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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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    Is Sine bounded? Is it linear?2011-04-12

1 Answers 1

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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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    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