§ L1 norm is greater than or equal to L2 norm

Pick two points A(x1,y1)A \equiv (x_1, y_1) and B(x2,y2)B \equiv (x_2, y_2), and suppose x1<x2x_1 < x_2 and y1<y2y_1 < y_2. So we imagine this as two sides of a triangle:
     B
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A
  • The L1 norm is x2x1+y2y1|x_2 - x_1| + |y_2 - y_1|. This is the distance on connecting to an origin OO:
  δx
O----B
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δy /
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A
  • The L2 norm is (x2x1)2+(y2y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, which is the distance of the vector ABAB, or the hypotenuse of the right angled triangle AOBAOB:
  δx
O----B
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δy / L2
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A
  • By triangle inequality, OA+OBABOA + OB \geq AB, hence L1=δx+δyL2L_1 = \delta_x + \delta_y \geq L_2