In R.D. Richtmyer, Principles of Advanced Mathematical Physics, p.85 an example is given of a continuous and square-integrable on $\bf{R}$ function, which is not bounded at infinity: $$f(x)=x^2\exp{(−x^8\sin^2{x})}.$$ Intuitively, it can be expected that $f(x)$ is square-integrable as its peaks become more and more narrow. But how can one rigorously prove that $f(x)$ is indeed square-integrable?
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asked Mar 22, 2020 at 13:30
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4$\begingroup$ By looking at the measure of the sets $\{ x:|f(x)|>c\}$. $\endgroup$– Alexandre EremenkoCommented Mar 22, 2020 at 13:59
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For $k=1,2,\dots$, let $$I_k:=\int_{|x-k\pi|<1/k}f(x)^2\,dx =\int_{|x-k\pi|<1/k}x^4\exp(-2x^8\sin^2 x)\,dx.$$ Then, as $k\to\infty$, $$I_k\asymp k^4\int_{|x-k\pi|<1/k}\exp\{-(2+o(1))(k\pi)^8\sin^2 x\}\,dx \\ =k^4\int_{|x|<1/k}\exp\{-(2+o(1))(k\pi)^8\sin^2 x\}\,dx \\ =k^4\int_{|x|<1/k}\exp\{-(2+o(1))(k\pi)^8 x^2\}\,dx \asymp1,$$ by substitution $u=(k\pi)^4 x$. (Here we use the usual notation: $A\asymp B$ meaning $A=O(B)$ and $B=O(A)$.) Hence, $$\int_{-\infty}^\infty f(x)^2\,dx\ge\sum_{k=1}^\infty I_k=\infty.$$ So, $f$ is actually not square-integrable.
Reasoning similarly (but using, say, $h_k:=1/k^{b/3}$ instead of $1/k$ in $|x-k\pi|<1/k$), one can see that for any real $a,b>0$, letting $$f(x):=|x|^a\exp(-|x|^b\sin^2x),$$ we have the following:
$f$ is continuous, but unbounded at $\infty$.
$f$ is square-integrable iff $2a-b/2<-1$.
To get this result, we also note that for $k=1,2,\dots$ $$\int_{h_k\le|x-k\pi|\le\pi}x^{2a}\exp(-2x^b\sin^2 x)\,dx \\ =O(k^{2a}\exp\{-(2+o(1))(k\pi)^b h_k^2\})=O(1/k^c)$$ for any real $c$.
In your example, we have $a=2$ and $b=8$, so that $2a-b/2=0\not<-1$, and so, your $f$ is not square-integrable.
answered Mar 22, 2020 at 14:14
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$\begingroup$ Just another way of seeing it: instead of cutting the integral with respect to $h_k$, one can first change $\sin(x)^2$ to $Cx^2$ (or $\varepsilon x^2$, depending on the direction of the estimate), then perform the change of variable $h_k\cdot y=x-k\pi$. Then the integral between ± something of order $1/h_k$ is less than the integral over $\mathbb R$, and more than the integral between ± something of order 1. $\endgroup$ Commented Mar 22, 2020 at 15:22
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$\begingroup$ Richtmyer says it is square-integrable, but it seems you are right (I have done some numerical calculations). The same statement is reproduced in arxiv.org/abs/quant-ph/9907069. Maybe there was a typo in Richtmyer. $\endgroup$ Commented Mar 22, 2020 at 15:23
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1$\begingroup$ @ZurabSilagadze : Perhaps, Richtmyer meant $a=1$ instead of $a=2$ (indeed, there is no reason to take $a=2$ to get an unbounded $f$, when $a=1$ would suffice). Then, still with $b=8$, we would have $2a-b/2=-2<-1$, just what is needed. $\endgroup$ Commented Mar 22, 2020 at 15:30
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$\begingroup$ @ Iosif Pinelis I think prefactor in your solution should be $(k\pi)^4$, not $k^4$. What is $c$ in the general case and why you cannot use $|x-k\pi|<1/k$ in this case also? $\endgroup$ Commented Mar 22, 2020 at 16:09
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1$\begingroup$ @ZurabSilagadze : (i) With the (standard, I think, and now explained) meaning of $\asymp$, positive real factors don't matter. (ii) As was stated, $c$ is any (constant) real number. (iii) I cannot use $h_k=1/k$ if $b\le2$, because I need $k^b h_k^2$ to be much greater than $\ln k$ for large $k$ -- in particular, to get the bound $O(1/k^c)$ for all real $c$. $\endgroup$ Commented Mar 22, 2020 at 16:25