Analysis(the norm) please help me quickly?

Question

x in R^2 let ||x|| =|x_1|+|x_2| show || || is norm on R^2 and describe geometrically ||x-y||=1 notice that( || || is the norm and | | is the absolute value)

Answer 1

So all you have to do here is show that your definition of ||x|| upholds the axioms of a norm, which are:

(i) ||x|| = 0 if and only if x = 0 (the zero vector)

(ii) The Triangle Inequality: ||x+y|| <= ||x|| + ||y|| for all x,y in R^2

(iii) Scaler Multiplication: ||ax|| = |a| ||x|| for all scalers a and for all x in R^2.

(i) Suppose ||x|| = 0, then we have |x_1| + |x_2| = ||x|| = 0 by definition. Since |x_1| and |x_2| are both positive, they must be 0, hence x = (x_1,x_2) = 0.

If x = 0, this is obvious.

(ii) This follows from the triangle inequality for absolute values;

||x+y|| = |x_1 + y_1| + |x_2 + y_2| <= |x_1| + |y_1| + |x_2| + |y_2| = ||x|| + ||y||.

(iii) ||ax|| = |a(x_1)| + |a(x_2)| = |a||x_1| + |a||x_2| = |a|(|x_1| + |x_2|) = |a| ||x||.

For the geometric part, if ||x-y|| = |x_1-y_1| + |x_2-y_2| = 1, this means that the sum of the difference of the horizontal and vertical components of x and y equal one.  So if you were to plot both x and y in R^2, and draw a line horizontally from x to the vertical position of y, then draw a vertical line going from the end of your first line to y, the length of the path of your pencil would be 1. After a little bit of thinking, you should see that all such y given an x would be a diamond with diagonal length 2 centered about x.

Hope this helps.