2

It's written in my book that $L:(X,|\cdot |_X)\to (Y,|\cdot |_y)$ is bounded if there is $C>0$ s.t. $$|L(x)|_Y\leqslant C|x|_X.$$

This definition seems very strange to me. Where is the motivation behind this definition ?

HelloWorld
  • 628
user330587
  • 1,624
  • 2
    The motivation is that the operator is then continuous if and only if it is bounded. – 5xum Jul 14 '17 at 09:28
  • Consider that a linear operator is much more special than an arbitrary function. And for a linear operator holds that it is continuous iff it is continuous at $0$ iff it is bounded. – Mundron Schmidt Jul 14 '17 at 09:33
  • Related: https://math.stackexchange.com/questions/635154/definition-of-a-bounded-operator-and-some-intuition-on-the-definition-of-the-nor, https://math.stackexchange.com/questions/2319967/motivation-behind-definition-of-bounded-linear-operator, https://math.stackexchange.com/questions/2173737/bounded-linear-operator-why-do-we-only-consider-points-in-the-unit-circle – Hans Lundmark Jul 14 '17 at 09:41

2 Answers2

3

The motivation: A linear operator $L$ is bounded if $L$ is bounded on the closed unit ball in $X$.

Fred
  • 77,394
1

The operator $L$ is bounded if there is $C>0$ s.t. $$\|L\|\leq C.$$ A good norm for $L$ is $$\|L\|=\sup_{x\in X}\frac{|L(x)|_Y}{|x|_X}.$$ Therefore, $L$ is bounded is equivalent to $$\exists C>0: \forall x\in X, |L(x)|_Y\leq C|x|_X.$$

Surb
  • 55,662