# Holder's inequality for integrals

1. Apr 18, 2012

### 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)

2. Apr 18, 2012

### DonAntonio

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

3. Apr 19, 2012

### brydustin

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

4. Apr 19, 2012

### 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....

5. Apr 19, 2012

### micromass

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

6. Apr 19, 2012

### brydustin

Actually I see the argument xy ≤ 1/2 (x^2 + y^2)
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.

7. Apr 19, 2012

### micromass

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

8. Apr 19, 2012

### DonAntonio

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

DonAntonio

9. Apr 19, 2012

### brydustin

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.

10. Apr 19, 2012

### micromass

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

Now integrate both sides, what do you get??