Show that the sets are closed and convex We consider the sets $K=\{ x \in \mathbb{R}^n: \sum_{j=1}^n |x_j|^2 \leq 1 \}, L=\{ x \in \mathbb{R}^n: \sum_{j=1}^n |x_j|^p \leq 1\}$ where $1 \leq p< +\infty, p \neq 2$ , and let $ z \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 $\leq 1$ ? 
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In the same way, we show that $L$ is closed.
In order to show that it is convex we need to show that $\sum_{j=1}^n |(1- \lambda)x_j+ \lambda y_j|^p \leq 1$.
How can we do this?
EDIT:
Suppose that we consider the set $K_2:=\{ x \in \mathbb{R}^n: \sum_{j=1}^n |x_j|^p \leq 1 \}$ where $0<p<1$.
In this case the set isn't convex.
How can we check if the (euclidean) distance of $z \in \mathbb{R}^n$ from $K$ is attained?
 A: For the bound of $\sum \mid x_j y_j\mid$, consider 
$$
\sum (\mid x_j\mid - \mid y_j\mid)^2=\sum \mid x_j
\mid^2+\sum \mid y_j\mid^2-2\sum \mid x_j y_j\mid\geq 0\implies 1\geq \sum \mid x_j y_j\mid
$$
A: To bound the sum $\sum_{j=1}^n|x_j||y_j|$, use the inequality $2ab \leqslant a^2 + b^2$ with $a = |x_j|$ and $b = |y_j|$.
To prove the convexity of $L$, use the following inequality:
$$|(1 - \lambda)x_j + \lambda y_j|^p \leq (1-\lambda)|x_j|^p + \lambda |y_j|^p$$
which just states that $f(x) = |x|^p$ is a convex function. The proof of this fact depends on what do you know about convex functions. For example, see the standard proof from this question that uses derivatives.
Alternatively, as suggested by Jendrik Stelzner, if you're familiar with the notion of the $p$-norm $\|\cdot\|_p$, the convexity of $L$ can be shown as follows. Note that $L = \{x \in \mathbb{R}^n \colon \|x\|_p \leq 1\}$ is a unit ball in $\mathbb{R}^n$ w.r.t. $p$-norm. Fix some $x, y \in L$ and $\lambda \in [0, 1]$. And consider $z = (1-\lambda)x + \lambda y$. From the trinalge inequality and the absolute homogenity for $p$-norm it follows:
$$\|z\|_p \leq \|(1-\lambda)x\|_p + \|\lambda y\|_p \leq |1-\lambda|\|x\|_p + |\lambda| \|y\|_p = 1 - \lambda + \lambda = 1,$$
where in the first inequality we use the triangle inequality and in the second we use the absolute homogenity of $\|\cdot\|_p$.
