Bounded Operators: Linearity & Inequality

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    Bounded Operators
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Discussion Overview

The discussion revolves around the properties of bounded operators in the context of linearity and inequalities. Participants explore the relationship between boundedness, linearity, and continuity, particularly in finite versus infinite-dimensional spaces.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that a linear operator T: X -> Y is bounded if there exists M>0 such that ||Tv||Y ≤ M*||v||X for all v in X.
  • Others argue that the inequality does not imply linearity, citing the example of the sine function, which is bounded but not linear.
  • One participant suggests that the inequality implies continuity, but clarifies that this does not lead to linearity.
  • There is a discussion about the requirements to show that an operator is a bounded linear operator, including proving linearity and demonstrating the mapping from X to Y.
  • A later reply notes that linearity implies boundedness in finite dimensions, indicating a distinction when considering infinite-dimensional spaces.

Areas of Agreement / Disagreement

Participants express disagreement regarding whether boundedness implies linearity, with some asserting that it does not while others clarify the implications of boundedness and continuity. The discussion remains unresolved on the broader implications of these properties.

Contextual Notes

Participants highlight that the relationship between boundedness and linearity may depend on the dimensionality of the vector spaces involved, particularly emphasizing the differences between finite and infinite dimensions.

mathplease
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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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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?
 
I think that it is automatically continuous. Is that what you wanted instead? As noted, this clearly doesn't imply linearity.
 
Jamma said:
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?
 
mathplease said:
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!
 
micromass said:
Yes, these are the things you need to show!

haha, thanks guys. much clearer now.
 
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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