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evinda

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We consider the set $K=\{ x \in \mathbb{R}^n: \sum_{j=1}^n |x_j|^2 \leq 1 \}$ and let $ \in \mathbb{R}^n$. Check if the euclidean distance of $z$ from $K$ is attained.

I want to show that $K$ is closed and convex. Then by a theorem, the distance will be attained by a unique point.

I have tried to show that $K$ is closed as follows.

Let $(\overline{x_m})$ be a sequence in $K$ such that $\overline{x_m} \to x \in \mathbb{R}^n$.

We will show that $x \in K$.

$\overline{x_m}=(x_{m1}, x_{m2}, \dots, x_{mn}), x=(x_1, x_2, \dots, x_n)$

Then $|x_{mi}-x_i| \leq ||\overline{x_m}-x||_2= \left( \sum_{j=1}^n |x_{mj}-x_j|^2\right)^{\frac{1}{2}} \to 0$.

Thus $x_{mi} \to x_i$ while $m \to +\infty$ for all $i=1, \dots, n$ with $\sum_{j=1}^n |x_{mj}|^2 \leq 1$ for all $m=1,2, \dots$

Thus $\sum_{j=1}^n |x_j|^2 \leq 1$ and so $\overline{x} \in K$.

In order to show that $K$ is convex , I have thought the following:

Let $x,y \in K, 0<\lambda<1$.

We will show that $(1- \lambda)x+ \lambda y \in K$.

We have $x=(x_1, \dots, x_n), y=(y_1, \dots, y_n)$ and so $(1- \lambda)x+ \lambda y=((1- \lambda)x_1+ \lambda y_1, \dots, (1-\lambda)x_n+ \lambda y_n )$

So it suffices to show that $\sum_{j=1}^n |(1- \lambda) x_j+ \lambda y_j|^2 \leq 1$.

$ |(1- \lambda) x_j+ \lambda y_j| \leq (1- \lambda) |x_j|+ \lambda |y_j|$

So $\sum_{j=1}^n |(1- \lambda) x_j+ \lambda y_j|^2 \leq \sum_{j=1}^n \left( (1- \lambda) |x_j|+ \lambda |y_j|\right)^2= \sum_{j=1}^n \left( (1- \lambda)^2 |x_j|^2+ 2 \lambda (1- \lambda) |x_j| |y_j|+\lambda^2 |y_j|^2\right)= (1- \lambda)^2 \sum_{j=1}^n |x_j|^2 + 2 \lambda (1- \lambda) \sum_{j=1}^n |x_j||y_j| + \lambda^2 \sum_{j=1}^n |y_j|^2$

What bound can we use for $\sum_{j=1}^n |x_j||y_j| $ to show that the above is $\leq 1$ ? (Thinking)