Holder's inequality for integrals

[brydustin]

Does anyone know a simple proof for holder's inequality?

I would be more interested in seeing the case of
|∫fg|≤ sqrt(∫f^2)*sqrt(∫g^2)

 

[DonAntonio]

Does anyone know a simple proof for holder's inequality?

I would be more interested in seeing the case of
|∫fg|≤ sqrt(∫f^2)*sqrt(∫g^2)

Be sure you can prove the following:

1) For any x,y\in\mathbb R\,\,,\,\,xy\leq\frac{1}{2}(x^2+y^2)

2) Now put \frac{f}{\sqrt{\int f^2}}:= x\,,\,\,\frac{g}{\sqrt{\int g^2 dx}}:=y\,\, in the above, integrate both sides and voila!, there you have your proof.

DonAntonio

 

[brydustin]

Be sure you can prove the following:

1) For any x,y\in\mathbb R\,\,,\,\,xy\leq\frac{1}{2}(x^2+y^2)

2) Now put \frac{f}{\sqrt{\int f^2}}:= x\,,\,\,\frac{g}{\sqrt{\int g^2 dx}}:=y\,\, in the above, integrate both sides and voila!, there you have your proof.

DonAntonio

Well the first part makes sense, I don't see why the second step follows.

 

[brydustin]

Also I don't see the first part either...

If max(x,y) = x.
Then xy <= x^2.
and y^2 <= xy
But the other part doesn't follow. I think your proof is lacking...

 

[micromass]

Think about (x-y)^2 and (x+y)^2.

 

[brydustin]

Think about (x+y)^2.

Actually I see the argument xy ≤ 1/2 (x^2 + y^2)
because if we start with x=y then the result is equality.
Now if we let min(x,y) = x. Such that x+ε=y as ε→0,
then xy doesn't decrease as fast as x^2. I.e. (y-ε)y < (y-ε)^2 = x^2
Okay, so the first part makes since, but I still don't see Holder's inequality.

 

[micromass]

What do you get if you fill in the x and y that DonAntonio suggested?

 

[DonAntonio]

Also I don't see the first part either...

If max(x,y) = x.
Then xy <= x^2.
and y^2 <= xy
But the other part doesn't follow. I think your proof is lacking...

xy\leq \frac{1}{2}(x^2+y^2)\Longleftrightarrow 0 \leq (x-y)^2 . I think something different is lacking...;)

DonAntonio

 

[brydustin]

What do you get if you fill in the x and y that DonAntonio suggested?

For the first part: Let x + ε= y. : ε≥0
y^2 - εy ≤ y^2 - εy + 2ε^2
y^2 - εy ≤ .5(2^2 - 2yε+ε^2)
(y-ε)y = xy ≤ .5[ (y-ε)^2 + y^2] = .5(x^2 + y^2)

The furthest I get for the second part is:
fg/[ (sqrt(∫g^2)*sqrt(∫f^2) ] ≤ .5 * [ (f^2/∫f^2) + (g^2/∫g^2) ]
Sorry, I don't see it.

 

[micromass]

For the first part: Let x + ε= y. : ε≥0
y^2 - εy ≤ y^2 - εy + 2ε^2
y^2 - εy ≤ .5(2^2 - 2yε+ε^2)
(y-ε)y = xy ≤ .5[ (y-ε)^2 + y^2] = .5(x^2 + y^2)

Why are you making it so difficult?? It's just basic algebra. What is (x-y)^2??

The furthest I get for the second part is:
fg/[ (sqrt(∫g^2)*sqrt(∫f^2) ] ≤ .5 * [ (f^2/∫f^2) + (g^2/∫g^2) ]
Sorry, I don't see it.

Now integrate both sides, what do you get??