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Minkowski's inequality! 
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#1
Jan2908, 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 CauchyBunyakovskySchwarz 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 


#2
Jan2908, 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>^½ so the CauchySchwarz inequality is <f,g> <= fg from what you get <f,g> <= fg 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 by cauchyschwarz <= (f^2+g^2+2fg)^½ = [(f+g)^2]^½ = f+g qed. 


#3
Jan2908, 04:05 PM

P: 230

i forgot a ^½ int the line
f+g = <f+g.f+g> it should be f+g = <f+g.f+g>^½ 


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