# Jensen's Formula and the number of zeros.

I am doing a question that requires me to find the number of zeros in a disk when $$f$$ is non-constant, bounded and analytic on $$D$$. I saw that the Wikipedia page of Jensen's formula has the following information:

Jensen's formula can be used to estimate the number of zeros of analytic function in a circle. Namely, if f is a function analytic in a disk of radius R centered at $$z_0$$ and if |f| is bounded by M on the boundary of that disk, then the number of zeros of f in a circle of radius r < R centered at the same point $$z_0$$ does not exceed Does anyone know how this formula is derived?

## 3 Answers

We can obviously assume $$z_0=0$$, we also assume that $$f$$ doesn’t vanish on the circle of radius $$r$$ (not a big deal, we can take $$r’> r$$ small enough going to $$r$$).

Let $$r < t < R$$. Let $$Z_t$$ be the “set” of zeroes of $$f$$ in the disk centered at $$z_0$$ with radius $$t$$ (counted with multiplicity). We will make $$t$$ go to $$R$$ – we can choose $$t$$ such that $$f$$ does not vanish on the circle centered at $$z_0$$.

The exact Jensen formula, combined with the inequality $$|f| \leq M$$, gives $$\log{\frac{M}{|f(0)|}} \geq -\sum_{z \in Z_t}{\log{\frac{|z|}{t}}}=\sum_{z \in Z_t}{\log{\frac{t}{|z|}}} \geq \sum_{z \in Z_r}{\log{\frac{t}{|z|}}} \geq \sum_{z \in Z_r}{\log{\frac{t}{r}}}\geq |Z_r|\log{\frac{t}{r}}.$$

Now just take $$t \rightarrow R$$.

Let $$f(z)$$ be analytic function such that $$f(0)\ne0$$, and let $$z_1,z_2,\dots,z_m$$ denote the zeros of $$f(z)$$ satisfying $$|z_k|. Then Jensen's formula gives

$$\int_0^1\log|f(Re^{2\pi it})|\mathrm dt=\log\left|f(0)\cdot{R\over z_1}\cdot{R\over z_2}\cdots{R\over z_m}\right|$$

Now, let $$|f(z)|\le M$$ for $$|z|\le R$$, then we have

$$\log\left|{R\over z_1}\cdot{R\over z_2}\cdots{R\over z_m}\right|\le\log M-\log|f(0)|$$

Let $$0, so we can classify $$z_1,z_2,\dots,z_m$$ into two classes:

$$\log\left|{R\over z_1}\cdot{R\over z_2}\cdots{R\over z_m}\right| =\sum_{|z_k|\le r}\log\left|R\over z_k\right|+\sum_{|z_j|>r}\log\left|R\over z_j\right|$$

Because $$\log|R/z_k|$$ is at least $$\log R/r$$, we have

$$N(r)\log\frac Rr\le\sum_{|z_k|\le r}\log\left|R\over z_k\right|$$

where $$N(r)$$ denotes the number of zeros of $$f(z)$$ satisfying $$|z|\le r$$. Now, by the transitivity of inequalities, we conclude that the number of zeros of $$f(z)$$ within $$|z|\le r$$ satisfies the following inequality:

$$N(r)\le{1\over\log R/r}\log\left|M\over f(0)\right|$$

Hope this will help you!

Lemma:

For an analytic $$f$$ with no zeros on $$|z|=r$$ and whose zeros inside $$|z|=r$$ are located at $$\left\{\alpha_k\right\}_{k=1}^n$$ with multiplicities, define $$g(z)=f(z)\prod_{k=1}^n\frac{r^2-\bar\alpha_kz}{r(z-\alpha_k)}\tag1$$ Then $$|g(z)|=|f(z)|$$ when $$|z|=r$$ and $$g$$ has no zeros in $$|z|\le r$$.

Proof:

The factors of $$z-\alpha_k$$ remove the zeros inside $$|z|=r$$, so $$g$$ has no zeros for $$|z|\le r$$.

If $$|z|=r$$, then \begin{align} \left|\frac{r^2-\bar\alpha z}{r(z-\alpha)}\right| &=\left|\frac{\bar z}r\frac{r^2-\bar\alpha z}{r(z-\alpha)}\right|\tag{2a}\\ &=\left|\frac{\bar zr^2-\bar\alpha|z|^2}{r^2(z-\alpha)}\right|\tag{2b}\\ &=\left|\frac{r^2(\bar z-\bar\alpha)}{r^2(z-\alpha)}\right|\tag{2c}\\[6pt] &=1\tag{2d} \end{align} Explanation:
$$\text{(2a)}$$: $$|\bar z|=|z|=r$$
$$\text{(2b)}$$: distribute $$\bar z$$ over the denominator
$$\text{(2c)}$$: collect the factors of $$|z|^2=r^2$$
$$\text{(2d)}$$: $$|\bar z-\bar\alpha|=|z-\alpha|$$

Thus, $$\prod\limits_{k=1}^n\frac{r^2-\bar\alpha_kz}{r(z-\alpha_k)}$$ has absolute value $$1$$ when $$|z|=r$$; and therefore, $$|g(z)|=|f(z)|$$ when $$|z|=r$$.

$$\large\square$$

Jensen's Formula

Taking the average over $$|z|=r$$ gives \begin{align} \frac1{2\pi}\int_0^{2\pi}\log\left|f\!\left(re^{i\theta}\right)\right|\,\mathrm{d}\theta &=\frac1{2\pi}\int_0^{2\pi}\log\left|g\!\left(re^{i\theta}\right)\right|\,\mathrm{d}\theta\tag{3a}\\ &=\log|g(0)|\tag{3b}\\ &=\log|f(0)|+\sum_{k=1}^n\left(\log(r)-\log|\alpha_k|\right)\tag{3c} \end{align} Explanation:
$$\text{(3a)}$$: the Lemma says $$|g(z)|=|f(z)|$$ when $$|z|=r$$
$$\text{(3b)}$$: the Lemma implies that $$\log|g(z)|$$ is harmonic for $$|z|\le r$$
$$\text{(3c)}$$: take the log of the absolute value of $$(1)$$ at $$z=0$$

$$(3)$$ is Jensen's Formula

Counting Zeros

If we know that all the zeros are inside $$|z|=r$$ and we know the average of $$\log|f|$$ on $$|z|=R$$, \begin{align} \frac1{2\pi}\int_0^{2\pi}\log\left|f\!\left(Re^{i\theta}\right)\right|\,\mathrm{d}\theta-\log|f(0)| &=\sum_{k=1}^n\left(\log(R)-\log|\alpha_k|\right)\tag{4a}\\ &\ge n(\log(R)-\log(r))\tag{4b} \end{align} So we have that $$n\le\frac1{\log(R)-\log(r)}\left(\frac1{2\pi}\int_0^{2\pi}\log\left|f\!\left(Re^{i\theta}\right)\right|\,\mathrm{d}\theta-\log|f(0)|\right)\tag5$$