# Is $\sum_{k\leqslant n} f'(k)f'(n-k) \asymp f'(n)f(n)$ when $f'$ is positive decreasing?

In this answer of a question of mine, the user Homegrown Tomato gave a nice argument that somewhat shows that

$$\int_{\substack{t+s\leqslant x \\ t,s \geqslant 0}} f'(t)f'(s)dtds \asymp f(x)^2.$$

I actually improved this statement (under the hypotheses of my question) and concluded that in fact it's asymptotic to $C\cdot f(x)^2$ for some positive real constant $C$, but I'm not sure whether that's beside the point of my actual question here or not.

I'm trying to estimate $\sum_{k\leqslant n} f'(k)f'(n-k)$ for $C^0$ decreasing $f'$, like $f'(x) = 1/\log{x}$, or $x^{-1},x^{-1/2},x^{-1/2}\log{x}$, etc. Unlike the case of $\sum_{k\leqslant n} f'(k)$, where to use the integral as an estimate is almost direct, I think something like

$$\sum_{k\leqslant n} f'(k)f'(n-k) \overset{?}\sim \int_{0\leqslant t\leqslant n} f'(t)f'(n-t)dt \tag{1}$$

must be true, but I was not able to prove. Well, still on that thought, the integration here is occuring on part of the boundary of a triangle, the same triangle of the integration on my previous question that I quoted above! So I thought in to try to relate the two questions, using something like Stokes or Reynolds' transport theorem, in order to get

$$\int_{0\leqslant t\leqslant x} f'(t)f'(x-t)dt \overset{?}\approx \frac{d}{dx}\int_{\substack{t+s\leqslant x \\ t,s \geqslant 0}} f'(t)f'(s)dtds \overset{?}\approx f'(x)f(x) \tag{2}.$$

So what I'm asking is: Is there a way to formalize (1) and (2), or at least one of them? And if no (or even if yes), there is another way (an easier, preferably haha) to show the statement of the question? In the worst-case scenario, a counterexample will be welcome hahaha

Concerning (1), you could use the integral convergence test: if $f'$ decreases not only towards $\infty$, but also towards $-\infty$, and $f'$ does not change sign, then $f'(\cdot) \space f'(n - \cdot)$ is a positive decreasing function of $\cdot$, which shows that $\sum \limits _{k=1} ^n = \sum \limits _{k=1} ^\infty - \sum \limits _{k=n+1} ^\infty$ behaves like $\int \limits _1 ^\infty - \int \limits _{n+1} ^\infty = \int \limits _1 ^n$.

Concerning (2), the integral in the middle term can be rewritten (using Fubini's theorem) as $\int \limits _0 ^x f'(s) \int \limits _0 ^{x-s} f'(t) \Bbb d t \Bbb d s$. Deriving with respect to $x$ one gets:

$$f'(x) \int \limits _0 ^0 f'(t) \Bbb d t - f'(0) \int \limits _0 ^x f'(t) \Bbb d t \cdot 0'+ \int \limits _0 ^x f'(s) \Big( f'(x-s) - f'(0) \cdot 0'\Big) \Bbb d s = \int \limits _0 ^x f'(s) f'(x-s) \Bbb d s ,$$

i.e. you get precisely the first equality in (2) (and it is true equality, not asymptotic behaviour).

(A note on the derivation with respect to $x$ if you feel puzzled:

$$\frac {\Bbb d} {\Bbb d x} \int \limits _{f(x)} ^{g(x)} F(x,t) \Bbb d t = F \big(x, g(x) \big) g'(x) - F \big(x,f(x) \big) f'(x) + \int \limits _{f(x)} ^{g(x)} \frac {\partial F} {\partial x} (x, t) \Bbb d t ,$$

i.e. you derive according to a rule similar to Leibniz's product rule: first you derive the integral and keep the integrand unchanged, next you keep the integral but derive the integrand.)

• Hmm, in the first case I was thinking in function defined on $(0,+\infty)$, like $1/\sqrt{x}$, I don't know if I can use that in this case, but I'll try! In (2) it seems pretty right, but I didn't verified yet (I have a little difficulty with derivation under integral sign haha), but what did you mean by $0'$? – Alufat Jun 10 '15 at 6:54
• @ChrisTáfula: Oh, $0'$ is an idiotic and pedantic way of reminding you that $0$ must be also derived - which, of course, gives you $0$, so those terms vanish too. – Alex M. Jun 10 '15 at 9:12