91Ó°ÊÓ

The infinity norm (also known as the maximum norm) of a vector \(\mathbf{x} = (x_1,x_2,\ldots,x_n) \) is defined as the maximum absolute value of its components: \[ \| \mathbf{x} \|_\infty = \max\{|x_1|,|x_2|,\ldots,|x_n|\}. \]

The Euclidean norm (also known as the 2-norm or L2-norm) of a vector \(\mathbf{x} = (x_1,x_2,\ldots,x_n) \) is defined as the square root of the sum of the squares of its components: \[ \| \mathbf{x} \|_2 = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}. \]

To show that \( \| \mathbf{x} \|_\infty \le \| \mathbf{x} \|_2 \), we'll reason as follows: 1. We observe that all of the components of the vector have a square that is non-negative, i.e., \( x_i^2 \ge 0 \). 2. Given that \( \| \mathbf{x} \|_\infty = \max\{|x_1|,|x_2|,\ldots,|x_n|\} \), all the squared components (\( x_i^2 \)) are smaller than or equal to the maximum squared component (\( (\| \mathbf{x} \|_\infty)^2 = \max\{|x_1|^2,|x_2|^2,\ldots,|x_n|^2\} \)). 3. Therefore, we can say that: \[ x_1^2 + x_2^2 + \cdots + x_n^2 \ge (\| \mathbf{x} \|_\infty)^ 2. \] 4. Taking the square root of both sides of the inequality, we arrive at the desired result: \[ \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} = \| \mathbf{x} \|_2 \ge \| \mathbf{x} \|_\infty. \] Thus, we have shown that the infinity norm of a vector is always less than or equal to its Euclidean norm.