L^p Spaces

  • Thread starter Oxymoron
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  • #1
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If I have a function [itex]f \in L^2([1,\infty))[/itex] then is it true that all I know about that function is that

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

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

[tex]g(x):=x^{-1}\circ f(x)[/tex]

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

[tex]\int_{[1,\infty)} |g|\mbox{d}\mu[/tex]

is finite? Is this all I have to do?
 
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Answers and Replies

  • #2
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If it is, then could I proceed as follows?:

I have to prove

[tex]\|g\|_1 < \infty[/tex]

but since [itex]g(x):=f(x)/x[/itex] then

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

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

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

[tex].\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}[/tex]

[tex].\quad\quad <_{\mbox{?}} \infty[/tex]

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

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

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

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

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

[tex]\sqrt{\int_1^{\infty}x^{-2}\mbox{d}\mu} = \sqrt{\int_1^{\infty}x^{-2}\mbox{d}\mu}[/tex]

[tex].\quad\quad = \sqrt{\lim_{t\rightarrow\infty}\int_1^t x^{-2}\mbox{d}\mu}[/tex]

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

[tex].\quad\quad = \sqrt{\lim_{t\rightarrow\infty}\left(1-\frac{1}{t}\right)}[/tex]

[tex].\quad\quad = \sqrt{1}[/tex]

[tex].\quad\quad = 1 < \infty[/tex]

Does this look right?

EDIT: Thanks StatusX for the correction.
 
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  • #3
StatusX
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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
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YES!!!! Thankyou! My error.
 
  • #5
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Therefore, since

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

we have that [itex]g \in L^1([1,\infty))[/itex]. Is this the correct way to show this sort of thing?
 
  • #6
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Well, assuming I have all that sorted I have one more question.

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

[tex]\sqrt{\int_0^{\infty} |f|^2\mbox{d}\mu}[/tex]

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

[tex]g(x):=\frac{f(x)}{x}[/tex]

is in [itex]L^1([0,\infty))[/itex]?

I figured that, once again, I have to prove that [itex]\|g(x)\|_1[/itex] 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
Oxymoron said:
[tex].\quad\quad = \int_{[1,\infty)}\left|\frac{f(x)}{x}\right|\mbox{d}\mu[/tex]

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

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

[tex]||fg||_1 \leq ||f||_2 ||g||_2[/tex]
From Holders inequality. So the rest of the proof follows anyway.... but that step wasn't right.
 
  • #8
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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.
 
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  • #9
StatusX
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Oxymoron said:
Well, assuming I have all that sorted I have one more question.

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

[tex]\sqrt{\int_0^{\infty} |f|^2\mbox{d}\mu}[/tex]

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

[tex]g(x):=\frac{f(x)}{x}[/tex]

is in [itex]L^1([0,\infty))[/itex]?

I figured that, once again, I have to prove that [itex]\|g(x)\|_1[/itex] 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?

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
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So you are saying that I will still be able to argue that [itex]g(x) \in L^1([0,\infty))[/itex], 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
StatusX
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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
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So do I begin just as I did before?

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

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

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

[tex].\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}[/tex]

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

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

So we write

[tex]\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[/tex]

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

[tex]\int_0^1\frac{1}{x^2}\mbox{d}\mu < \infty[/tex]

Calculate the limit:

[tex]\int_0^1 \frac{1}{x^2}\mbox{d}\mu = \lim_{t\rightarrow 0^+}\int_t^1\frac{1}{x^2}\mbox{d}\mu[/tex]

[tex].\quad\quad = \lim_{t\rightarrow 0^+}\left.\frac{1}{x}\right|_t^1[/tex]

[tex].\quad\quad = \lim_{t\rightarrow 0^+}\left(\frac{1}{t} - 1\right)[/tex]

[tex].\quad\quad = \infty[/tex]

which just goes to show that since

[tex]\lim_{t\rightarrow 0^-}\frac{1}{t} \neq \lim_{t\rightarrow 0^+}\frac{1}{t}[/tex]

you dont get a limit at 0. Therefore the limit does not exist at all. I need another way.
 
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  • #13
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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
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I dont see where I could insert a differentiability argument into the above calculation to get [itex]\int_0^{\infty}x^{-2}\mbox{d}\mu[/itex] to be finite.
 
  • #15
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Posted by StatusX:

In fact, look at the definition of the derivative at x=0 and this should be really obvious.

Could I possibly just say:

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

[tex]\lim_{x\rightarrow 0}f(x) = f(0)[/tex]

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

[tex]f'(0) = \lim_{x\rightarrow 0}\frac{f(x) - f(0)}{x-0}[/tex]

[tex].\quad = \lim_{x\rightarrow 0}\frac{f(x)}{x}[/tex] since f(0) = 0 by hypothesis.

But this tells me nothing.
 
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  • #16
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Posted by 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.

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 [itex]mx[/itex] 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
StatusX
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Oxymoron said:
Could I possibly just say:

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

[tex]\lim_{x\rightarrow 0}f(x) = f(0)[/tex]

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

[tex]f'(0) = \lim_{x\rightarrow 0}\frac{f(x) - f(0)}{x-0}[/tex]

[tex].\quad = \lim_{x\rightarrow 0}\frac{f(x)}{x}[/tex] since f(0) = 0 by hypothesis.

But this tells me nothing.

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
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Of course!!! Perfect. Thankyou.
 

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