Minimise $\sum \left\lvert x_j \right\rvert^2 + \sum \left\lvert y_j \right\rvert^2$ subject to the constraints $\sum x_j = 1$ and $\sum y_j = 0$.

Question

Let $X := \mathbb{C}^n$ with the inner product defined by $$ \big\langle \left( x_1, \ldots, x_n \right) \, , \, \left( y_1, \ldots, y_n \right) \big\rangle := \sum_{j=1}^n x_j \overline{y_j}. $$ Let the set $M$ be defined by $$ M := \left\{ \, \left( x_1, \ldots, x_n \right) \in \mathbb{C}^n \colon \sum_{j=1}^n x_j = 1 \, \right\}. $$ Then how to find a vector (or vectors) of minimum norm in $M$?

In other words, how to solve the following version of the above minimisation problem?

Let $x_1, \ldots, x_n, y_1, \ldots, y_n$ be some real variables. Then how to minimize $$ f \left( x_1, \ldots, x_n, y_1, \ldots, y_n \right) := \sum_{j=1}^n \left\lvert x_j \right\rvert^2 + \sum_{j=1}^n \left\lvert y_j \right\rvert^2, $$ subject to the constraints $$ \sum_{j=1}^n x_j = 1 \qquad \mbox{ and } \qquad \sum_{j=1}^n y_j = 0? $$

I've not much of an idea of how to even start my solution. Hence no attempt shown.

I'd appreciate an answer that is as detailed and elementary as possible.

Answer 1

The minimum value is $\frac1 n$. By Cauchy Scwarz inequlity $|\sum x_i|^{2}\leq n\sum |x_i|^{2}$ which shows that $\sum |x_i|^{2}+\sum |y_i|^{2} \geq \frac 1 n$. This lower bound is attained when $x_i=\frac 1 n$ for each $i$ and $y_i=0$ for each $i$.