I haven’t blogged here for quite a while, I hope this blog has not been abandoned. Anyway this is the first post since I have been in Australia. This is actually a note I prepared when I was in undergraduate year two, about the power mean inequality (I’ve forgotten about this note and found it in my computer recently). More precisely these are all the exercises in a chapter of a book whose name I’ve forgotten. The stuff is quite elementary but perhaps some of you are interested. If I have to make this note now, I think I will organize it in a different way, since it seems that this is organized quite loosely. But since I do not have plenty of time, I just do little formatting and put it here.
where is a vector of positive weights with total mass of and is a vector of positive real numbers. Unless otherwise stated, we will hereafter assume this condition for x and p. As we will see, there is a reasonable definition for the formula (1) at and at so that the map is continuous at .
Proposition 1 (The Geometric Mean as a Limit)
The geometric mean can be regarded as the limit of the general mean,
This quantity is defined to be , so that the map is continuous at .
Taking the limit into and then applying exponential, we can get the result.
By Cauchy’s inequality and the splitting trick , we have
Taking the -th root on both sides, we get (3). We deduce that for all ,
Taking the limit ,
Siegel’s doubling relation (3) and the following plot of the two-term power mean for
suggests the following inequality:
Theorem 3 (Power Mean Inequality) The mapping is a nondecreasing function on , i.e.
The equality holds if and only if
Proof: For , by Jensen’s inequality for the map with ,
With this inequality and the substitutions , gives
so taking -th root gives the desired inequality in this case. Moreover, the strict convexity of for ensures that the equality holds if and only if .
For , we have , then by the above,
For . As , by (2),
It is easy to show that We also have the continuity relations
For and all ,
As , as , so for all ,
By for and by the first continuity relation, the second continuity relation follows.
Proposition 4 (Harmonic Means and Recognizable Sums)
Suppose are positive and denotes their sum,
Proof: By AM-HM inequality,
Exercise 1 (Polya’s Minimax Characterization) Suppose you have to guess for an unknown x in the interval , it is possible to minimize the largest possible relative error of your guess. That is, for
there is a such that
Proof: Firstly, for a given , we can differentiate with respect to to see that only when , which is when the relative error is zero. Therefore for to attains its maximum, must be equal to or . Therefore
Let , which is the harmonic mean of and with equal weights. It is easy to see that is increasing for and is decreasing for . Therefore attains its minimum at .
Proposition 5 (The Geometric Mean as a Minimum) The geometric mean has the representation
where is the region of defined by
Proof: Assume for all , by AM-GM inequality,
The AM-GM inequality also tells us that the equality holds if and only if then and . So choosing , we obtain the result.
This result can be used to prove that geometric mean is superadditive, i.e.
Proposition 6 (More on the Method of Halves)
Proof: By ,
So we get the result.
Substitute in the identity and using the formula , we have the identity
Proposition 7 (A Niven-Zuckerman Lemma for -th Powers) Consider a sequence of -tuples of nonnegative real numbers , . Suppose there is a for which
and suppose for some ,
Then for all ,
Proof: By the power mean inequality, as , for all ,
By Proposition 3 we know that this equality holds if and only if
But as , so , i.e.