I have a problem I have been trying to integrate using by parts, but I am a bit stuck. We have that $x \geq 0$ and assume $f$ is a smooth function (so we have no issues at $0$) and compact in some interval $f \in C^{[a,b]}(R_{+})$, with $0<a<b<\infty$.
The equation is the following:
\begin{align} \|(Bf(x))\|_{L^2_{(R_+)}} & =\left(\int_0^\infty \left\lvert\frac{1}{x} \ \int_0^x f(t) \, dt\right\rvert^2 \ dx\right)^{1/2} \\ &\leq\left(\int_0^\infty \left\lvert\frac{1}{x}\right\rvert^{2} \ \left\lvert \int_0^x f(t) \, dt\right\rvert^2 \ dx\right)^{1/2} \\ &=\left(\int_0^\infty \frac{d}{dx} \left\lvert\dfrac{-1}{x}\right\rvert \ \left\lvert \int_0^x f(t) \ dt\right\rvert^2 \ dx\right)^{1/2} \\ &\leq \left(\int_0^\infty \frac{d}{dx}\left\lvert\dfrac{-1}{x}\right\rvert \ \int_0^x \left\lvert f(t) \right\rvert^{2} \ dt \ dx\right)^{1/2} \end{align}
Now to solve this I have tried setting $dv=\dfrac{d}{dx} \dfrac{-1}{x} dt$ and $v=\dfrac{d}{dx}\dfrac{-1}{x}$ (as $x \geq 0$ we can drop the absolute values), with $u=|f(t)|^{2}$ and $du=2\lvert f(t) \rvert dt$.
My other attempt was setting my $dv=\dfrac{d}{dx}\dfrac{-1}{x} dx$ and $v=\dfrac{-1}{x}$ (The dx's cancel when we integrate dv/dx) and setting $u=\int_0^x \lvert f(t) \rvert^2 dt$ and $du=2f(x)\int_0^x \lvert f(t) \rvert \ dt \ dx$. This was to try and solve the whole integral at once. This was the more promising of the two but I can't get an answer that doesn't seem to diverge/become unbounded.
Can someone please tell me where I am going wrong? What should I be setting my $u, du, v$ & $dv$ parts as?
The end result is to prove this is bounded, but with Holder's inequality and density arguments this part should be easy. This intermediate step I am just stuck at. Thank you for any help given!
$$\int_{0}^{\infty}\dfrac{d}{dx}\lvert\dfrac{1}{x}\rvert \ \int_{0}^{x}\lvert f(t) \rvert^{2} \ dt \ dx$$
Define $K(x)=\int_{0}^{x}\lvert f(t) \rvert^{2} \ dt $. You know what is $K'(x)$ right?(though not required!)
$$\int_0^\infty{d\over dx}\left(|{1\over x}|K(x)\right)dx=\int_0^\infty d\left(|{1\over x}|K(x)\right)=\left[{K(x)\over |x|}\right]_0^\infty$$
If you know $f$ Find $K(x)$ and you are done I think!