Get ready Part 32

The key observation :

Consider the complexe plane, $\eta > \varepsilon > 0$ and $z_0\in\mathbb{C}$ . Then, $\underset{|\alpha - z_0|\geq \eta}{\inf} \left( \underset{|z-z_0|\leq \varepsilon}{\min} \left|\dfrac{z-\alpha}{\alpha}\right|\right) = C(z_0, \ \eta,\ \varepsilon) > 0$

Proof the observation

Indeed, if $|\alpha - z_0|\gg 1$ , then $\underset{|z-z_0|\leq \varepsilon}{\min} \left|\dfrac{z-\alpha}{\alpha}\right| \approx 1 \geq 1/2$ . Therefore we are bring back to a compact set : $\eta \leq |\alpha-z_0|\leq R$ for some $R>\eta$ (we call this compact set $K$ ).

But because $\alpha \longmapsto \underset{|z-z_0|\leq \varepsilon}{\min} \left|\dfrac{z-\alpha}{\alpha}\right|$ is continuous, on the compact set $K$ , it is bounded and reaches its bounds. Now observes that for every $\alpha\in K$ , $\underset{|z-z_0|\leq \varepsilon}{\min} \left|\dfrac{z-\alpha}{\alpha}\right| > 0$ (In fact $\geq \dfrac{\eta-\varepsilon}{R}$ )

Therefore, $\underset{\alpha \in K }{\inf} \left( \underset{|z-z_0|\leq \varepsilon}{\min} \left|\dfrac{z-\alpha}{\alpha}\right|\right) = \underset{\alpha \in K}{\min} \left( \underset{|z-z_0|\leq \varepsilon}{\min} \left|\dfrac{z-\alpha}{\alpha}\right|\right) > 0$

Which ends the proof of the observation.

Remark : the denpence on $z_0$ is hidden in $R$ .

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Now we are ready to answer our problem. Let $P$ polynomial of degree 1999.

Without loss of generality (by homogeneity), we can suppose $P(0)=1$ .

Therefore $P=\displaystyle\prod_{k=1}^{1999} \dfrac{X-\alpha_k}{\alpha_k}$ where the $\alpha_k$ 's are the complexe roots of $P$ .

Consider 2000 following disks : $B(-1+\frac{2k-1}{2000}, \ \frac{1}{2000})$ for $k=1, \ 2, \cdots, \ 2000$ . Where $B(z,r)$ is the disk center at $z$ with radius $r$ . The disks are disjoint and there are only 1999 roots so, at least one of them does not contain a root (let's say the $m^{th}$ ball). Call it center $z_m$ and apply the key observation with $\varepsilon=\frac{1}{8000}$ and $\eta=\frac{1}{4000}$ :

$|P(x)| \geq C(z_m, \ \frac{1}{4000},\ \frac{1}{8000})^{1999} \hspace{1cm} \forall x\in [z_m-\frac{1}{8000}, \ z_m + \frac{1}{8000}]$

So, $|P(x)| \geq \underset{ h = 1, \ 2, \cdots, \ 2000}{\min} C(z_h, \ \frac{1}{4000},\ \frac{1}{8000})^{1999} = Q( \frac{1}{4000},\ \frac{1}{8000})^{1999} >0 \hspace{1cm} \forall x\in [z_m-\frac{1}{8000}, \ z_m + \frac{1}{8000}]$

Now, $Q( \frac{1}{4000},\ \frac{1}{8000})^{1999} =: R_{1999}$ no more depends on $P$ at all.

Finally, $\displaystyle\int_{-1}^1 |P| \geq \displaystyle\int_{z_m-\frac{1}{8000}}^{z_m + \frac{1}{8000}} |P| \geq \dfrac{R_{1999} }{4000} > 0$

So, $|P(0)|=1 \leq \dfrac{4000}{R_{1999}} \displaystyle\int_{-1}^1 |P|$

I maybe wrong but we just need to set the constant variable to be less than zero since the RHS and LHS are always positive(can be proven easily).