Weighted sum of absolute values is a norm

Question

How can I prove that $p: \mathbb{R}^n \rightarrow \mathbb{R}_{\geqslant0}$ \begin{equation} p(\mathbf{x}) = \sum_{i=1}^{n}w_i\lvert x_i\rvert,\quad w_i>0 \end{equation} is a norm.

I immediately answered my own question and the proof to be verified is in my own answer. I apologize if this is not a good way to pose the proof verification questions.

Answer 1

According to the definition of the norm the function

1. has to satisfy the triangle inequality \begin{equation} p(\mathbf{x}+\mathbf{y}) = \sum_{i=1}^{n}w_i\lvert x_i + y_i\rvert \leqslant \sum_{i=1}^{n}w_i\lvert x_i\rvert + \sum_{i=1}^{n}w_i\lvert y_i\rvert = p(\mathbf{x})+p(\mathbf{y}) \end{equation}
2. has to be absolutely homogeneous \begin{equation} p(a\mathbf{x})= \sum_{i=1}^{n}w_i\lvert ax_i\rvert = \sum_{i=1}^{n}w_i\lvert a\rvert\lvert x_i\rvert = \lvert a\rvert p(\mathbf{x}) \end{equation}
3. and $p(\mathbf{x})=0 \Rightarrow \mathbf{x}=\mathbf{0}$.

In order to prove the third point lets prove the the contrapositive $\mathbf{x}\neq \mathbf{0} \Rightarrow p(\mathbf{x}) > 0$. If any of the elements of the vector is different than zero $x_i\neq0$ then its weighted absolute value is positive $w_i\lvert x_i\rvert >0$. The function is a sum of non-negative values which can not be equal to zero if one of its elements is positive. This proves the contrapositive.

Answer 2

Hint

Since $||x||_1$ is the famous $1$-norm and $x\to wx$ is a linear transformation where $w=\text{diag}(w_1,\cdots ,w_n)$, hence the result.