|Jan29-08, 10:51 AM||#1|
I need to proove the Minkowski's inequality for integrals.
I am taking a course in analysis.
[ int(f+g)^2 ] ^(1/2) =< [int(f^2)]^(1/2) + [int(g^2)]^(1/2)
now we are given that both f and g are Riemann integrable on the interval.
So by the properties of Riemann integrals, so is f^2,g^2 and fg.
We are also given a hint to expand the integral on the left and then use the Cauchy-Bunyakovsky-Schwarz inequality (now this i've already prooved in a previous exercice using the discriminant).
I was trying to expand the left side but i don't know what to do with the squared root, moreover i was trying to expand regardless the squared root and then at the end take a squared root but it still hasn't worked..
I need help =)
|Jan29-08, 04:03 PM||#2|
form what you wrote i assume you have an inner product given by
<f,g> = int fg dx
and the induced norm
|f| = <f,f>^½
so the Cauchy-Schwarz inequality is
|<f,g>| <= |f||g|
from what you get
<f,g> <= |f||g|
so starting with
|f+g| = <f+g.f+g> = (int (f+g)^2)^½ = (int (f^2+g^2+2fg)^½
= (int f^2 + int g^2 + int 2fg)^½ = (<f,f>+<g,g>+2<f,g>)^½
<= (|f|^2+|g|^2+2|f||g|)^½ = [(|f|+|g|)^2]^½ = |f|+|g|
|Jan29-08, 04:05 PM||#3|
i forgot a ^½ int the line
|f+g| = <f+g.f+g>
it should be
|f+g| = <f+g.f+g>^½
|Similar Threads for: Minkowski's inequality!|
|Proof this inequality using Chebyshev's sum inequality||Calculus & Beyond Homework||1|
|Minkowski's Inequality||Calculus & Beyond Homework||6|
|minkowski's 4 dimensional world||Special & General Relativity||30|
|Equality holds in Minkowski's Inequality when||Calculus||2|
|Minkowski's geometric||Special & General Relativity||1|