Lehmer mean

Let $p$ be a real number.  Lehmer mean of the positive numbers $a_1,\mathrm{\dots },a_n$ is defined as

$L_p(a_1,\mathrm{\dots },a_n):={\displaystyle \frac{a_1^p+\mathrm{\dots }+a_n^p}{a_1^{p-1}+\mathrm{\dots }+a_n^{p-1}}}.$(1)

This definition fulfils both requirements set for means (http://planetmath.org/Mean3).  In the case of Lehmer mean of two positive numbers $a$ and $b$ we see for  $a\leqq b$  that

$$a=\frac{a^p+a{b}^{p-1}}{a^{p-1}+b^{p-1}}\leqq \frac{a^p+b^p}{a^{p-1}+b^{p-1}}\leqq \frac{a^{p-1}b+b^p}{a^{p-1}+b^{p-1}}=b.$$

The Lehmer mean of certain numbers is the greater the greater is the parametre $p$, i.e.

$$L_p(a_1,\mathrm{\dots },a_n)\geqq L_q(a_1,\mathrm{\dots },a_n)\mathit{\hspace{1em}}\forall p>q.$$

This turns out from the nonnegativeness of the partial derivative of $L_p$ with respect to $p$; in the case  $n=2$  it writes

$$\frac{\partial L_p}{\partial p}=\frac{a^{p-1}{b}^{p-1}(a-b)(\ln a-\ln b)}{{(a^{p-1}+b^{p-1})}^2}\geqq \mathrm{\hspace{0.33em}0}.$$

Thus in the below list containing special cases of Lehmer mean, the is the least and the contraharmonic the greatest (cf. the comparison of Pythagorean means).

E.g. for two arguments $a$ and $b$, one has

• $L_0(a,b)={\displaystyle \frac{2ab}{a+b}}$,  harmonic mean,

• $L_{1/2}(a,b)=\sqrt{ab}$,  geometric mean,

• $L_1(a,b)={\displaystyle \frac{a+b}{2}}$,  arithmetic mean,

• $L_2(a,b)={\displaystyle \frac{a^2+b^2}{a+b}}$,  contraharmonic mean.

Note.  The least (http://planetmath.org/LeastNumber) and the greatest of the numbers (http://planetmath.org/GreatestNumber)$a_1,\mathrm{\dots },a_n$ may be regarded as borderline cases of the Lehmer mean, since

$$\underset{p\to -\mathrm{\infty }}{lim}{L}_p(a_1,\mathrm{\dots },a_n)=\min \{a_1,\mathrm{\dots },a_n\},\underset{p\to +\mathrm{\infty }}{lim}{L}_p(a_1,\mathrm{\dots },a_n)=\max \{a_1,\mathrm{\dots },a_n\}.$$

For proving these equations, suppose that there are exactly $k$ greatest (resp. least) ones among the numbers and that those are  $a_1=\mathrm{\dots }=a_k$.  Then we can write

$$L_p(a_1,\mathrm{\dots },a_n)=\frac{a_1^p\left[k+{\left(\frac{a_{k+1}}{a_1}\right)}^p+\mathrm{\dots }+{\left(\frac{a_n}{a_1}\right)}^p\right]}{a_1^{p-1}\left[k+{\left(\frac{a_{k+1}}{a_1}\right)}^{p-1}+\mathrm{\dots }+{\left(\frac{a_n}{a_1}\right)}^{p-1}\right]}.$$

Letting  $p\to +\mathrm{\infty }$  (resp. $p\to -\mathrm{\infty }$),  this equation yields

$$L_p(a_1,\mathrm{\dots },a_n)⟶a_1.$$