I know that $ \int_0^\infty \frac{|f(x)|^2}{x} < \infty $, which condition should I have to prove $ \int_0^\infty \frac{|f(x)|^2}{x^2} < \infty $?

Question

I'm working with the wavelet transform, and I'm facing a problem proving that

$$ \int_0^\infty \frac{|f(x)|^2}{x^2}\mathrm{d}x $$

converges.

The only thing I know is that $f(x)$ is the Fourier transform of some analytical mother wavelet which indicates that

$$ \int_0^\infty \frac{|f(x)|^2}{x}\mathrm{d}x < \infty $$.

I have no idea about what conditions should be added to prove that and the ideas to prove that.

Thanks for helping.

Answer 1

The integral converges if and only if $\int_1^\infty |f(1/t)|^2 dt$ is finite. Here are two examples of conditions that ensure this holds:

• $f(0)=0$ and $f$ is differentiable at $0$.
• $\lim_{x\to 0^+}|f(x)|/x^{\rho}$ exists for some $\rho>1/2$.

Proof that the integral converges if and only if $\int_1^\infty |f(1/t)|^2 dt<\infty$:

Write

$$\int_0^\infty \frac{|f(x)|^2}{x^2}dx = \int_0^1 \frac{|f(x)|^2}{x^2}dx + \int_1^\infty \frac{|f(x)|^2}{x^2}dx $$ The latter integral converges by comparison: $|f(x)|^2/x^2\leq |f(x)|^2/x$ holds for every $x\geq 1$. The first integral is improper because the integrand could be singular at $0$, so we’re interested in when $$\lim_{t\to 0^+}\int_t^1 \frac{|f(x)|^2}{x^2}dx\text{ exists}$$ But applying the substitution $x = 1/u$, we see that $$\int_t^1 \frac{|f(x)|^2}{x^2}dx= \int_1^{\frac{1}{t}}\left|f\left(\frac{1}{u}\right)\right|^2du$$ Since $1/t\to\infty$ as $t\to 0^+$, the claimed equivalence immediately follows.