Bounded Operators: Linearity & Inequality

[mathplease]

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

ll Tv ll_Y \leq M*ll v ll_X for all v in X

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

 

[micromass]

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

|\sin(x)|\leq |x|

But the sine function is not linear...

Was this what you meant?

 

[Jamma]

I think that it is automatically continuous. Is that what you wanted instead? As noted, this clearly doesn't imply linearity.

 

[mathplease]

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?

 

[micromass]

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!

 

[mathplease]

Yes, these are the things you need to show!

haha, thanks guys. much clearer now.

 

[Jamma]

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.