Inequality

[Camillo]

Let be : $p,q > 0 $ such that $ p+q = 1 $ .
Prove that exists $ c > 0 $ such that $AA x in [ -pi; pi]$ :

$ |pe^(iqx) + q e^(-ipx) | <= e^(-cx^2) $

[nicola de rosa]

[quote="Camillo"]Let be : $p,q > 0 $ such that $ p+q = 1 $ .
Prove that exists $ c > 0 $ such that $AA x in [ -pi; pi]$ :
avevo commesso un errore, perciò ho tolto la mia precedente soluzione

[Luca.Lussardi]

Ma lo scopo di questi post non era quello di rispondere in inglese?

[nicola de rosa]

Ma lo scopo di questi post non era quello di rispondere in inglese?

excuse me. I have edited

[Luca.Lussardi]

Ok, thanks.

[Camillo]

Let be : $p,q > 0 $ such that $ p+q = 1 $ .
Prove that exists $ c > 0 $ such that $AA x in [ -pi; pi]$ :
avevo commesso un errore, perciò ho tolto la mia precedente soluzione

Ok we wait for your updated solution.

[Thomas]

First of all, sorry for the big number of mistakes I'm going to write ...

As a firt step, it's easy to check that we can rewrite the first member this way:

$ (|pe^(iqx) + q e^(-ipx) |)^2 =(1-q)^2+q^2+2q(1-q)cosx$

with $0<q<1$

so the equation, elevating both members (the quantities are all positive):

$ (1-q)^2+q^2+2q(1-q)cosx<=e^(-(2cx^2))$

taking the logaritm of both sides we need to find $c/2$ that respects:

$(log( (1-q)^2+q^2+2q(1-q)cosx))/(-x^2)>=2c$

remembering that $x$ was in $[ -pi; pi]$, if we show that the function at the left member is limited, we have finished.

But this is true because it is obviously continuos in $[-pi,pi]\0$ and in $0$ the limit exists and is finished (with de l'hopital theorem it's also easy to evaluate it and the condition $0<q<1$ assures that this limit is positive)...
furthermore, the function is always positive.. we can so conclude that the minimum existes and is positive...


is it right???

[Camillo]

Are you sure that the function $log[(1-q)^2+q^2+2q(1-q)cosx]/(-x^2) $ has a minimum in $ x = 0 $ ?

Anyhow there is a quicker way using triangle inequality....

[Thomas]

I didn't say there is a minimum in 0, but that there is a minimum (positive) in $[-\pi,\pi]$... it can be in zero (anyway, in 0 the function is not defined but can be extended by continuity), but we don't care...

But I had to check that the limit in zero existed!! ... in fact Weiestrass theorem tells us that there is a minimum only if the function is continuos...

done that, I observede that the minimum is positive and this proved the existence of such a $c$...

[Camillo]

I didn't say there is a minimum in 0, but that there is a minimum (positive) in $[-\pi,\pi]$... it can be in zero (anyway, in 0 the function is not defined but can be extended by continuity), but we don't care...

But I had to check that the limit in zero existed!! ... in fact Weiestrass theorem tells us that there is a minimum only if the function is continuos...

done that, I observede that the minimum is positive and this proved the existence of such a $c$...

OK correct, I misunderstood your explanation .
Try also with the triangle inequality, it is very nice