Norm: The Rule for Measuring Length

How do we measure the length of a vector? The measuring rule is the norm.

The rule has three parts. A length is never negative, and only the zero vector has length 0. Scaling a vector scales its length by the same factor. And the length of a sum never exceeds the sum of the lengths. That last rule is the triangle inequality. Change the ruler and the length changes: the block-counting ruler, the straight-line ruler, and the king-move ruler all differ.

For example, in a πŸ™ grid city, go one block east and one block north. The straight line is one diagonal, but on blocks you must walk twice. A chess king goes diagonally in one move. The same move measures 2, about 1.41, and 1 under the three rulers.

βˆ₯xβˆ₯β‰₯0,βˆ₯xβˆ₯=0β€…β€ŠβŸΊβ€…β€Šx=0\|x\| \ge 0, \quad \|x\| = 0 \iff x = 0
βˆ₯axβˆ₯=∣aβˆ£β€‰βˆ₯xβˆ₯\|a x\| = |a| \, \|x\|
βˆ₯x+yβˆ₯≀βˆ₯xβˆ₯+βˆ₯yβˆ₯\|x + y\| \le \|x\| + \|y\|
βˆ₯xβˆ₯1=∣x1∣+∣x2∣,βˆ₯xβˆ₯2=x12+x22,βˆ₯xβˆ₯∞=max⁑(∣x1∣,∣x2∣)\|x\|_1 = |x_1| + |x_2|, \quad \|x\|_2 = \sqrt{x_1^2 + x_2^2}, \quad \|x\|_\infty = \max(|x_1|, |x_2|)

Examples & Exercises

For the vector \x = (3, 4)\, compute \\\|x\\|_1\, \\\|x\\|_2\, and \\\|x\\|_\\infty\ each.

βˆ₯xβˆ₯1=∣x1∣+∣x2∣,βˆ₯xβˆ₯2=x12+x22,βˆ₯xβˆ₯∞=max⁑(∣x1∣,∣x2∣)\|x\|_1 = |x_1| + |x_2|, \qquad \|x\|_2 = \sqrt{x_1^2 + x_2^2}, \qquad \|x\|_\infty = \max(|x_1|, |x_2|)
The vector (1.5,2)(1.5, 2), that is (3,4)(3, 4) halved β€” by homogeneity the lengths halve too.
Show proof

  1. βˆ₯xβˆ₯1=∣3∣+∣4∣=7\|x\|_1 = |3| + |4| = 7.

  2. βˆ₯xβˆ₯2=32+42=25=5\|x\|_2 = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.

  3. βˆ₯xβˆ₯∞=max⁑(∣3∣,∣4∣)=4\|x\|_\infty = \max(|3|, |4|) = 4. The same vector measures 7, 5, and 4 under the three rulers.

Prove completely, from the norm axioms, the reverse triangle inequality ∣βˆ₯xβˆ₯βˆ’βˆ₯yβˆ₯βˆ£β‰€βˆ₯xβˆ’yβˆ₯|\|x\| - \|y\|| \le \|x - y\|.

∣βˆ₯xβˆ₯βˆ’βˆ₯yβˆ₯βˆ£β‰€βˆ₯xβˆ’yβˆ₯|\|x\| - \|y\|| \le \|x - y\|
x=(1,1)x = (1, 1), y=(0.5,0.5)y = (0.5, 0.5) β€” parallel, so the difference of lengths shows up directly.
Show proof

  1. Since x=(xβˆ’y)+yx = (x - y) + y, the triangle inequality gives βˆ₯xβˆ₯≀βˆ₯xβˆ’yβˆ₯+βˆ₯yβˆ₯\|x\| \le \|x - y\| + \|y\|. Rearranging, βˆ₯xβˆ₯βˆ’βˆ₯yβˆ₯≀βˆ₯xβˆ’yβˆ₯\|x\| - \|y\| \le \|x - y\|.

  2. Swapping the roles of xx and yy gives βˆ₯yβˆ₯βˆ’βˆ₯xβˆ₯≀βˆ₯yβˆ’xβˆ₯\|y\| - \|x\| \le \|y - x\|. But βˆ₯yβˆ’xβˆ₯=βˆ₯(βˆ’1)(xβˆ’y)βˆ₯=βˆ₯xβˆ’yβˆ₯\|y - x\| = \|(-1)(x - y)\| = \|x - y\|, so βˆ₯yβˆ₯βˆ’βˆ₯xβˆ₯≀βˆ₯xβˆ’yβˆ₯\|y\| - \|x\| \le \|x - y\| holds too.

  3. Combining the two inequalities, ∣βˆ₯xβˆ₯βˆ’βˆ₯yβˆ₯∣=max⁑(βˆ₯xβˆ₯βˆ’βˆ₯yβˆ₯,βˆ₯yβˆ₯βˆ’βˆ₯xβˆ₯)≀βˆ₯xβˆ’yβˆ₯|\|x\| - \|y\|| = \max(\|x\| - \|y\|, \|y\| - \|x\|) \le \|x - y\|. This is the reverse triangle inequality.

On a vector space with a norm βˆ₯β‹…βˆ₯\|\cdot\|, put d(x,y)=βˆ₯xβˆ’yβˆ₯d(x, y) = \|x - y\|. Prove completely that dd satisfies all the metric axioms.

d(x,y)=βˆ₯xβˆ’yβˆ₯d(x, y) = \|x - y\|
x=(1,1)x = (1, 1), y=0y = 0 β€” the distance between two points is the norm of their difference.
Show proof

  1. Non-negativity and separation: by the first norm axiom, d(x,y)=βˆ₯xβˆ’yβˆ₯β‰₯0d(x, y) = \|x - y\| \ge 0, and d(x,y)=0d(x, y) = 0 is equivalent to βˆ₯xβˆ’yβˆ₯=0\|x - y\| = 0, which is equivalent to xβˆ’y=0x - y = 0, that is, x=yx = y.

  2. Symmetry: by the second norm axiom, d(y,x)=βˆ₯yβˆ’xβˆ₯=βˆ₯(βˆ’1)(xβˆ’y)βˆ₯=βˆ£βˆ’1∣βˆ₯xβˆ’yβˆ₯=βˆ₯xβˆ’yβˆ₯=d(x,y)d(y, x) = \|y - x\| = \|(-1)(x - y)\| = |-1| \|x - y\| = \|x - y\| = d(x, y).

  3. Triangle inequality: by the third norm axiom, d(x,z)=βˆ₯xβˆ’zβˆ₯=βˆ₯(xβˆ’y)+(yβˆ’z)βˆ₯≀βˆ₯xβˆ’yβˆ₯+βˆ₯yβˆ’zβˆ₯=d(x,y)+d(y,z)d(x, z) = \|x - z\| = \|(x - y) + (y - z)\| \le \|x - y\| + \|y - z\| = d(x, y) + d(y, z). All four axioms hold, so dd is a metric.

3. Sequences