Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

L^p Spaces

  1. Nov 13, 2006 #1
    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?
     
    Last edited: Nov 13, 2006
  2. jcsd
  3. Nov 13, 2006 #2
    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.
     
    Last edited: Nov 13, 2006
  4. Nov 13, 2006 #3

    StatusX

    User Avatar
    Homework Helper

    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.
     
  5. Nov 13, 2006 #4
    YES!!!! Thankyou! My error.
     
  6. Nov 14, 2006 #5
    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?
     
  7. Nov 14, 2006 #6
    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?
     
  8. Nov 14, 2006 #7
    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.
     
  9. Nov 14, 2006 #8
    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
  10. Nov 14, 2006 #9

    StatusX

    User Avatar
    Homework Helper

    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).
     
  11. Nov 14, 2006 #10
    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?)
     
  12. Nov 14, 2006 #11

    StatusX

    User Avatar
    Homework Helper

    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.
     
  13. Nov 14, 2006 #12
    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.
     
    Last edited: Nov 15, 2006
  14. Nov 14, 2006 #13
    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.
     
  15. Nov 14, 2006 #14
    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.
     
  16. Nov 14, 2006 #15
    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.
     
    Last edited: Nov 15, 2006
  17. Nov 15, 2006 #16
    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...
     
  18. Nov 15, 2006 #17

    StatusX

    User Avatar
    Homework Helper

    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.
     
  19. Nov 15, 2006 #18
    Of course!!! Perfect. Thankyou.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: L^p Spaces
  1. For 1 < p < oo l^p (Replies: 1)

  2. L^p derivative (Replies: 1)

Loading...