Foreverlive

I have about 50kpp in FV server and would like to trade for the equivalent(value) here in Ayonae Ro. Please PM or reply if you are interested.

And of course, if you can solve a math problem of mine, I will offer give you 10kpp. Prove:

For real-valued functions f, and g,

Int_0^1 Int_0^1 |f(x)g(y) + f(y)g(x)|dxdy >= (Int_0^1 |f(x)|dx)(Int_0^1 |g(y)|dy).

where Int_a^b := Integral from a to b,
and a >= b means "a greater or equal to b."


Thanks.

Cyane300

I PM"d ya

Xapp

Assume f,g>0, integrable.

Claim: Int_0^1 Int_0^1 |f(x)g(y)+f(y)g(x)|dxdy > Int_0^1 |f(x)|dx * Int_0^1 |g(y)|dy.

But since f,g>0 absolute value notation is superfluous.

Then, Int_0^1 Int_0^1 [f(x)g(y)+f(y)g(x)] dxdy > Int_0^1 f(x)dx * Int_0^1 g(y)dy.

Since f and g are univariate functions with exogenous variables x and y we may pass the appropriate functions outside the integrals,

Int_0^1 [ g(y) Int_0^1 f(x) dx + f(y) Int_0^1 g(x) dx ] dy > Int_0^1 f(x)dx * Int_0^1 g(y)dy.

Using the same rule (since the integral of functions of x will not have any y variables we may pass them outside the integral w.r.t. y) we have,

Int_0^1 f(x)dx * Int_0^1 g(y)dy + Int_0^1 g(x)dx * Int_0^1 f(y)dy > Int_0^1 f(x)dx * Int_0^1 g(y)dy.

But there are identical terms on each side of this inequality. It reduces to:

Int_0^1 g(x) Int_0^1 f(y)dy > 0, which is true. Q.E.D.

Foreverlive

Hi Xapp,

Your answer works if we restrict f, g to be non-negative. When we don't make that restriction. Basically,

Int[Int(f(x)g(y)dx]dy = Int f(x)dx * Int g(y)dy >= 0.

Let me know if you have progress for the case f, g being arbitrary real valued function (riemann integrable can be assumed).

P.S. also post here if you are interested in what others thought of it so far.
http://forums.ayonae.ro/showthread.php?t=5243


Thanks.
Foreverlive.

Foreverlive

Also note that for real numbers a and b, we can not say

|a + b| = a + b so "distributing" the dx dy will cause trouble as well.

Palarran

Yeah, the problem is dealing with the regions where either 1 or 3 of f(x), f(y), g(x), and g(y) are negative, so that |f(x)g(y) + f(y)g(x)| < |f(x)g(y)|.

Foreverlive

Cy,

I PM'd back.


Thanks,
Foreverlive.

Nikium

Is the offer swapping PP from server to server still going?

Foreverlive

got my pp transfer already. thx though.

Cados Evilsbane

Ack! Not math, argh =(

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