a bit careful in finding a correct generalization of R^2? What notions have you tried thus far? –  Yemon Choi Jan 28 '13 at 16:20
    
Maybe it also helps if you indicate what do you need such a result for. –  András Bátkai Jan 28 '13 at 16:45

1 Answer 1

up vote 8 down vote accepted

Say $\mu$ is a Borel probability measure on a connected set $A$ in a topological space. Let $f : A \to \mathbb R$ be continuous. Then the mean value $\int_A f\;d\mu$ is equal to $f(a)$ for some $a \in A$. Proof: the mean value is between the sup of all values and the inf of all values, so (by connectedness) it is a value of the function.

Of course the desired result fails for non-connected sets. Even in the two-point group we get a counterexample.

share|improve this answer
    
Excellent answer, thank you. But please note that your proof works in general, while I require more: if there is an underlying group structure and the integral is translation-invariant, do I get any extra information about the points where the mean value is attained? One minor remark: one must have $\mu (A)<\infty$ in order for the above proof to work. –  Alex M. Jan 28 '13 at 20:40
    
One seems to need $\mu(A)=1$ in fact! Either that or Gerald forgot to divide by $\mu(A)$; both are trivial ways to fix things. –  user30035 Jan 28 '13 at 22:10
3  
$\mu(A)=1$ is implied: $\mu$ was a probability measure on $A$. –  Vince Jan 29 '13 at 1:32
    
How would one extend the proof presented by Gerald Edgar above to the case of complex-valued functions? Separating into the real and imaginary part is a wrong approach, since this would give two different points. –  Alex M. Feb 7 '13 at 17:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.