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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?

Thanks!

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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).2011-04-11
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    Is Sine bounded? Is it linear?2011-04-12

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  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?2011-04-11
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    What definition of bounded are you using exactly for nonlinear operators?2011-04-11
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    @Jonas: Are you asking me?2011-04-11
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    @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?2011-04-11
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    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.2011-04-11
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    I would call an operator which is Lipschitz continuous just Lipschitz continuous, this (in the context of differential operators) is a useful condition (since it is usually a necessary ingredient to show uniqueness of solutions) just by itself. I might call a non-linear operator from one topological space to another bounded if it maps compact sets to compact sets, but there are a lot of different (and useful) notions. I didn't really think it would be helpful to go into any particular one, but good question.2011-04-11
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    Most interesting (for me) non-linear operators are in no sense bounded, however. This may (*may*) be why it is not so well clarified in the literature.2011-04-11
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    Glen, thanks. But if you do not give a definition of bounded operator, what is the meaning of the claim in your answer that the example you gave is a bounded operator? You must have some definition in mind there, right? Sending compact sets to compact sets is an interesting criterion. With that definition, every continuous function would be bounded, but also many discontinuous functions.2011-04-12
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    @Jonas Good point. Yes, that's the criterion I had in mind when I wrote that. I'll edit. It's not so fantastic a criterion, but it is one I've seen used.2011-04-12
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    Hi @Tim. In the definition I gave, something like $Tx = x^2$ is bounded, whereas $Tx = \frac{1}{x(x-1)}$ is not (with the usual topology on $\mathbb{R}$). This might help you see some of the problems with that definition.2011-04-12