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Minkowski's inequality!

by A_I_
Tags: inequality, minkowski
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A_I_
#1
Jan29-08, 10:51 AM
P: 137
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 =)
Thanks,

Joe
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mrandersdk
#2
Jan29-08, 04:03 PM
P: 230
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>^

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)^ = (int (f^2+g^2+2fg)^
= (int f^2 + int g^2 + int 2fg)^ = (<f,f>+<g,g>+2<f,g>)^
= (|f|^2+|g|^2+2<f,g>)^

by cauchy-schwarz

<= (|f|^2+|g|^2+2|f||g|)^ = [(|f|+|g|)^2]^ = |f|+|g|

qed.
mrandersdk
#3
Jan29-08, 04:05 PM
P: 230
i forgot a ^ int the line

|f+g| = <f+g.f+g>

it should be

|f+g| = <f+g.f+g>^


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