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Chapter 5: Problem 13

Show that $$\|\mathbf{x}\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|$$ defines a norm on \(\mathbb{R}^{n}\)

Short Answer

Expert verified
The function \(\|\mathbf{x}\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|\) defines a norm on \(\mathbb{R}^n\) because it satisfies the three properties of a norm: non-negativity, homogeneity, and triangle inequality. These properties are proven by showing \(\|\mathbf{x}\|_{1} \geq 0\) and \(\|\mathbf{x}\|_{1} = 0\) if and only if \(\mathbf{x} = 0\), \(\|\alpha \mathbf{x}\|_{1} = |\alpha| \|\mathbf{x}\|_{1}\) for all \(\mathbf{x} \in \mathbb{R}^n\) and \(\alpha \in \mathbb{R}\), and \(\|\mathbf{x} + \mathbf{y}\|_{1} \leq \|\mathbf{x}\|_{1} + \|\mathbf{y}\|_{1}\) for all \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^n\).

Step by step solution

01

Property 1: Non-negativity

Since the absolute value function \(|x_i|\) is non-negative for any real number \(x_i\), the sum of the absolute values of the components of \(\mathbf{x}\) will also be non-negative, hence: $$\|\mathbf{x}\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right| \geq 0$$ If \(\|\mathbf{x}\|_{1} = 0\), then the sum of the absolute values of the components of \(\mathbf{x}\) is zero. This means that each component of \(\mathbf{x}\) must be zero, i.e., \(\mathbf{x} = 0\). Conversely, if \(\mathbf{x} = 0\), then each component of \(\mathbf{x}\) is zero, and thus: $$\|\mathbf{x}\|_{1}=\sum_{i=1}^{n}\left|0\right| = 0$$ So we have that \(\|\mathbf{x}\|_{1} \geq 0\), and \(\|\mathbf{x}\|_{1} = 0\) if and only if \(\mathbf{x} = 0\).
02

Property 2: Homogeneity

Let \(\alpha \in \mathbb{R}\). Then, for any \(\mathbf{x} \in \mathbb{R}^n\) we have: \begin{align*} \|\alpha \mathbf{x}\|_{1} &= \sum_{i=1}^{n}\left|\alpha x_{i}\right| \\ &= \sum_{i=1}^{n} |\alpha| \cdot |x_{i}| \\ &= |\alpha| \sum_{i=1}^{n} |x_{i}| \\ &= |\alpha| \|\mathbf{x}\|_{1} \end{align*} Thus, the homogeneity property is satisfied.
03

Property 3: Triangle inequality

Let \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^n\). Then, \begin{align*} \|\mathbf{x} + \mathbf{y}\|_{1} &= \sum_{i=1}^{n}|x_i + y_i| \\ &\leq \sum_{i=1}^{n}(|x_i| + |y_i|) & \text{ (using triangle inequality for absolute values)} \\ &= \sum_{i=1}^{n}|x_i| + \sum_{i=1}^{n}|y_i| \\ &= \|\mathbf{x}\|_{1} + \|\mathbf{y}\|_{1} \end{align*} Hence, the triangle inequality property is also satisfied. Since all three properties of a norm are satisfied, we conclude that the function \(\|\mathbf{x}\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|\) defines a norm on \(\mathbb{R}^{n}\).

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