# Continuous linear transformation

1. Aug 20, 2007

### Andy_ToK

T is a linear transformation from R^m->R^n, prove that T is continuous.

I have proved that there's always a positive real number C that |T(x)|<=C|x|. How shall I proceed then?
Thanks~

2. Aug 20, 2007

### quasar987

May I ask how you have shown that?

3. Aug 20, 2007

### Andy_ToK

x is an element of R^m
|T(x)|=|sum_i(xi*T(ei))|<=sum_i{xi(|T(ei)|)}
let A=max{|T(e1)|,|T(e2)|...|T(em)|)
then |T(x)|<=A*sum_i(xi)<A/sqrt(m)*sqrt(sum_i(xi^2))=A/sqrt(m)*|x| (AM<RMS)

sorry, i don't know how to use latex here

4. Aug 20, 2007

### quasar987

1) You forgot an absolute value sign around xi in line 2:

|T(x)|=|sum_i(xi*T(ei))|<=sum_i{|xi|(|T(ei)|)}

2) How did you get

sum_i|xi|<[1/sqrt(m)]*sqrt(sum_i(xi^2)) ?

Surely this is not true. Take for instance the case when all but one of the x_i are non-vanishing. The inequality becomes

|xi|<[1/sqrt(m)]*|xi| <==> sqrt(m)<1

You had good ideas in your attempt but I suggest working directly with the definition of continuity. Given e>0 and x0 , try to find a d>0 such that |x-x0|<d ==> |T(x)-T(x0)|<e.

Last edited: Aug 20, 2007
5. Aug 20, 2007

### ZioX

Well, can you show that it is Lipschitz?

6. Aug 21, 2007

### Andy_ToK

1.Thanks for pointing that out, it should be the absolute value.
2.sorry i made a mistake there, it should be
sum_i|xi|<sqrt(m)*sqrt(sum_i(xi^2)) by AM<RMS

7. Aug 21, 2007

### ZioX

Here is the standard way of doing it:

For any linear transformation we define the norm of it as $||T||=\sup_{|x|<1}|T(x)|$. Your arguments show that this norm always exists for any linear map.

Then we have $|T(x-y)|\le||T|||x-y|$. Therefore $T$ is Lipschitz, hence uniformly continuous. If you don't know anything about Lipschitz and uniform continuity you can just let $|x-y|<\epsilon/||T||$ and we have our desired result.

8. Aug 21, 2007

### quasar987

Oh, ok. I did not know this inequality, but it is true as long as you change that < for a <=, because obviously for m=1, we have an equality. What does AM and RMS stand for?

What you've proven in showing |T(x)|<=C|x| is that T is Lipschitz continuous at 0. Usually, we say that a function f is Lipschitz at x0 if there exists numbers c and M such that |x-x0|<c ==> |f(x)-f(x0)| $\leq$ M|x-x0|. A function that is Lipschitz is necessarily continuous because given e>0, choose d=min{c,e/M}.

So, can you extend your result and show that T is Lipschitz everywhere?

Last edited: Aug 21, 2007
9. Aug 21, 2007

### quasar987

Shouldn't there be a sqrt{m} also on the RHS as in Andy's AM<RMS?

10. Aug 21, 2007

### Andy_ToK

oops, it should be <=, thanks.
AM=arithmetic mean
RMS=root mean square AM<=RMS (the equality holds when |x1|=|x2|....=|xm|)

11. Aug 21, 2007

### Andy_ToK

Thanks quasar987 and ZioX
I forgot to use T(x)-T(y)=T(x-y) for the linear transformation...
let d=e/C then, |x-y|<d --> |T(x)-T(y)|=|T(x-y)<cd=e

12. Aug 27, 2007

### ZioX

I'll explain the logic a bit.

Define $||T||=\sup_{|x|<1}|T(x)|$.

Now the previous argument shows that there exists a C such that $|T(x)| \le C|x|$.

Now, if we restrict $|x| < 1$ we get $|T(x)| \le C|x| < C$ so that |T(x)| has an upper bound, hence a supremum.

Now, for continuity, it just becomes a matter of verifying that $0 \le |T(x)-T(y)|=|T(x-y)|<\epsilon$ whenever $0 \le |x-y|< \min\{\epsilon / ||T||,1\}$.