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It is known that if $f$ is a continuous function on $\mathbb{R}$ and $\int_\mathbb{R}|f(x)|dx<\infty$ then $f$ is not necessarily bounded.

However, Barbalat’s Lemma tells that if $f$ is uniformly continuous and integrable then $f$ is not only bounded but also vanishes at infinity.

Now suppose that $f$ is uniformly continuous and bounded on $\mathbb{R}$ and $\int_\mathbb{R}|f(x)x^a|dx<\infty$ for some $a>0$. Can we say that $|f(x)x^b|$ is bounded for some $b\in(0,a]$?

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    $\begingroup$ Isn't the product of uniformly continuous functions uniformly continuous? So shouldn't Barbalat's Lemma also apply for $|f(x) x^a|$ ? $\endgroup$
    – hal4math
    Mar 11, 2022 at 14:58
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    $\begingroup$ The product of uniformly continuous functions is not necessarily uniformly continuous (see, e.g., math.stackexchange.com/questions/2685175/…? ) $\endgroup$
    – Gerome
    Mar 11, 2022 at 15:02
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    $\begingroup$ Ups, right. Also $x^2$ wouldn't even be uniformly continuous. Sorry for that. $\endgroup$
    – hal4math
    Mar 11, 2022 at 15:04
  • $\begingroup$ But are you sure you want $a >0$? Since in that case $x^a$ might become complex on $\mathbb{R}$. $\endgroup$
    – hal4math
    Mar 11, 2022 at 15:20

1 Answer 1

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Here is a counterexample. Consider the function $$ f(x) = \sum_{n\ge 2} \bigg(\frac1{\ln n} - \frac{1}{ \ln |\ln |x - n||}\bigg)\mathbf 1_{[n-e^{-n},n+e^{-n}]}(x), $$ where $1/\ln |\ln 0| := 0$. It is straightforward to check that this function is non-negative and uniformly continuous: its modulus of continuity is $\omega(h) = \frac{1}{\ln |\ln h|},h\in [0,e^{-2}]$. Further, for any $a>0$ $$ \int_{\mathbb R} f(x) |x|^a dx \le \sum_{n\ge 2} \int_{n-e^{-n}}^{n+e^{-n}}\frac{|x|^a dx}{\ln n} \le 4\sum_{n\ge 2} \frac{n^a}{e^n \ln n} <\infty. $$ However, for any $a>0$, $$ f(n) n^a = \frac{n^a}{\ln n} \to \infty, n\to\infty. $$


Some intuition behind the example.

We wish to take some function $g$ which rapidly decreases to zero from its maximum in zero and then replicate it along the real line (taking the very same function is important in view of the uniform continuity requirement), i.e. to consider $g(x-a_n)$ with $a_n\to \infty$. We also need to lower the replicas down to zero (so that the integral is finite), i.e., to consider $\big(g(x - a_n) - g(0) + b_n\big)^+$ with $b_n$ decreasing to $0$ from above, but not too fast, since we need $f(x)|x|^a$ to be unbounded.

For $a_n$, the simplest is to take $a_n = n$; after that it's a matter of choosing $g$, which should decrease from its maximum much faster than any power (in my example, $g(x) = (1 - 1/\ln|\ln x|)^+ )$, and $b_n$, which should decrease slower than any power (in my example, $b_n = 1/\ln n$).

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  • $\begingroup$ This is an amazing example! May I ask how did you come up with it? Is it well-known? More importantly I saw where you work. Don't have much words but I hope you and your loved ones are doing okay! 🇺🇦 $\endgroup$
    – hal4math
    Mar 12, 2022 at 0:11
  • $\begingroup$ Thanks a lot for the answer! If I may, I'd like to ask whether it is true that under the same conditions $\int_\mathbb{R}|f^2(x)x^b|dx<\infty$ for some $b>a$? Thanks again and all the best to you and your country! $\endgroup$
    – Gerome
    Mar 12, 2022 at 5:55
  • $\begingroup$ @Gerome: Actually, this function satisfies $\int_{\mathbb R} |f(x)|^b |x|^a dx<\infty$ for all $a>0$ and $b>0$. Note also that the bigger $b$ is, the more intergal is likely to be finite. So asking about finiteness of integral with $f^2$ makes little sense, when that with $f$ is finite. $\endgroup$
    – zhoraster
    Mar 12, 2022 at 9:23
  • $\begingroup$ @hal4math, thank you for the support! The example is unlikely to be well-known, but I would say that the technique for constructing it is quite standard. I edited the post to describe the idea; hopefully, it will be useful for you. Actually, I enjoy constructing such kind of examples and have written several articles centered around 'constructions'. $\endgroup$
    – zhoraster
    Mar 12, 2022 at 9:46
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    $\begingroup$ @Gerome, oh, I see. Please post a separate question about that. $\endgroup$
    – zhoraster
    Mar 12, 2022 at 9:54

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