A question about average deviation of given $n$ complex numbers

This question just came to my mind and I have no idea as to how to approach it. Let $$z_1,z_2,\dots,z_n$$ be $$n$$ be any complex numbers in the unit disc $$|z| \leq 1.$$ Consider a complex function on the unit disc wiith real values $$f(z)=\sum_{i=1}^n \frac{|z-z_i|}{n}.$$ My questions:

• Does there exist a $$z \in |w| \leq 1$$ so that $$f(z)=|z|$$?
• If not can we make $$f(z)$$ arbitrarily close to $$|z|$$ for some $$z \in |w| \leq 1$$?
• What about the maximum and the minimum value of $$f(z)$$?

Lots of thanks for any responces\hints\suggestions

• If $f(z)$ can be made arbitrarily close to $|z|$, then shouldn’t compactness of the closed unit disc and continuity of $f$ guarantee that equality is attained somewhere? Mar 28, 2022 at 9:47
• @yes indeed.I thought that might be easier approach to reach to the possibility of equality Mar 28, 2022 at 9:58
• The minimum of the expression is a very famous problem called the Fermat-Weber location problem. See here for some history ssabach.net.technion.ac.il/files/2015/12/BS2015.pdf Mar 28, 2022 at 11:22
• The notation $z\in|z|\leq 1$ should probably be more like $z\in\{w:|w|\leq 1\}$. Mar 28, 2022 at 15:21
• Yes ,it should be .Thanks for pointing it out Mar 28, 2022 at 17:23

The answer is no. E.g., let $$n=3$$ and $$z_j=e^{i(j-1)2\pi/3}$$ for $$j=1,2,3$$. Then $$f(z)>|z|+15/100>|z|$$ if $$|z|\le1$$.

This counterexample generalizes to any $$n\ge3$$. Indeed, take any $$n\ge3$$ and let $$z_j=e^{i(j-1)2\pi/n}$$ for $$j=1,\dots,n$$. Then $$f(z)=\frac1n\,\sum_{j=1}^n f_j(z),$$ where $$f_j(z):=|z-z_j|$$. Note that for each $$j$$ the function $$f_j$$ is convex and, moreover, $$f_j$$ is strictly convex on any straight line not through the point $$z_j$$. Since $$n\ge3$$, there is no straight line passing through all the points $$z_1,\dots,z_n$$. So, the function $$f=\frac1n\,\sum_{j=1}^n f_j$$ is strictly convex.

Also, for any real $$a$$ and $$b$$, the derivative of $$f$$ at $$0$$ along the vector $$(a,b)=a+ib$$ is \begin{aligned}&\frac d{ds}\,f(s(a+ib))\Big|_{s=0} \\ &=-\frac1n\,(a,b)\cdot\sum_{j=1}^n \Big(\cos\frac{(j-1)2\pi}n,\sin\frac{(j-1)2\pi}n\Big) \\ &=-\frac1n\,(a,b)\cdot(0,0)=0, \end{aligned} where $$\cdot$$ denotes the dot product.

Thus, the strictly convex function $$f$$ has a unique minimum at $$0$$, and the minimum value of $$f$$ is $$1$$. That is, $$f(0)=1 for all $$z$$ such that $$0<|z|\le1$$.

So, for all $$z$$ such that $$0<|z|\le1$$, we have $$f(z)>1\ge|z|$$ and hence $$f(z)\ne|z|$$. Also, $$f(0)=1\ne0$$. We conclude that there is no $$z$$ such that $$|z|\le1$$ and $$f(z)=|z|$$, which proves the claim.

On the other hand, the answer is yes for $$n=1$$ and $$n=2$$.

• @losif Pinels ,you have taken the points on the unit circle and not general points in the unit disc.Could you kindly clarify ? Mar 29, 2022 at 13:36
• @sajjadveeri : You asked if for any $n$ points in the unit disk there is some $z$ in the unit disk with $f(z)=|z|$. My answer proves that this is not true; that is, for some $n$ points in the unit disk there is no $z$ in the unit disk with $f(z)=|z|$. If you had asked if for some $n$ points in the unit disk there is some $z$ in the unit disk with $f(z)=|z|$, then the answer would have trivially been yes -- by letting e.g. $z_j=0$ for all $j$ and then choosing $z=0$. I hope you can understand this logic. Mar 29, 2022 at 13:59
• Previous comment continued: Try not to use vague terms such as "general points"; use quantifiers "for all/any" and "there exist(s)" instead. Mar 29, 2022 at 13:59
• Thank you for your edifying comment.Valuable Advice noted! Mar 29, 2022 at 14:19