The power mean inequality and some related topics

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.

The quantities that provide the upper bound in AM-GM inequality are special cases of the general means

\displaystyle  M_t=M_t [\textbf{x};\textbf{p}]\equiv \{\sum_{k=1}^n p_k x_k{}^t \}^{1/t} \ \ \ \ \ (1)

where {\textbf{p}=(p_1, \ldots, p_n)} is a vector of positive weights with total mass of {p_1+\ldots +p_n=1} and {\textbf{x}=(x_1, \ldots, x_n)} 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 {t=\pm \infty} and at {t=0} so that the map {t\mapsto M_t} is continuous at {t=0}.

Proposition 1 (The Geometric Mean as a Limit)

The geometric mean can be regarded as the limit of the general mean,

\displaystyle  \lim_{t\rightarrow 0}\{\sum_{k=1}^n p_k x_k{}^t\}^{1/t}=\prod_{k=1}^n x_k{}^{p_k}.

This quantity is defined to be {M_0}, so that the map {t\mapsto M_t} is continuous at {t=0}.  

Proof: Consider

\displaystyle  \begin{array}{rcl}  \lim_{t\rightarrow 0}\log(M_t)&=&\lim_{t\rightarrow 0}\frac{\log(\sum_{k=1}^n p_k x_k{}^t)} {t}\\ &=& \lim_{t\rightarrow 0}\log(\sum_{k=1}^n p_k x_k{}^t \log(x_k)) \textrm{ \quad(L'Hospital's rule)}\\ &=& \log(\sum_{k=1}^n p_k \log(x_k)) \end{array}

Taking the limit into {\log} and then applying exponential, we can get the result. \Box

Proposition 2 (Power Mean Bound for the Geometric Mean)

\displaystyle  \prod_{k=1}^n x_k{}^{p_k} \leq \{\sum_{k=1}^n p_k x_k{}^t\}^{1/t} \textrm{ for all }t>0. \ \ \ \ \ (2)



We aim at the inequality (Siegel’s doubling relation)

\displaystyle  M_t\leq M_{2t} \textrm{ for all }t>0. \ \ \ \ \ (3)

By Cauchy’s inequality and the splitting trick {p_k x_k{}^t=p_k{}^\frac{1}{2}p_k{}^\frac{1}{2}x_k{}^t}, we have

\displaystyle  M_t{}^t=\sum_{k=1}^n p_k x_k{}^t =\sum_{k=1}^n p_k{}^\frac{1}{2}p_k{}^\frac{1}{2}x_k{}^t \leq (\sum_{k=1}^n p_k){}^\frac{1}{2} (\sum_{k=1}^n p_k x_k{}^{2t}){}^\frac{1}{2}=M_{2t}{}^t

Taking the {t}-th root on both sides, we get (3). We deduce that for all {t>0},

\displaystyle  M_{t/2^j} \leq M_{t/2^{j-1}} \leq \ldots \leq M_{t/2} \leq M_t

Taking the limit {j\rightarrow \infty},

\displaystyle \prod_{k=1}^n x_k{}^{p_k} =M_0=\lim_{j\rightarrow \infty}M_{t/2^j}\leq M_t\leq \{\sum_{k=1}^n p_k x_k{}^t\}^{1/t} \textrm{ for all }t>0.


Siegel’s doubling relation (3) and the following plot of the two-term power mean {(px^t+(1-p)y^t)^{1/t}} for { x=1, y=2}

suggests the following inequality:

Theorem 3 (Power Mean Inequality) The mapping {t\mapsto M_t} is a nondecreasing function on {\mathbb{R}}, i.e.

\displaystyle \{\sum_{k=1}^n p_k x_k{^s}\}^{1/s} \leq \{\sum_{k=1}^n p_k x_k{^t}\}^{1/t} \textrm{ for }-\infty<s<t<\infty

The equality holds if and only if {x_1=\ldots =x_n.}  

Proof: For {0<s<t}, by Jensen’s inequality for the map {x\mapsto x^p} with {p>1},

\displaystyle \{ \sum _{k=1}^n p_k x_k\}^p \leq \sum_{k=1}^n p_k x_k{}^p.

With this inequality and the substitutions {y_k=x_k{}^s}, {p=t/s>1} gives

\displaystyle  \{ \sum _{k=1}^n p_k y_k\}^{t/s} \leq \sum_{k=1}^n p_k y_k{}^{t/s},

so taking {t}-th root gives the desired inequality in this case. Moreover, the strict convexity of {x \mapsto x^p} for {p>1} ensures that the equality holds if and only if {x_1=\ldots=x_n}.
For {s<t<0}, we have {0<-t<-s}, then by the above,

\displaystyle  \begin{array}{rcl}  \{\sum_{k=1}^n p_k (x_k^{-1})^{-t}\}^{-1/t} &\leq& \{\sum_{k=1}^n p_k (x_k^{-1})^{-s}\}^{-1/s} \\ \Rightarrow \{\sum_{k=1}^n p_k x_k{}^s\}^{1/s} &\leq& \{\sum_{k=1}^n p_k x_k{}^t\}^{1/t} \end{array}

For {s<0<t}. As {-s>0}, by (2),

\displaystyle  M_0=\prod_{k=1}^n x_k{}^{p_k} \leq \{\sum_{k=1}^n p_k (x_k^{-1})^{-s}\}^{-1/s} \Rightarrow \{\sum_{k=1}^n p_k x_k{}^s\}^{1/s} \leq \prod_{k=1}^n x_k{}^{p_k}. \ \ \ \ \ (4)

As {t>0}, together with (2) gives the results. For the remaining cases {0=s<t} and {s<t=0}, they have been covered by (2) or (4) already.
Finally, as the figure suggests, we define

\displaystyle M_{-\infty}[\textbf{x}, \textbf{p}]=\min_k x_k \textrm{ and } M_\infty[\textbf{x}, \textbf{p}]=\max_k x_k

It is easy to show that {M_{-\infty}[\textbf{x}, \textbf{p}]\leq M_t[\textbf{x}, \textbf{p}]\leq M_\infty[\textbf{x}, \textbf{p}] \textrm{ for }t\in \mathbb{R}.} We also have the continuity relations

\displaystyle \lim_{t\rightarrow \infty}M_t[\textbf{x}, \textbf{p}]=M_\infty[\textbf{x}, \textbf{p}] \textrm{ and }\lim_{t\rightarrow -\infty}M_t[\textbf{x}, \textbf{p}]=M_{-\infty}[\textbf{x}, \textbf{p}]

For {t>0} and all {1\leq k \leq n},

\displaystyle  p_k x_k{}^t \leq M_t{}^t \leq M_\infty{}^t, \text{ so } p_k{}^{\frac{1}{t}} x_k \leq M_t \leq M_\infty

As {p_k>0}, {p_k{}^{1/t}\rightarrow 1} as {t \rightarrow \infty}, so for all {k},

\displaystyle x_k \leq \lim_{t\rightarrow \infty}M_t \leq M_\infty


\displaystyle  \max_k x_k\leq \lim_{t\rightarrow \infty}M_t \leq M_\infty


\displaystyle \lim_{t\rightarrow \infty}M_t=M_\infty \textrm{ as }\max_k x_k=M_\infty.

By {M_{-t}(x_1, \ldots, x_n; \textbf{p})=M_t{}^{-1}(1/x_1, \ldots, 1/x_n; \textbf{p})} for {t<0} and by the first continuity relation, the second continuity relation follows. \Box

Proposition 4 (Harmonic Means and Recognizable Sums)

Suppose {x_1, \ldots, x_n} are positive and {S} denotes their sum,

\displaystyle \frac{n^2}{2n-1}\leq \frac{S}{2S-x_1}+\ldots +\frac{S}{2S-x_n}


Proof: By AM-HM inequality,

\displaystyle  \begin{array}{rcl}  \sum_{k=1}^n \frac{S}{2S-x_k}&\geq& n^2/(\frac{\sum_{k=1}^n (2S-x_k)}{S})\\ &=& n^2/(\frac{2nS-S}{S})\\ &=& \frac{n^2}{2n-1}. \end{array}


Exercise 1 (Polya’s Minimax Characterization) Suppose you have to guess for an unknown x in the interval {[a, b]\subset (0, \infty)}, it is possible to minimize the largest possible relative error of your guess. That is, for

\displaystyle F(p)=\max_{x\in[a, b]}\{\frac{|p-x|}{x}\},

there is a {p^*} such that

\displaystyle F(p^*)=\min_p F(p)=\min_p \max_{x\in[a, b]}\{\frac{|p-x|}{x}\}.


Proof: Firstly, for a given {p}, we can differentiate {g(x)=\frac{(p-x)^2}{x^2}} with respect to {x} to see that {g'(x)=0} only when {x=p}, which is when the relative error is zero. Therefore for {\frac{|p-x|}{x}} to attains its maximum, {x} must be equal to {a} or {b}. Therefore

\displaystyle F(p)=\max_{x\in[a, b]}\{\frac{|p-x|}{x}\}=\max\{\frac{p-a}{a}, \frac{b-p}{b}\}.


\displaystyle  \frac{p-a}{a}\geq \frac{b-p}{b} \Leftrightarrow p\geq \frac{2ab}{a+b}\in [a, b] \textrm{ and }\frac{p-a}{a}\leq \frac{b-p}{b} \Leftrightarrow p\leq \frac{2ab}{a+b}\in [a, b].

Let {H=\frac{2ab}{a+b}}, which is the harmonic mean of {a} and {b} with equal weights. It is easy to see that {\frac{p-a}{a}} is increasing for {p\geq H} and {\frac{b-p}{b}} is decreasing for {p\leq H}. Therefore {F(p)} attains its minimum at {p=H=\frac{2ab}{a+b}}.

Proposition 5 (The Geometric Mean as a Minimum) The geometric mean has the representation

\displaystyle \{ \prod_{k=1}^n a_k \}^{1/n}=\min\{\frac{1}{n} \sum_{k=1}^n a_k x_k:(x_1,\ldots, x_n)\in D \},

where {D} is the region of {\mathbb{R}^n} defined by

\displaystyle D=\{(x_1, \ldots, x_n):\prod_{k=1}^n x_k=1, x_k\geq 0, k=1, \ldots , n\}.


Proof: Assume {a_k\neq 0} for all {k}, by AM-GM inequality,

\displaystyle \frac{1}{n} \sum_{k=1}^n a_k x_k\geq (\prod_{k=1}^n a_k x_k)^{1/n}=(\prod_{k=1}^n a_k)^{1/n} \textrm{ as } (x_1, \ldots, x_n)\in D.

The AM-GM inequality also tells us that the equality holds if and only if {a_1x_1=\ldots =a_nx_n=c } then {x_k=c/a_k} and {1=c^n/(a_1 a_2\ldots a_n)\Rightarrow c=(a_1 a_2\ldots a_n)^{1/n}}. So choosing {x_k=\frac{(a_1 a_2\ldots a_n)^{1/n}}{a_k}}, we obtain the result. \Box

This result can be used to prove that geometric mean is superadditive, i.e.

\displaystyle  \begin{array}{rcl}  & &\{ \prod_{k=1}^n a_k \}^{1/n}+\{ \prod_{k=1}^n b_k \}^{1/n}\\ &=&\min\{\frac{1}{n} \sum_{k=1}^n a_k x_k:(x_1,\ldots, x_n)\in D \}+\min\{\frac{1}{n} \sum_{k=1}^n b_k x_k:(x_1,\ldots, x_n)\in D \}\\ &\leq& \min\{\frac{1}{n} \sum_{k=1}^n (a_k+b_k) x_k:(x_1,\ldots, x_n)\in D \}\\ &=& \{ \prod_{k=1}^n (a_k+b_k) \}^{1/n}. \end{array}

Proposition 6 (More on the Method of Halves)

\displaystyle \frac{\sin x}{x}=\prod_{k=1}^\infty \cos(x/2^k).


Proof: By {\sin(x)=2\sin(x/2)\cos(x/2)},

\displaystyle  \frac{\sin x}{2 \sin(x/2)}\cdot\frac{\sin(x/2)}{2\sin(x/2^2)}\ldots \frac{\sin(x/2^{n-1})}{2\sin(x/2^n)}=\frac{\sin(x)}{2^n \sin(x/2^n)}=\prod_{k=1}^n \cos(x/2^k)


\displaystyle \prod_{k=1}^\infty \cos(x/2^k)=\lim_{n\rightarrow\infty} \frac{\sin(x)}{2^n \sin(x/2^n)}


\displaystyle \lim_{n\rightarrow \infty }\frac{\sin(x/2^n)}{1/2^n}=x.

So we get the result. \Box

Substitute {x=\pi/2} in the identity and using the formula {\cos(x/2)=\sqrt{\frac{1+\cos x}{2}}}, we have the identity

\displaystyle \frac{2}{\pi}=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\ldots

Proposition 7 (A Niven-Zuckerman Lemma for {p}-th Powers) Consider a sequence of {n}-tuples of nonnegative real numbers {(a_{1k}, a_{2k}, \cdots, a_{nk})}, {k=1, 2, \ldots}. Suppose there is a {\mu \geq 0} for which

\displaystyle a_{1k}+a_{2k}+\cdots+a_{nk}\rightarrow n\mu \textrm{ as }k\rightarrow \infty,

and suppose for some {1<p<\infty},

\displaystyle a_{1k}{}^p+a_{2k}{}^p+\cdots+a_{nk}{}^p\rightarrow n\mu^p \textrm{ as }k\rightarrow \infty.

Then for all {1 \leq j \leq n},

\displaystyle \lim_{k\rightarrow \infty}a_{jk}=\mu.


Proof: By the power mean inequality, as {p>1}, for all {k},

\displaystyle  (\sum_{i=1}^n \frac{a_{ik}{}^p}{n})^{1/p}\geq \sum_{i=1}^n \frac{a_{ik}}{n}.


\displaystyle \mu=\lim_{k\rightarrow \infty}(\sum_{i=1}^n \frac{a_{ik}{}^p}{n})^{1/p}\geq \lim_{k\rightarrow \infty}\sum_{i=1}^n \frac{a_{ik}}{n}=\mu.


\displaystyle (\sum_{i=1}^n \frac{(\lim_{k\rightarrow \infty}a_{ik})^p}{n})^{1/p}=\sum_{i=1}^n \frac{\lim_{k\rightarrow \infty}a_{ik}}{n}=\mu.

By Proposition 3 we know that this equality holds if and only if

\displaystyle \lim_{k\rightarrow \infty}a_{ik}=c \textrm{ for all }i.

But {a_{1k}+\cdots+a_{nk}\rightarrow n\mu} as {k \rightarrow \infty}, so {c=\mu}, i.e.

\displaystyle \lim_{k\rightarrow \infty} a_{ik}=\mu \textrm{ for all }i.


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