Does $\int_0^\infty |f(x)| \, dx$ and $\int_0^\infty |f'(x)| \, dx$ being finite imply that $\lim_{x \to \infty} xf(x) = 0$?

(Context: I am working through an analytic number theory textbook. In a lemma, the book assumes the conditions on $f$ and $f'$. The term $xf(x)$ appears from an integration by parts, but is tossed out, which must mean that $\lim_{x\to\infty} xf(x) = 0$. However, I've tried to prove this and was not successful. I also cannot think of any counterexamples.)


Here's a construction of a counterexample. The idea is based on a simple construction for series that is given on p. 350 of the centenary edition of G. H. Hardy's A Course of Pure Mathematics. The idea in Hardy's example is to construct a sequence $\{a_n\}$ of positive numbers such that $\sum a_n < \infty$ but $na_n \not\to 0$ as $n\to\infty$; to do this he simply puts $a_n = 1/n^2$ unless $n$ is a perfect square, in which case he puts $a_n = 1/n$.

To adapt this idea, choose a smooth function $\rho$, with $0\leq \rho \leq 1$, that is supported in $(-2,2)$ and that satisfies $\rho(x)=1$ for all $x \in (-1,1)$. Let $M$ be a bound for $|\rho'|$. Put $$ \begin{align*} \eta(x) & = \sum_{n = 2}^\infty \rho(x - n^2). \end{align*} $$ Basically, $\eta$ is identically $1$ in a neighborhood of each squared integer, and vanishes at each $x$ for which $(x-2,x+2)$ contains no squared integer. In fact, since the supports of $\rho(x-n^2)$ are disjoint, we have $0\leq \eta\leq 1$ and $|\eta'|\leq M$.

Finally, define $$ f(x) = {\eta(x)\over 1 + x}. $$ for $x \geq 0$. Obviously $f(x)\geq 0$ for all $x$. Also, $f$ is integrable, since $$ \begin{align} \int_0^\infty f(x)\,dx &\leq \sum_{n = 2}^\infty \int_{n^2 - 2}^{n^2 + 2}{dx\over 1 + x} \leq \sum_{n = 2}^\infty{4\over n^2 - 1}. \end{align} $$

Now $$ \begin{align} f'(x) & = {\eta'(x)\over 1+x} + O\left({1\over (1+x)^2}\right) \end{align} $$ as $x \to \infty$. Since $|\eta'|\leq M$, we get $$ \begin{align*} \int_0^\infty{|\eta'(x)|\over 1+x}\,dx & = \sum_{n = 2}^\infty \int_{n^2-2}^{n^2+2}{|\eta'(x)|\over 1+x}\,dx \leq M\sum_{n = 2}^\infty {4\over n^2-1}, \end{align*} $$ and it follows that $f'(x)$ is integrable over $(0,\infty)$ as well.

On the other hand, $xf(x) \not\to0$ as $x\to\infty$, since $$ n^2f(n^2) = {n^2\over 1 + n^2} \to 1 $$ as $n\to\infty$. Thus $f$ provides a counterexample.

  • $\begingroup$ Last line: $f(n^2)=1/(n^2+1)$, so $n^2f(n^2)=n^2/(n^2+1)\rightarrow 1$. $\endgroup$ – i707107 May 22 '13 at 6:42
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    $\begingroup$ @i707107 Fixed! Thanks for the correction. $\endgroup$ – Nick Strehlke May 22 '13 at 6:57
  • $\begingroup$ Thanks for your answer! I have seen some people making mistake like: "For positive sequence $a_n$, $\sum a_n$ converges if and only if $na_n\rightarrow 0$", but those two are irrelevant. (Counterexample in both ways) $\endgroup$ – i707107 May 22 '13 at 14:56

One way to think about problems of such type is in terms of linear interpolation. Define your function $f(n)=a_n>0$ let $f$ to be linear between $[n,n+1].$ Then, $f\in L_1$ is equivalent to $\sum_{n=1}a_n<\infty$ whereas $f'\in L_1$ is equivalent to $\sum_{n=1}|a_{n+1}-a_n|<\infty.$ Now it is clear that the standard example of series mentioned by Nick satisfies both conditions. The only thing left is to "smoothen" $f$ at the corners to make sure that $f'$ exists.


To estimate $xf(x)$ we consider $$\int^{a}_{0} f(x)dx=xf(x)|^{a}_{0}-\int^{a}_{0}xf'(x)$$ We know the first term is finite since $f\in L^{1}$, so if $\lim_{a\rightarrow \infty}af(a)=0$ we must require $\int^{\infty}_{0}xf'(x)$ be finite as well. To be more precise we need $$\int^{\infty}_{0}f(x)dx=-\int^{\infty}_{0}xf'(x)$$because the middle term goes to $0$.

However, it is not clear to me how to find a counter example using the above criterion. The condition that $xf'(x)$ must be integrable is a quite strong one, but seems to be okay for all elementary test functions I come up with. On the other hand I also do not know how to prove it using $f,f\in L^{1}$. Hope this 25 cents might be helpful.


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