# Bounded operators

## Main Question or Discussion Point

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

ll Tv llY $$\leq$$ 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:

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

But the sine function is not linear...

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

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?

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!

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.