In Iwaniec's "Spectral methods of automorphic forms" he says (page 128) that a (smooth and of bounded variation) $f$ can be written through Kontorovitch-Lebedev inversion if $f$ satisfies \[ f^j(x)\ll \frac {1}{(x+1)^\alpha },\hspace {10mm}f(0)=0\hspace {20mm}(1)\] for some $\alpha >2$ and for $j=0,1,2$. With this he then goes on to prove a trace formula. The same conditions are given on page 411 of Iwaniec and Kowalski's "Analytic Number Theory". In the Appendix of the first book, he says the inversion is fine if \[ \int _{0}^\infty |f(x)|\left (\frac {1}{\sqrt x}+\frac {|\log x|}{x}\right )dx<\infty \hspace {20mm}(2).\] But this is weaker than (1) right? (Unless I've made a mistake, I'll go through it now to try confirm). So my question is: what purpose do the conditions (1) serve?
Let's replace (1) by just \[ f^j(x)\ll \frac {1}{(x+1)^\alpha }\hspace {20mm}(3)\] for $j=0,1$ and some $\alpha >1$. I'll try showing this is already enough for (2).
Since $f(x)\ll 1/x$ for large $x$ and $f(x)\ll 1$ for small $x$ showing (2) reduces to showing \[ \int _0^{1/2}\frac {|f(x)\log x|dx}{x}<\infty .\] This integral is (I'm going to ignore modulus signs on $f$) \[ \left (\frac {f(x)(\log x)^2}{2}\right )_0^{1/2}-\frac {1}{2}\int _0^{1/2}f'(x)(\log x)^2dx\] and, since $f'(x)\ll 1$ for small $x$, the integral here is $\ll t\left ((\log t)^2+\log t+1\right )$ so our work amounts to having to prove \[ \lim _{x\rightarrow 0}f(x)(\log x)^2=\lim _{x\rightarrow 0}x(\log x)^2=0\hspace {10mm}(4).\] Now (I'm going to ignore constants) \[ \lim _{x\rightarrow \infty }x(\log x)^3=\lim _{x\rightarrow 0}\frac {(\log x)^2/x}{1/x^2}=\lim _{x\rightarrow 0}x(\log x)^2=...=\lim _{x\rightarrow 0}x(\log x)^0=0\] so the second limit in $(4)$ is ok and the first is \[ \lim _{x\rightarrow 0}\frac {xf'(x)}{1/(\log x)^3}\ll \lim _{x\rightarrow 0}x(\log x)^3\] so also ok.