I am reposting the following question that I asked in the MSE site here.
As I mentioned there, The following generalizations of the Riemann-Lebesgue lemma are rather well known (Kahane, C. S., Generalizations of the Riemann-Lebesgue and Cantor-Lebesgue lemmas, Czechoslovak Mathematical Journal, Vol 30 (1980), No. 1, 108-117):
Theorem I: Suppose $g\in L_\infty([0,\infty),m)$. A necessary and sufficient condition for \begin{align} I(f; g):=\lim_{\lambda\rightarrow\infty}\int^\infty_0 f(t) g(\lambda t)\,dt\tag{0}\label{zero} \end{align} to exists for every $f\in L_1([0,\infty),m)$ is that $g$ has the mean value property, i.e. \begin{align} I(g)=\lim_{T\rightarrow\infty}\frac1T\int^T_0 g(t)\,dt\tag{1}\label{one} \end{align} exists. Furthermore, when this is the case, \begin{align} I(f; g)=\Big(\int^\infty_0 f(t)\,dt\Big) I(g)\tag{2}\label{two} \end{align}
This results extends easily to $L_1(\mathbb{R},m)$ as follows:
Corollary I: Suppose $g\in L_\infty(\mathbb{R},m)$. A necessary and sufficient condition for \begin{align} I(f; g):=\lim_{\lambda\rightarrow\infty}\int_\mathbb{R} f(t) g(\lambda t)\,dt\tag{3}\label{three} \end{align} to exists for every $f\in L_1(\mathbb{R},m)$ is that $g_-(t)=g(-t)\mathbb{1}_{[0,\infty)}(t)$ and $g_+(t)=g(t)\mathbb{1}_{[0,\infty)}(t)$ have the mean value property. If this is the case, \begin{align} I(f; g)=\Big(\int^0_{-\infty} f(t)\,dt\Big) I(g_-) + \Big(\int^\infty_0 f(t)\,dt\Big) I(g_+)\tag{4}\label{four} \end{align}
Perhaps the most common version of these results is the case where $g\in L_\infty(\mathbb{R},m)$ and $g$ is $P$-periodic function ($g(x+P)=g(x)$ for all $x$ and some $P>0$). Then \eqref{four} has the form \begin{align} \lim_{\lambda\rightarrow\infty}\int_{\mathbb{R}}f(t)g(\lambda t)\,dt=\Big(\frac{1}{P}\int^P_0 g(t)\,dt\Big)\int_\mathbb{R} f(t)\,dt\tag{5}\label{five} \end{align}
The assumption that $g\in L_\infty(\mathbb{R},m)$ can be relaxed by considering integrals over finite intervals and some duality assumptions:
Theorem II: Suppose $g\in L^{loc}_q([0,\infty),m)$ , $q>1$, and let $0\leq a< b<\infty$. For the limit \begin{align} I(f; g,[a,b]):=\lim_{\lambda\rightarrow\infty}\int^b_a f(t) g(\lambda t)\,dt\tag{6}\label{six} \end{align} to exists for every $f\in L_p([a,b],m)$, $\frac1p+\frac1q=1$, it is necessary and sufficient that
- (i) $\frac1T\int^T_0|g(t)|^q\,dt=O(1)$ as $T\rightarrow\infty$,
- (ii) $g$ has the mean value property \eqref{one}.
If (i) and (ii) hold, then the limit \eqref{six} takes the form \begin{align} I(f; g, [a,b])=\Big(\int^b_a f(t)\,dt\Big) I(g)\tag{7}\label{seven} \end{align}
The validity of the results presented above is based on duality between $L_p$ spaces. Holder's inequality provides uniform bounds on some linear operators, uniform boundedness principle and density arguments then give the desired results.
Question: I would like to know if there is a counter example to Theorem I if the assumption $g\in L_\infty([0,\infty),m)$ is relaxed while maintaining the integrability assumption for $f$. To be more precise and to simplify matters,
Are there measurable functions $g$ and $f$, such that,
- $g$ is $1$-periodic, $g\in L_1([0,1],m)\setminus L_\infty([0,1],m)$,
- $f\in L_1(\mathbb{R},m)$
- for any $\lambda>0$, $t\mapsto f(t) g(\lambda t)\in L_1(\mathbb{R},m),$
but either $I(f; g)=\lim_{\lambda\rightarrow\infty}\int^\infty_0 f(t) g(t\lambda)\,dt$ does not exists (as a real number), or if it does, $I(f; g)\neq I(g) \int^\infty_0 f$.
If any body knowns a counter example or may be able to build one, that would also be fantastic!
