Wikipedia:Reference desk/Archives/Mathematics/2015 July 28

= July 28 =

"Root mean-n-power"
Let A be a (finite) sequence of b nonnegative real numbers with nth term denoted by an. Define $$F(m)= \sum^b_{k = 0} {a_k}^m$$.

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 $$\lim_{m\to\infty} (F(m)/n)^{1/m} = \max(A) $$ 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, $$G(m) = \int_c^d f(x)^m dx $$; 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 $$\lim_{m\to\infty}(G(m)/(d - c))^{1/m}$$ 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 $$\epsilon>0$$ there is $$\delta>0$$ such that $$|t-x|<\delta\implies f(t)\ge M-\epsilon$$. Then $$G(m) = \int_c^df(t)^m\ dt \ge \int_{x-\delta}^{x+\delta}f(t)^m\ dt \ge \int_{x-\delta}^{x+\delta}(M-\epsilon)^m\ dt = 2\delta (M-\epsilon)^m$$. So $$(G(m)/(d-c))^{1/m} \ge (M-\epsilon)(2\delta/(d-c))^{1/m}$$. Since the second term goes to 1 as $$m\to\infty$$, $$\lim_{m\to\infty}(G(m)/(d-c))^{1/m} \ge M-\epsilon$$. This is true for every $$\epsilon$$ so $$\lim_{m\to\infty}(G(m)/(d-c))^{1/m} \ge M$$. 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,&mu;), $$\left(\int_X|f|^m\,d\mu\right)^{1/m}$$ tends to the essential supremum of f as $$m\to\infty$$. (It's also true for an integrable function, without the assumption that the measure space is finite.)  S ławomir  Biały  11:22, 29 July 2015 (UTC)