I'm reading by calculus lectures and there is the following task:
How can we prove that if $f$ is even, strictly monotonically decreasing on $[0, \infty)$ and $f \in C^1(\mathbb{R})$ then it's Fourier transform is $o(\frac{1}{y})$ for $y \to \infty$?
First attempt:
I've tried to multiply it by $y$, write complex exponent using Euler's formula (if we do it we can say that one of integrals is zero). and now I need to prove that $yF[f(x)](y) \to 0$ as $y \to \infty$. if we would use Riemann–Lebesgue lemma we could say that $F[f(x)](y) \to 0$ as $y \to \infty$. but the problem is that we have $y$ factor and it goes to infinity.
Second attempt:
I've tried to use formula for Fourier Transform of a derivative: $F[f'] = (iy)F[f]$. Now we need to show that $F[f'] \to 0$ as $y \to \infty$ and it follows from Riemann–Lebesgue lemma. but for doing all this we need to prove that $f \in L_1(\mathbb{R})$ and $f' \in L_1(\mathbb{R})$ and we only know that $f$ is even and decreasing... I don't know how to use parity and other given information (decreasing and $C_1(\mathbb{R}))$ to prove that $f' \in L_1(\mathbb{R})$.
could you please help me solve the task?