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Consider $f_n(x) = \min_{|z|=x} \Re \sum_{j=1}^{n} \frac{z^j}{j}$, a real function of positive variable $x>0$.

I am interested in lower bounds on $f_n(x)$. Specifically, I ask: what lower bounds can be given on $f_n(x)$ in the regime where $x=1+O(1/n)$ and $n$ tends to $\infty$?

Trivially, $f_n(x) \ge - \sum_{j=1}^{n} \frac{x^j}{j}$ which has order of magnitude $-\log n$ (times a constant) in the aforementioned regime. However, equality cannot be achieved here, since $z^j$ cannot equal to $-x^j$ for both $j=1$ and $j=2$. I don't even know if $f_n(x)$ is eventually negative.

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    $\begingroup$ $f_n(x)$ is indeed eventually negative, moreover $\limsup f_n(x) \le -\log(2) < 0$. Indeed, $f_n(x)$ is a harmonic function, thus it satisfies the minimum principle. On the other hand for $x\in (-1, 1)$ we have $f_n(x)\to -\log(1-x)$, so taking $x = -1 + \varepsilon$ gives the bound. I'm not sure if the $\limsup$ is $-\infty$ though, wouldn't be surprised if it is bounded. $\endgroup$
    – Aleksei Kulikov
    Commented Dec 8, 2021 at 20:43
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    $\begingroup$ Take a look at Lemma 2 in math.dartmouth.edu/~carlp/pvrev.pdf . $\endgroup$
    – Lucia
    Commented Dec 9, 2021 at 5:28
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    $\begingroup$ From Lucia’s reference and a simple comparison of the values at $z$ and $z/|z|$ I think it’s not hard to give a constant lower bound. $\endgroup$
    – Aleksei Kulikov
    Commented Dec 9, 2021 at 8:45
  • $\begingroup$ Thank you both. @AlekseiKulikov If this is indeed possible I would accept this as an answer. (Keep in mind I am not a complex analyst, so I am not sure I followed your first comment.) $\endgroup$
    – Ofir Gorodetsky
    Commented Dec 9, 2021 at 9:01
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    $\begingroup$ Surely $f_n(x)$ is negative in the desired regime, as the value of the sum is negative at $z=-x$, due to Leibnitz. $\endgroup$
    – Ilya Bogdanov
    Commented Dec 10, 2021 at 14:23

1 Answer 1

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So, we have a function $u_n(z) = \Re \sum_{j = 1}^n \frac{z^j}{j}$. As a real part of an analytic function, it is a harmonic function. We are interested in its behaviour on the circle $|z| = x = 1 + \frac{c}{n}$, where $c > 0$ is some constant and we want to estimate the minimal value of $u_n$ on it. Let me start with the upper bound from my comment:

$u_n$ is a harmonic function, thus it satisfies the minimum principle in the form $$\min_{|z| \le x} u_n(x) = \min_{|z| = x} u_n(x)$$ (if you only heard of the maximum principle, then it is it applied to $-u_n$).

Thus, we have $f_n(x) \le u_n(-1)$. On the other hand $u_n(-1)\to -\log(2)$, so $\limsup f_n(x) \le -\log(2)$ (in reality, $u_n(-1) \le -\log(2) + \frac{1}{n}$, so actually $f_n(x) \le -\log(2) + \frac{1}{n}$).

Now for the lower bound. According to the Lemma 2 from the link provided by Lucia, we have $f_n(1) \ge -\log(2) - \frac{2}{n}$. For $z$ with $|z| = x$ let us denote $w = \frac{z}{|z|}$. We have

$$u_n(z) \ge u_n(w) - |u_n(z) - u_n(w)| \ge -\log(2) - \frac{2}{n} - |u_n(z)-u_n(w)|.$$

So, if we can give a uniform upper bound on $|u_n(z)-u_n(w)|$ then we get a uniform lower bound on $f_n(x)$. We have $$u_n(z) - u_n(w) = \sum_{j = 1}^n \frac{z^j-w^j}{j} = \sum_{j=1}^n w^j \frac{x^j-1}{j}.$$

Since $x = 1 + \frac{c}{n}$ for $j\le n$ we have $1\le x^j\le 1 + \frac{Cj}{n}$ for some $C$ depending on $c$ (something like $C = e^c$ should work). Plugging this in and taking the absolute values (remember that $|w| = 1$) we get $$|u_n(z) - u_n(w)| \le \sum_{j = 1}^n \frac{C}{n} = C,$$ which proves the desired bound. Note though that the final constant $D$ in $f_n(x) \ge -D$ depends on $c$ even if we are concerned with only the big values of $n$. I don't know if it can be made approaching $\log(2)$ (it definitely can't be smaller by the minimum principle argument from above).

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