# $f(x)$ is positive, continuous, monotone and integrable in (0,1]. Is $\lim_{x \rightarrow 0} xf(x) = 0$?

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!

• You assumed $f$ decreasing so $f(x)\leq f(t)$ and $0\leq xf(x)\leq \int_0^xf(t)dt$. – Davide Giraudo Feb 24 '12 at 22:02
• 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. – ro44 Feb 24 '12 at 22:08

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 $$\lim_{b\rightarrow0^+}\int_0^b f(x)\,dx=0.$$

• Ah! Thank you. That's the finishing touch I needed. – ro44 Feb 24 '12 at 22:41

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,

• Don't you need to specify that $c(x)<x$? – Quinn Culver Feb 24 '12 at 22:59
• @bgins : I don't need this ; $\int_0^x f(t)\, dt \le \int_0^1 f(t)\, dt$ by positivity of $f$. Therefore all my integrals exist and they converge towards $0$. – Patrick Da Silva Feb 24 '12 at 23:09
• @Quinn : I thought it was clear. I'll edit it if it pleases you. – Patrick Da Silva Feb 24 '12 at 23:09
• yes, I see now, very nice! – bgins Feb 25 '12 at 7:19

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 $xf(x) \to 0$.

• Hey Patrick, that seems incorrect. You're assuming an inequality on the entire interval, so the negation should be "there exist a point such that the inequality doesn't hold" in that interval. – ro44 Feb 24 '12 at 22:50
• Yes, quite right. See the edit. .. .well -- perhaps that's wrong too. hmm – Patrick Feb 24 '12 at 22:54
• Added an edit. Looks okay to me now. – Patrick Feb 24 '12 at 23:06
• 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. – Patrick Da Silva Feb 24 '12 at 23:27

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.

• Your answer is way too descriptive : If $\lim_{x \to 0^+} \int_x^1 f(t) \, dt = L$ exists, then $\lim_{\delta \to 0} \lim_{x \to 0} \int_x^{\delta} f(t) \, dt = \lim_{\delta \to 0^+} \left( L - \int_{\delta}^1 f(t) \, dt \right) = L-L = 0$. – Patrick Da Silva Feb 24 '12 at 23:24
• @PatrickDaSilva: How does your reasoning show that for $|x-\delta|\to 0$ also $\int_x^\delta f(t)dt \to 0$? – savick01 Feb 24 '12 at 23:29
• I am not saying that. I am saying that the improper integral $\int_0^{\delta} f(t) \, dt$ goes to $0$ as $\delta$ goes to zero. Since the integral $\int_0^{\delta}$ is defined as the limit $\lim_{x \to 0} \int_x^{\delta} = \lim_{x \to 0} \int_x^1 - \int_{\delta}^1$, I am done by using linearity of existing limits. – Patrick Da Silva Feb 25 '12 at 0:16
• @PatrickDaSilva: So you only want to replace the 4 first sentences of the second paragraph with those limits. It's OK. However I wasn't answering the question - rather explaining something. And for me the triangle inequality is much more understandable and intuitive than iterated limits. – savick01 Feb 25 '12 at 0:24
• I do not want to replace the the first $4$, I want to replace the whole thing ; if you believe expanding definitions explicitly make things clearer, then try to notice why my proof is exactly your proof using limit properties. Perhaps you will like my proof better afterwards ; I did not "find" my proof, I only translated yours using limit notation. – Patrick Da Silva Feb 25 '12 at 1:01

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.