Proving That a Function is in L^1([1,$\infty)$)

[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?

 

[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.

 

[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.

 

[Oxymoron]

YES! Thankyou! My error.

 

[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?

 

[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?

 

[ObsessiveMathsFreak]

.\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|

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.

 

[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.

 

[StatusX]

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?

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).

 

[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?)

 

[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.

 

[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 don't get a limit at 0. Therefore the limit does not exist at all. I need another way.

 

[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.

 

[Oxymoron]

I don't 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.

 

[Oxymoron]

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

\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.

 

[Oxymoron]

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 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...

 

[StatusX]

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.

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.

 

[Oxymoron]

Of course! Perfect. Thankyou.