91Ó°ÊÓ

A norm, denoted by \(\| \cdot \|\), is said to be induced by an inner product if it satisfies the following properties for all vectors \(u, v, w\) in the vector space and scalar \(c\): 1. Positivity: \(\|u\| \geq 0\) and \(\|u\| = 0\) if and only if \(u = 0\). 2. Linearity: \(\|cu\| = |c| \|u\|\). 3. Triangle inequality: \(\|u + v\| \leq \|u\| + \|v\|\).

For the given norm, $$\left\|\left(x_{1}, x_{2}\right)\right\|=\left|x_{1}\right|+\left|x_{2}\right|,$$ we can observe that both \(\left|x_{1}\right|\) and \(\left|x_{2}\right|\) are non-negative. Hence their sum is also non-negative. Moreover, if \(\left|x_{1}\right| = 0\) and \(\left|x_{2}\right| = 0\), then \((x_{1}, x_{2}) = (0, 0)\). Conversely, if \((x_{1}, x_{2}) = (0, 0)\), then \(\left|x_{1}\right| = 0\) and \(\left|x_{2}\right| = 0\). Therefore, the positivity property holds.

We want to show that for any scalar \(c\) and any vector \((x_{1}, x_{2})\), the following holds: $$\left\|\left(cx_{1}, cx_{2}\right)\right\| = |c| \left\|\left(x_{1}, x_{2}\right)\right\|.$$ The left-hand side is given by \(\left|cx_{1}\right| + \left|cx_{2}\right|\). We know that for any scalar \(c\) and real numbers \(a\), the following identity holds: \(\left|c a\right| = |c| |a|\). Thus, we get \(\left|cx_{1}\right|+\left|cx_{2}\right| = |c| |x_{1}| + |c| |x_{2}| = |c|(|x_{1}|+|x_{2}|)\). So the linearity property holds.

Let \((x_{1}, x_{2})\) and \((y_{1}, y_{2})\) be any two vectors in \(\mathbf{R}^{2}\). Then we need to show that $$\left\|\left(x_{1} + y_{1}, x_{2} + y_{2}\right)\right\| \leq \left\|\left(x_{1}, x_{2}\right)\right\| + \left\|\left(y_{1}, y_{2}\right)\right\|.$$ Evaluating both sides, we have: $$\left| x_{1} + y_{1} \right| + \left| x_{2} + y_{2} \right| \overset{?}\leq \left| x_{1} \right| + \left| x_{2} \right| + \left| y_{1} \right| + \left| y_{2} \right|.$$ Now, due to the triangle inequality for real numbers, we know that \(\left| x_{1} + y_{1} \right| \leq \left| x_{1} \right| + \left| y_{1} \right|\) and \(\left| x_{2} + y_{2} \right| \leq \left| x_{2} \right| + \left| y_{2} \right|\). Adding these inequalities together, we get the desired inequality, and therefore the triangle inequality also holds. Since all the properties of a norm induced by an inner product are satisfied, we can conclude that there exists an inner product on \(\mathbf{R}^{2}\) such that the associated norm is given by $$\left\|\left(x_{1}, x_{2}\right)\right\|=\left|x_{1}\right|+\left|x_{2}\right|.$$