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July 28

"Root mean-n-power"

Let A be a (finite) sequence of b nonnegative real numbers with nth term denoted by a_n. Define $F(m)=\sum _{k=0}^b{a_k}^m$.

For m = 1, $\frac{F(1)}{b}$ is simply the ordinary mean, while $\sqrt{\frac{F(2)}{b}}$ is the root mean square. We know that $\underset{m\to \mathrm{\infty }}{\lim }(F(m)/n{)}^{1/m}=\max (A)$ where max(A) is the biggest number of A, as can be demonstrated by L'hôpital's rule, assuming there is a unique maximum (i.e. there are not two numbers for max(A)).

My question is then whether I can claim the same in the continuous case. Let f(x) be a continuous function on the positive real line that doesn't take negative values. Then, over an interval [c, d] somewhere on the positive real line with d > c, $G(m)=\underset{c}{\overset{d}{\int }}f(x{)}^{m}dx$; d could possibly be infinite, in which case it must be assumed that f vanishes sufficiently quickly for the integral to always converge. Unfortunately, my technique using L'hôpital's rule fails to resolve the question $\underset{m\to \mathrm{\infty }}{\lim }(G(m)/(d-c){)}^{1/m}$ because in a quotient of integrals (as I obtain) you cannot cancel factors in the integrands. I do know that integrals are defined as the limit of finite Riemann sums, each of which can have the discrete method for F applied, but I am also wary of the interchange of limiting operations.--Jasper Deng (talk) 19:16, 28 July 2015 (UTC)

See also Power mean inequality. --JBL (talk) 19:39, 28 July 2015 (UTC)

Thanks, I had been looking for the name of the discrete case. And it turns out that it's not even necessary for max(A) to be unique.--Jasper Deng (talk) 19:44, 28 July 2015 (UTC)

Since f is continuous on a closed interval it has a maximum M and some x for which $f(x)=M$ (I'll assume $x\in (c,d)$ but the proof is the same if x is c or d). For every $ϵ>0$ there is $\delta >0$ such that $|t-x|<\delta ⟹f(t)\ge M-ϵ$. Then $G(m)=\underset{c}{\overset{d}{\int }}f(t{)}^{m}dt\ge \underset{x-\delta }{\overset{x+\delta }{\int }}f(t{)}^{m}dt\ge \underset{x-\delta }{\overset{x+\delta }{\int }}(M-ϵ{)}^{m}dt=2\delta (M-ϵ{)}^m$. So $(G(m)/(d-c){)}^{1/m}\ge (M-ϵ)(2\delta /(d-c){)}^{1/m}$. Since the second term goes to 1 as $m\to \mathrm{\infty }$, $\underset{m\to \mathrm{\infty }}{\lim }(G(m)/(d-c){)}^{1/m}\ge M-ϵ$. This is true for every $ϵ$ so $\underset{m\to \mathrm{\infty }}{\lim }(G(m)/(d-c){)}^{1/m}\ge M$. It is obviously not greater, so it is equal to the maximum. -- Meni Rosenfeld (talk) 07:48, 29 July 2015 (UTC)

Thanks. What an elegant proof that didn't even rely on interchanging limits!--Jasper Deng (talk) 08:29, 29 July 2015 (UTC)

More generally, a similar proof will show that for any measurable function f on a finite measure space (X,μ), ${(\underset{X}{\int }|f{|}^{m}d\mu )}^{1/m}$ tends to the essential supremum of f as $m\to \mathrm{\infty }$. (It's also true for an integrable function, without the assumption that the measure space is finite.) Sławomir
Biały 11:22, 29 July 2015 (UTC)