# Estimating $\int_0^x f(x-t)f'(t)dt$

I'm attempting to estimate $$\int_0^x f(x-t)f'(t)dt$$ in terms of a simple asymptotic expression with an error term for some 'well-behaved' functions, namely $$f = O(x)$$, of class $$C^1$$ or higher, with $$f$$ (asymptotically) increasing and $$f'$$ (asymptotically) monotonic. (e.g. $$\sqrt[3]{x}$$, $$\frac{x}{\log{x}}$$, $$\sqrt{x}\log{x}$$).

If $$f$$ is not positive or well-defined in $$(0,x)$$ we can just change the domain of integration to $$(\alpha, x-\alpha)$$, this change doesn't modify any asymptotic result anyway.

Example. For $$f(x) = \sqrt{x}$$, it's not hard to show that $$\int_0^x f(x-t)f'(t)dt \sim \frac{\pi}{4}x$$.

More generally, I suspect that $$\int_0^x f(x-t)f'(t)dt = Cf(x)^2 + O(f(x))$$, for some constant $$C$$ depending on $$f$$, but the only thing that I have sketched a proof so far was that $$\int_0^x f(x-t)f'(t)dt \asymp f(x)^2$$, as follows:

Firstly, by the order of $$f$$, holds $$f(x-t) \gg f(x)-f(t)$$ as $$x\to\infty$$. So:

$$\int_0^x f(x-t)f'(t)dt \gg f(x)\int_0^x f'(t)dt - \int_0^x f(t)f'(t)dt \gg f(x)^2$$.

Secondly, given that $$f$$ is increasing,

$$\int_0^x f(x-t)f'(t)dt < f(x)\int_0^{x/2} f'(t)dt + \int_{x/2}^x f(t)f'(t)dt \ll f(x)^2$$.

The problem is that I have no clue how to prove (or if I have to add more hypothesis over $$f$$ in order to prove) my suspicion. I would appreciate any suggestion.

To simplify things, I assume that $f$ is increasing and $f'$ is decreasing* on $(0,\infty)$; otherwise shift the variable. Begin with $$\begin{split} \int_0^x f(x-t)f'(t)\,dt &= \int_0^x \left(f(0)+\int_0^{x-t}f'(s)\,ds\right)f'(t)\,dt \\&=f(0)f(x) + \iint_{s+t\le x}f'(s)f'(t)\,ds\,dt \end{split}$$ (The variables of integration are always nonnegative here).
Also, $f(x)^2= -f(0)^2+2f(0)f(x)+(f(x)-f(0))^2$ where the last term is $$(f(x)-f(0))^2 = \left(\int_0^{x}f'(t)\,dt\right)^2 = \iint_{s,t\le s}f'(s)f'(t)\,ds\,dt$$ Since $f'$ is nonnegative and $\{(s,t):s+t\le x\}\subset \{(s,t):s,t\le x\}$, it follows that $$\int_0^x f(x-t)f'(t)\,dt \le f(x)^2 + O(f(x)) \tag{1}$$ On the other hand, $\{(s,t):s+t\le x\}\supset \{(s,t):s,t\le x/2\}$, which leads to $$\int_0^x f(x-t)f'(t)\,dt \ge f(x/2)^2 + O(f(x)) \tag{2}$$ And since $f'$ is decreasing*, $f(x)-f(x/2)\le f(x/2)-f(0)$, which implies $f(x)\le 2f(x/2)+f(0)$, hence $f(x)^2\le 4f(x/2)^2 + O(f(x))$. So, (1) and (2) imply $$\frac14 f(x)^2 + O(f(x)) \le \int_0^x f(x-t)f'(t)\,dt \le f(x)^2 + O(f(x)) \tag{3}$$
I don't believe that the limit $\frac{1}{f(x)^2}\int_0^x f(x-t)f'(t)\,dt$ exists without additional assumptions on $f$. The problem is that $f'$ may decrease somewhat irregularly, so that the integrals of $f'(s)f'(t)$ over a square and its left-lower diagonal half don't have to be asymptotically proportional.
(*) You did not say that $f'$ is decreasing, only that it's monotone. But if $f'$ is increasing, then due to the assumption $f(x)=O(x)$ we have $f'(x)\to L<\infty$, hence $f(x)\sim Lx$, from where the conclusion $\int_0^x f(x-t)f'(t)\,dt\sim \frac12 L^2 x^2\sim \frac12 f(x)^2$ follows.