# L^p Spaces

1. Nov 13, 2006

### Oxymoron

If I have a function $f \in L^2([1,\infty))$ then is it true that all I know about that function is that

$$\sqrt{\int_{[1,\infty)}|f|^2\mbox{d}\mu} < \infty$$

I ask this because I am asked to show that if I have a function as above, then the function $g(x):= f(x)/x$ is in $L^1([1,\infty))$. Now, is it true that g is simply a composition of functions from $L^2([1,\infty))$?

$$g(x):=x^{-1}\circ f(x)$$

If I want to prove that a function is in $L^1([1,\infty))$ then do I have to show that

$$\int_{[1,\infty)} |g|\mbox{d}\mu$$

is finite? Is this all I have to do?

Last edited: Nov 13, 2006
2. Nov 13, 2006

### Oxymoron

If it is, then could I proceed as follows?:

I have to prove

$$\|g\|_1 < \infty$$

but since $g(x):=f(x)/x$ then

$$\|g\|_1 = \int_{[1,\infty)}|g|\mbox{d}\mu$$

$$.\quad\quad = \int_{[1,\infty)}\left|\frac{f(x)}{x}\right|\mbox{d}\mu$$

$$.\quad\quad = \left| \int_{[1,\infty)}f(x)x^{-1}\,\mbox{d}\mu\right|$$

$$.\quad\quad \leq \sqrt{\int_{[1,\infty)}f(x)^2\,\mbox{d}\mu}\sqrt{\int_{[1,\infty)}x^{-2}\,\mbox{d}\mu}\quad\quad\mbox{By Cauchy-Schwarz}$$

$$.\quad\quad <_{\mbox{?}} \infty$$

But this last inequality implies that both $\sqrt{\int_{[1,\infty)}f(x)^2\,\mbox{d}\mu}$ and $\sqrt{\int_{[1,\infty)}x^{-2}\,\mbox{d}\mu}$ be finite. And since we already know that

$$\sqrt{\int_{[1,\infty)}f(x)^2\,\mbox{d}\mu} < \infty$$

because, by hypothesis, $f \in L^2([1,\infty))$, it suffices to show that

$$\sqrt{\int_{[1,\infty)}x^{-2}\,\mbox{d}\mu} < \infty$$

in order to show that $g \in L^1([1,\infty))$. Then

$$\sqrt{\int_1^{\infty}x^{-2}\mbox{d}\mu} = \sqrt{\int_1^{\infty}x^{-2}\mbox{d}\mu}$$

$$.\quad\quad = \sqrt{\lim_{t\rightarrow\infty}\int_1^t x^{-2}\mbox{d}\mu}$$

$$.\quad\quad = \sqrt{\lim_{t\rightarrow\infty}\, \left. \frac{1}{x}\right|_1^t}$$

$$.\quad\quad = \sqrt{\lim_{t\rightarrow\infty}\left(1-\frac{1}{t}\right)}$$

$$.\quad\quad = \sqrt{1}$$

$$.\quad\quad = 1 < \infty$$

Does this look right?

EDIT: Thanks StatusX for the correction.

Last edited: Nov 13, 2006
3. Nov 13, 2006

### StatusX

If you're using the Cauchy Schwarz inequality, then you want that to be 1/x^2, not x, and that does have a finite integral.

4. Nov 13, 2006

### Oxymoron

YES!!!! Thankyou! My error.

5. Nov 14, 2006

### Oxymoron

Therefore, since

$$\|g\|_1 \leq \sqrt{\int_{[1,\infty)}f(x)^2\,\mbox{d}\mu}\sqrt{\int_{[1,\infty)}x^{-2}\,\mbox{d}\mu} < \infty$$

we have that $g \in L^1([1,\infty))$. Is this the correct way to show this sort of thing?

6. Nov 14, 2006

### Oxymoron

Well, assuming I have all that sorted I have one more question.

Suppose I have a function, $f$ from $L^2([0,\infty))$, that is

$$\sqrt{\int_0^{\infty} |f|^2\mbox{d}\mu}$$

is finite. Suppose further that this function is differentiable at 0, that is, f is continuous at 0. Suppose even further that $f(0) = 0$. Then how would I prove that

$$g(x):=\frac{f(x)}{x}$$

is in $L^1([0,\infty))$?

I figured that, once again, I have to prove that $\|g(x)\|_1$ is finite, but this time when I went to integrate I found that I would have a zero in the denominator! What am I to do?

7. Nov 14, 2006

### ObsessiveMathsFreak

Woah there! This step is certainly not valid. I don't think you need cauchy schwartz. If I remember correctly;

$$||fg||_1 \leq ||f||_2 ||g||_2$$
From Holders inequality. So the rest of the proof follows anyway.... but that step wasn't right.

8. Nov 14, 2006

### Oxymoron

Hmmm, Holder's inequality...of course! You can say that because 1/2 + 1/2 = 1. Which means I have the inequality straight away, but I assume the rest of my calculation is correct.

Yes it is.

Last edited: Nov 15, 2006
9. Nov 14, 2006

### StatusX

You'll need to use the differentiability, since for example, 1/ln(x) approaches 0 as x->0, but is not differentiable there, and the integral of 1/(xln(x)) diverges. That f is differentiable means that it looks like a line near the origin, and this will cancel the singularity (you'll need to make this more precise).

10. Nov 14, 2006

### Oxymoron

So you are saying that I will still be able to argue that $g(x) \in L^1([0,\infty))$, even though there is a singularity at x=0, because I have differentiability at x=0 PLUS the function is 0 at that point.

So if it was not differentiable at x=0 I could not argue this? Is it the differentiability alone which allows me to do this problem (or that and the fact that f(0) = 0?)

11. Nov 14, 2006

### StatusX

Right, it's because the function looks like m*x near the origin for some finite m, so f(x)/x goes as m, and stays finite. In fact, look at the definition of the derivative at x=0 and this should be really obvious.

12. Nov 14, 2006

### Oxymoron

So do I begin just as I did before?

We have to prove that $\|g\|_1 < \infty$. But since $g(x):=f(x)/x$, then

$$\|g\|_1 = \int_{[0,\infty)}|g|\,\mbox{d}\mu$$

$$.\quad\quad = \int_{[0,\infty)}\left|\frac{f(x)}{x}\right|\,\mbox{d}\mu$$

$$.\quad\quad\leq \sqrt{\int_{[0,\infty)}f(x)^2\,\mbox{d}\mu}\sqrt{\int_{[0,\infty)}x^{-2}\,\mbox{d}\mu}\quad\quad\mbox{By Holder's Inequality}$$

Then since $\int f(x)^2\mbox{d}\mu < \infty$ because $f \in L^2([0,\infty))$ by hypothesis, it suffices to show that

$$\sqrt{\int_{[0,\infty)}x^{-2}\,\mbox{d}\mu} < \infty$$

So we write

$$\int_0^{\infty}\frac{1}{x^2}\mbox{d}\my = \int_0^1\frac{1}{x^2}\mbox{d}\mu + \int_1^{\infty}\frac{1}{x^2}\mbox{d}\mu$$

and we already know that $\int_1^{\infty}\frac{1}{x^2}\mbox{d}\mu$ is finite, so it suffices to show that

$$\int_0^1\frac{1}{x^2}\mbox{d}\mu < \infty$$

Calculate the limit:

$$\int_0^1 \frac{1}{x^2}\mbox{d}\mu = \lim_{t\rightarrow 0^+}\int_t^1\frac{1}{x^2}\mbox{d}\mu$$

$$.\quad\quad = \lim_{t\rightarrow 0^+}\left.\frac{1}{x}\right|_t^1$$

$$.\quad\quad = \lim_{t\rightarrow 0^+}\left(\frac{1}{t} - 1\right)$$

$$.\quad\quad = \infty$$

which just goes to show that since

$$\lim_{t\rightarrow 0^-}\frac{1}{t} \neq \lim_{t\rightarrow 0^+}\frac{1}{t}$$

you dont get a limit at 0. Therefore the limit does not exist at all. I need another way.

Last edited: Nov 15, 2006
13. Nov 14, 2006

### tuananh

Here is an counter-example: define f(x) = 1 if x in [0,1], f(x)=0 eslewhere
Apparently, f is square-integrable on [0,infty) but f(x)/x is not.

14. Nov 14, 2006

### Oxymoron

I dont see where I could insert a differentiability argument into the above calculation to get $\int_0^{\infty}x^{-2}\mbox{d}\mu$ to be finite.

15. Nov 14, 2006

### Oxymoron

Could I possibly just say:

Since f(x) is differentiable at x=0 it must be continuous as x=0. Therefore

$$\lim_{x\rightarrow 0}f(x) = f(0)$$

In other words, if f(x) is differentiable at 0 then

$$f'(0) = \lim_{x\rightarrow 0}\frac{f(x) - f(0)}{x-0}$$

$$.\quad = \lim_{x\rightarrow 0}\frac{f(x)}{x}$$ since f(0) = 0 by hypothesis.

But this tells me nothing.

Last edited: Nov 15, 2006
16. Nov 15, 2006

### Oxymoron

It is true that as x approaches 0 the tangent line approaches a vertical line, but a vertical line is an indication that the function is not differentiable, which is a contradiction.

How to conclude that the function looks like $mx$ near the origin for some finite m. This means that near the origin the tangent line intersects the function graph always at some finite point m. But as f(x)/x approaches the origin the tangent line becomes vertical and we cannot say that it intersects at some finite m, but at infinity! So, f(x)/x goes as m, yes, but it does not stay finite. Unless I am mistaken...

17. Nov 15, 2006

### StatusX

That tells you everything! Remember, f'(0) is finite, so now you have that f(x)/x is bounded in some neighborhood of 0. The rest of R won't be a problem because the integral of 1/x^2 will be finite there.

18. Nov 15, 2006

### Oxymoron

Of course!!! Perfect. Thankyou.

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