4
$\begingroup$

I'm having trouble with this question from an example test.

We have a positive function $f(x)$ that's monotone, continuous and integrable in $(0,1]$. Is $\lim_{x \rightarrow 0} xf(x) = 0$?

Progress

The only problematic case seems to be when $f(x)$ is unbounded and monotonic decreasing. For that case, I found out that $xf(x)=\int_{0}^{x} f(x)dt$ and that $0\leq xf(x)\leq \int_{0}^{x} f(t)dt$. From here I'm not sure how to go on.

Thanks!

  • 0
    Thank you Davide, I've already written that in the progress section. The problem is, since $\int_{0}^{x} f(t)dt$ is not necessarily a 'regular' integral I don't know if it's continuous at 0. It seems like it should be, but we haven't proved that in the course.2012-02-24

5 Answers 5

3

If $f$ is improperly integrable over $[0,1]$, then $\lim\limits_{a\rightarrow0^+}\int_a^1 f(x)\,dx=L$ for some finite number $L$.

Note, for any $b$ in $(0,1)$: $\eqalign{ L =\int_0^1f(x)\,dx = \int_0^bf(x)\,dx +\int_b^1f(x)\,dx\cr } $ Now, letting $b\rightarrow0^+$, we have that $ \int_b^1f(x)\,dx$ converges to $L$, which implies that $ \lim_{b\rightarrow0^+}\int_0^b f(x)\,dx=0. $

  • 0
    Ah! Thank you. That's the finishing touch I needed.2012-02-24
2

If $f$ is increasing, then $\lim_{x \to 0^+} f(x) = C < +\infty$ exists, so that the result is trivial.

If $f$ is decreasing, $c < x$ implies $f(c) \ge f(x)$. Now by the mean value theorem for integrals, for every $0 < x < 1$, there exists $0 < c(x) < x$ such that $ \int_0^x f(t) \, dt = f(c(x))x. $ This gives $ 0 \le x f(x) \le x f(c(x)) = \int_0^x f(t) \, dt \to 0. $

Hope that helps,

  • 1
    yes, I see now, very nice!2012-02-25
1

If there is some $\beta > 0$ and $L>0$ so that $xf(x) \ge L$ on $(0,\beta)$, then $f(x) \ge L/x$ on $(0,\beta)$ and hence $ \int_0^\beta f(x) dx $ doesn't exist. Otherwise for every $L$ there is a decreasing sequence $\{u_{L,n}\} \to 0$ such that $u_{L,n}f(u_{L,n}) < L$. Hence by monotonicity, $xf(x) < L$ on $(0,u_{L,1})$.

Now let $L\to 0$. It follows that $xf(x) \to 0$.

  • 0
    I didn't comment because I thought you "understood" that. But indeed this is what was needed to complete the proof. Although you don't need $u_{L,n}$, you can just choose $u_n$ such that u_n f(u_n) < \frac 1n, that's enough.2012-02-24
0

Just a longer comment - I'll show that improper integrals are continuous.

Improper integral on $(0,1]$ is just the limit of regular integrals on $[\delta,1]$. If the limit $L$ exists, then for $\delta < \delta_0$ this integral is $\varepsilon$-close to L. From the triangle inequality we know, that two integrals on $[\delta,1]$ and [\delta',1] are $2\varepsilon$-close (note that \delta,\delta' might also be zero here, then we get just $\varepsilon$). It means that when we integrate on an interval included in $[0,\delta_0]$, then the result is smaller than $2\varepsilon$. Moreover on $[\delta_0,1]$ the function is bounded by some $M$ (from its integrability). So, if we take an interval $[a,b]$ shorter than $\min(\delta_0,\varepsilon/M)$, then (by triangle inequality) the integral is smaller than $2\varepsilon + \varepsilon=3\varepsilon$, which shows the continuity of an improper integral.

  • 0
    @PatrickDaSilva: 1. But the whole thing is not the answer to the question [and the whole thing is more than you write]. It solves a problem mentioned in one of the comments. 2. I can see the equivalence between those two (4 sentences and your calculations), but I still believe that at low level (it's calculus here!) it is sometimes clearer to write something longer and with little notation.2012-02-25
0

As you pointed out $0\leq x f(x) \leq \int_{0}^{x} f(t)\, dt$. But $F(x):= \int_{0}^{x}f(t)\, dt$ is continuous, so $\lim_{x\to 0} F(x) = F(0) = 0$, and the result follows from the squeeze theorem.