Bounded operators

  • Thread starter mathplease
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  • #1
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Main Question or Discussion Point

a linear operator T: X -> Y is bounded if there exists M>0 such that:

ll Tv llY [tex]\leq[/tex] M*ll v llX for all v in X

conversely, if i know this inequality is true, is it always true that T: X ->Y and is linear?
 
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Answers and Replies

  • #2
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No, if this inequality is true, then your function is not necessarily linear. For example:

[tex]|\sin(x)|\leq |x|[/tex]

But the sine function is not linear...

Was this what you meant?
 
  • #3
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I think that it is automatically continuous. Is that what you wanted instead? As noted, this clearly doesn't imply linearity.
 
  • #4
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I think that it is automatically continuous. Is that what you wanted instead? As noted, this clearly doesn't imply linearity.
i see, yes. so in general, to show that something is a bounded linear operator from X to Y, you need to show the inequality, prove linearity and show that its a mapping from X to Y?
 
  • #5
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i see, yes. so in general, to show that something is a bounded linear operator from X to Y, you need to show the inequality, prove linearity and show that its a mapping from X to Y?
Yes, these are the things you need to show!
 
  • #6
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Yes, these are the things you need to show!
haha, thanks guys. much clearer now.
 
  • #7
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Note that linearity implies boundedness over finite dimensions- the linear maps are just the matrices. You only really need to take the boundedness into account when you are working over infinite-dimensional vector spaces.
 

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