What is the Limit of a Lebesgue Integral as t Approaches Infinity?

[xeno_gear]

Homework Statement

Suppose f \in L^1((0,\infty), dm). Prove that
\[ L := \lim_{t\to\infty} \frac{1}{t}\int_0^t x\, f(x)\, dm(x) = 0. \]

Homework Equations

dm(x) represents integration with respect to Lebesgue measure.

The Attempt at a Solution

I know how to do this if f is in Lp and not just L1 as long as the "x" and "1/t" aren't there,.. I tried a similar approach here but can't do any better than this:
<br /> \begin{align*}<br /> \left|\int_0^t xf(x)\, dm(x)\right| &amp;\le \int_0^\infty |xf(x) \chi_{[0,t]}(x)|\, dm(x)\\<br /> &amp;\le \|f\|_1 \cdot \|x \chi_{[0,t]}(x)\|_\infty\\<br /> &amp;= \|f\|_1 \cdot t.<br /> \end{align*}<br />
The last inequality is Hölder with p=1 and q = \infty.
But then L \le \lim_{t\to\infty} \|f\|_1, which only shows that L is finite but not necessarily zero. Thanks in advance!

 

[micromass]

I would try some partial integration... Have you seen the formula of partial integration for Lebesgue-integrals?

 

[xeno_gear]

I haven't. This is a qualifying exam problem from 4 years ago so I'm not sure what their class did, although generally anything goes as long as we justify it. The formula is the same but we need to throw in hypotheses about integrability, right?

 

[micromass]

Well, the formula for integration by parts is just a little bit more complicated for Lebesgue-integrals. I don't think you could just apply the classical formula.

However, maybe you do the following:

\int_0^t{xf(x)dm(x)}=\int_0^t{f(x)\int_0^x{1dm(u)}dm(x)

Now apply Fubini-Tonelli...

 

[xeno_gear]

I get:

<br /> \begin{align*}<br /> \int_0^t xf(x)\, dm(x) &amp;= \int_0^t f(x) \int_0^x 1 \, dm(u) \, dm(x)\\<br /> &amp;= \int_0^t \int_0^x f(x) \, dm(u) \, dm(x)\\<br /> &amp;= \int_0^x \int_0^t f(x) \, dm(x) \, dm(u)\\<br /> &amp;= \int_0^x \, dm(u) \int_0^t f(x) \, dm(x)\\<br /> &amp;= x \int_0^t f(x) \, dm(x),<br /> \end{align*}<br />
which will give the desired result, but that just doesn't seem right since that's essentially just pulling the "x" out of the integral. Did I do something wrong? Should that "u" be a "t" instead? If it helps at all, the qual for 2 or 3 semesters later asks the exact same question but with x^2 and 1/t^2 instead of x and 1/t.

 

[micromass]

No, you just switched the integrals. I mean, you basically did the following:

\int_0^t\int_0^x=\int_0^x\int_0^t

This is incorrect. If you switch the integrals, you have to change your ranges somehow...

 

[xeno_gear]

argh, of course! I do that too often. Will try again, thanks!

 

[xeno_gear]

Obviously I need to brush up on multiple integrals.. but anyway, I get this:

<br /> \begin{align*}<br /> \int_0^t xf(x)\, dm(x) &amp;= \int_0^t \int_0^x f(x) \, dm(u) \, dm(x)\\<br /> &amp;= \int_0^t \int_u^t f(x) \, dm(x) \, dm(u)<br /> \end{align*}<br />
which I'm pretty sure is right but then I'm not sure what else to do >:/
I know the integrals can't be separated because of the "u" in the lower limit of the inner integral. I tried this:
<br /> \begin{align*}<br /> \int_0^t xf(x)\, dm(x) &amp;= \int_0^t \int_u^t f(x) \, dm(x) \, dm(u)\\<br /> &amp;= \int_0^t \left[\int_0^t f(x) \, dm(x) - \int_0^u f(x)\,dm(x)\right] \, dm(u)\\<br /> &amp;= t \int_0^t f(x) \, dm(x) - \int_0^t \int_0^u f(x)\,dm(x) \, dm(u),<br /> \end{align*}<br />
but then I just kind of run into the same problem with the 2nd integral on the RHS there..

 

[micromass]

OK, we're not going to get there with this method. I'm sorry, I really thought it would work.

But what does work is the classical method: first check it for simple functions, then positive functions, and finally integrable functions.

Each of these steps should be quite easy...

 

[xeno_gear]

oh yeah, "bootstrapping" as they call it in McDonald & Weiss. I tried that on the analogous problem (with the x^2 and 1/t^2) a week or so ago and couldn't get it to work.. but now I see why! Thanks for all your help!

 

[xeno_gear]

OK, so I couldn't get it doing that either.. Had trouble with characteristic functions and boundedness of sets blahblahblah.. But I ended up trying this other thing in the analysis bag o' tricks that I forgot about.. I'll put it here in case anyone else has a similar question later and finds this thread.

Let f \in L^1 ((0,+\infty), dm). Then \int_0^{+\infty} |f(x)| \, dm &lt; +\infty which means for any \varepsilon &gt; 0 there is a T &gt; 0 such that
\[ \int_T^{+\infty} |f(x)| \, dm(x) &lt; \varepsilon \].

I'll leave the details out but if we split it up like:

<br /> \[ <br /> \frac{1}{t} \int_0^t xf(x)\, dm(x) = <br /> \underbrace{\frac{1}{t} \int_0^T xf(x)\, dm(x)}_{:= I_1} +<br /> \underbrace{\frac{1}{t} \int_T^t xf(x)\, dm(x)}_{:= I_2}<br /> \]<br />

Then we can show |I_1| \to 0 and |I_2| &lt; \varepsilon as t \to +\infty.