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The Wikipedia article for bounded operator is all about linear bounded operator. I was wondering

  1. Can a bounded operator be non-linear? If yes, how is this defined?
  2. Is a bounded operator generally assumed to be linear?


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2. Yes, many people use "operator" and "linear operator" to mean the same (it is tiresome to repeat "linear" every time). –  wildildildlife Apr 11 '11 at 19:07
Is Sine bounded? Is it linear? –  scineram Apr 12 '11 at 11:19
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1 Answer

up vote 4 down vote accepted
  1. Yes, a bounded operator can be nonlinear. There are a lot of useful notions of `bounded non-linear operator'. One is that for an operator between topological spaces that the image of compact sets is compact. The operator $Tx = 1/(1-x)$ is bounded on $[0,\infty)$ under this definition, but so are a lot of nasty operators. It depends on what you are trying to get out of your operator.

  2. No, one should always prove this.

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Thanks! So a bounded operator is defined such that the norm of the image of the operator on a vector is bounded by a constant times the norm of the vector? –  Tim Apr 11 '11 at 18:44
What definition of bounded are you using exactly for nonlinear operators? –  Jonas Meyer Apr 11 '11 at 19:20
@Jonas: Are you asking me? –  Tim Apr 11 '11 at 19:40
@Tim: Oops, I didn't think I had posted that. I was going to elaborate on my question, then decided not to post it, then must have accidentally posted it anyway. Actually, I meant to ask @Glen. The point of asking is that boundedness for linear operators can be defined in several equivalent ways, the equivalence being due to linearity, and to me (having little experience with the theory of nonlinear operators) it isn't clear what boundedness should mean for nonlinear operators. For example, should it mean Lipschitz continuous? –  Jonas Meyer Apr 11 '11 at 19:46
Hi @Jonas. I thought about giving a way of defining the boundedness of a non-linear operator, but this would have been dishonest no matter which way I spun it. The truth is that there are lots of norms one may place on such operators and that each have certain properties. –  Glen Wheeler Apr 11 '11 at 21:03
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