I got this question in my real analysis class and I have no idea how to start this...We learned this in metric spaces. I know what's Young's Inequality and Convexity mean. But for this question, I don't know why take the average of x when x has infinite many elements equals to the maximum of all the elements. Any thought?
http://d.yimg.com/hd/answers/i/cadea74cdbc14600871fdbf4f80d5e78_A.jpeg?a=answers&mr=0&x=1380121948&s=c1d27a07acb28460bec3dc27927d7576
http://d.yimg.com/hd/answers/i/cadea74cdbc14600871fdbf4f80d5e78_A.jpeg?a=answers&mr=0&x=1380121948&s=c1d27a07acb28460bec3dc27927d7576
-
Let x = (x1,...,xd) ∈ R^d. By definition of ||x||_∞, |xj| ≤ ||x||_∞ for each j. Thus, for every p > 1,
(||x||_p)^p = |x1|^p + ••• + |xd|^p ≤ (||x||_∞)^p + ••• + (||x||_∞)^p = d(||x||_∞)^p.
So ||x||_p ≤ d^(1/p) • ||x||_∞ for all p > 1. This implies lim sup ||x||_p ≤ ||x||_∞.
On the other hand, since ||x||_∞ = max{|x1|,...,|xd|}, for every e > 0, there exists k ∈ {1,...,d} such that ||x||_∞ < |xk| + e. Since |xk| ≤ ||x||_p for all p ≥ 1, we have ||x||_∞ < ||x||_p + e. Thus ||x||_∞ ≤ lim inf ||x||_p + e. Since e was arbitrary, ||x||_∞ ≤ lim inf ||x||_p.
In summary, we have lim sup ||x||_p ≤ ||x||_∞ ≤ lim inf ||x||_p. Therefore lim ||x||_p = ||x||_∞.
(||x||_p)^p = |x1|^p + ••• + |xd|^p ≤ (||x||_∞)^p + ••• + (||x||_∞)^p = d(||x||_∞)^p.
So ||x||_p ≤ d^(1/p) • ||x||_∞ for all p > 1. This implies lim sup ||x||_p ≤ ||x||_∞.
On the other hand, since ||x||_∞ = max{|x1|,...,|xd|}, for every e > 0, there exists k ∈ {1,...,d} such that ||x||_∞ < |xk| + e. Since |xk| ≤ ||x||_p for all p ≥ 1, we have ||x||_∞ < ||x||_p + e. Thus ||x||_∞ ≤ lim inf ||x||_p + e. Since e was arbitrary, ||x||_∞ ≤ lim inf ||x||_p.
In summary, we have lim sup ||x||_p ≤ ||x||_∞ ≤ lim inf ||x||_p. Therefore lim ||x||_p = ||x||_∞.