Given $f$ is a smooth function and $b_r = \sum_{d \mid r} a_d\mu(\frac{r}{d})$ with $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$. Then if (and only if) $f(k)$ is a global maxima in the region $c$ to $d$ for all $a_r \in R_{> 0}$ (where $R_{> 0}$ is the set of all positive reals) then:

$$ (d-c) f(k) \lim_{n \to \infty} \sum_{r=1}^n \frac{a_r}{n} - \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r} {r^s}\right)\int_c^d f(x)dx > 0 $$

Obviously this is too powerful a formula. My best guess is the proof is wrong? On the off chance I haven't made a mistake then is it pragmatic to use this formula to find the maxima?


Let's say I have a function which has a maxima $f(b)$ in the region $c$ and $d$. Now, let's assume there is a solution $f(k)$ which I'm uncertain if it's a global maxima. Then,

$$ a_i(f(k) - f(l)) > 0$$

where $l$ is an arbitrary number and $a_i$ is a fudge factor which enables the inequality to still be true even when our assumption ($f(k)$ is not a global maxima) is false: $f(k) - f(c) < 0$ by simply becoming a negative number. Now consider:

$$ a_1(f(k) - f(c + \frac{1}{n})) > 0$$ $$ a_2(f(k) - f(c + \frac{2}{n})) > 0$$ $$ a_3(f(k) - f(c + \frac{3}{n})) > 0$$ $$ \vdots $$ $$ a_n(f(k) - f(c + \frac{n(d-c) }{n})) > 0$$

Adding the above (and removing the point $f(b) - f(b)$:

$$ \sum_{r=1}^n a_r ( f(k) - f(c + \frac{r}{n})) > 0 $$

Multiplying by $\frac{(d-c) }{n}$ both sides and taking $n \to \infty$:

$$ \lim_{n \to \infty} \sum_{r=1}^n a_r \frac{(d-c) }{n}( f(k) - f(c + \frac{r}{n})) > 0 $$

We can integrate the following using this formula which claims:

Let $b_r = \sum_{d \mid r} a_d\mu(\frac{r}{d})$.

Claim: If $\lim_{n \to \infty} \frac{\log^2(n)}{n}\sum_{r=1}^n |b_r| = 0$ and $f$ is smooth, then but, for any smooth $f$, that $$\lim_{n \to \infty} \sum_{r=1}^n a_rf\left(\frac{rk}{n}\right)\frac{k}{n} = \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r} {r^s}\right)\int_0^k f(x)dx$$


$$ (d-c) f(k) \lim_{n \to \infty} \sum_{r=1}^n \frac{a_r}{n} - \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r} {r^s}\right)\int_c^d f(x)dx > 0 $$

If there exists any solution where all $a_i > 0$ and

$$ (d-c) f(k) \lim_{n \to \infty} \sum_{r=1}^n \frac{a_r}{n} - \left(\lim_{s \to 1} \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{a_r} {r^s}\right)\int_c^d f(x)dx < 0 $$

then $f(k)$ is not a global maxima.

  • $\begingroup$ I have some questions. (i) What is $m$ in the definition of $b_r$ and the statement of the claim you borrowed later on? (ii) What exactly is your claim? The "iff" part makes it confusing, should than be an "if"? If so, you can just use the fact that $f(k)$ is a maxima when proving the claim (you seem to be trying to avoid using this). Also, could you clarify what you mean by "for all possible $a_r>0$" in the initial claim? $\endgroup$
    – Besfort
    Apr 17, 2020 at 23:01
  • $\begingroup$ @Besfort (i) the $m$ is a typo it should be $r$ (now edited). (ii) The claim is if $f(k)$ is not a maxima then it forces some $a_i$ to be negative (or the opposite). I've also replaced iff with if and only if. (iii) $a_r \in R_{ > 0}$ where $ R_{> 0}$ is the reals greater than $0$. $\endgroup$ Apr 17, 2020 at 23:17
  • $\begingroup$ For $a_n \ge 0$, then $\lim_{N\to \infty} \frac1N \sum_{n=1}^N a_n=A$ exists implies that $\lim_{s\to 1^+} (s-1)\sum_{n=1}^\infty a_n n^{-s} = A$ $\endgroup$
    – reuns
    Apr 18, 2020 at 5:37


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