91Ó°ÊÓ

Let \(\|\cdot\|_{1}\) be the \(\ell^{1}\) norm on \(\mathbf{R}^{n}\), defined by \[\|x\|_{1} = \sum_{i=1}^{n} |x_{i}|\] for any \(x = (x_{1}, x_{2}, \dots, x_{n}) \in \mathbf{R}^{n}\). 1. Non-negativity: Since the absolute value of each component of \(x\) is non-negative, their sum is also non-negative. So, \(\|x\|_{1} \geq 0\). If \(\|x\|_{1} = 0\), this means \(\sum_{i=1}^{n} |x_{i}| = 0\). Since each term in the sum is non-negative, we must have \(|x_{i}| = 0\) for each \(i\). Thus, \(x_{i} = 0\) for all \(i\) and \(x = 0\). 2. Scalability: Given any \(\alpha \in \mathbf{R}\) and \(x \in \mathbf{R}^{n}\), the \(\ell^{1}\) norm of the scaled vector is \[\begin{aligned} \|\alpha x\|_{1} &= \sum_{i=1}^{n} |(\alpha x)_{i}| \\ &= \sum_{i=1}^{n} |\alpha x_{i}| \\ &= |\alpha|\sum_{i=1}^{n} |x_{i}| \\ &= |\alpha| \|x\|_{1}. \end{aligned}\] 3. Triangle inequality: For any \(x, y \in \mathbf{R}^{n}\), we can see that \[\begin{aligned} \|x + y\|_{1} &= \sum_{i=1}^{n} |x_{i} + y_{i}| \\ &\leq \sum_{i=1}^{n} (|x_{i}| + |y_{i}|) \\ &= \sum_{i=1}^{n} |x_{i}| + \sum_{i=1}^{n} |y_{i}| \\ &= \|x\|_{1} + \|y\|_{1}. \end{aligned}\] Hence, \(\|\cdot\|_{1}\) satisfies all three properties of a norm and is therefore a norm on \(\mathbf{R}^{n}\).

Let \(\|\cdot\|_{\infty}\) be the \(\ell^{\infty}\) norm on \(\mathbf{R}^{n}\), defined by \[\|x\|_{\infty} = \max_{1\leq i \leq n} |x_{i}|\] for any \(x = (x_{1}, x_{2}, \dots, x_{n}) \in \mathbf{R}^{n}\). 1. Non-negativity: Since the \(\ell^{\infty}\) norm is the maximum of the absolute values of the components of \(x\), it must be non-negative, i.e., \(\|x\|_{\infty} \geq 0\). If \(\|x\|_{\infty} = 0\), then the maximum absolute value among the components of \(x\) is zero, which means all the components of \(x\) must be zero. So, \(x = 0\). 2. Scalability: Let \(\alpha \in \mathbf{R}\) and \(x \in \mathbf{R}^{n}\). Then \[\begin{aligned} \|\alpha x\|_{\infty} &= \max_{1\leq i \leq n} |(\alpha x)_{i}| \\ &= \max_{1\leq i \leq n} |\alpha x_{i}| \\ &= |\alpha|\max_{1\leq i \leq n} |x_{i}| \\ &= |\alpha| \|x\|_{\infty}. \end{aligned}\] 3. Triangle inequality: Let \(x, y \in \mathbf{R}^{n}\). Then, for any \(1 \leq i \leq n\), we have \[|x_{i} + y_{i}| \leq |x_{i}| + |y_{i}|.\] Since this inequality holds for all components, we can take the maximum on both sides and get \[\|x + y\|_{\infty} = \max_{1\leq i \leq n} |x_{i} + y_{i}| \leq \max_{1\leq i \leq n} (|x_{i}| + |y_{i}|) \leq \|x\|_{\infty} + \|y\|_{\infty}.\] Thus, \(\|\cdot\|_{\infty}\) satisfies all three properties of a norm and is therefore a norm on \(\mathbf{R}^{n}\).