# Thread: Finite dimensional normed spaces

1. ## Finite dimensional normed spaces

Let X and Y be normed spaces. Suppose that $dimX<\infty$. Then every linear operator $T\colon X\rightarrow Y$ is continuous.

How to prove the previous corollary?

2. ## Re: Finite dimensional normed spaces

First, what is this a "corollary" to? Second, which definition of "continuous" are you using?

3. ## Re: Finite dimensional normed spaces

Sorry, it is corollary to: Any two norms on a finite-dimensional linear space are equivalent.

Definition: Let X and Y be metric spaces. An operator $f:X\rightarrow Y$ is said to be continuous at a point $a\in X$ if, for every $\epsilon >0$, there exists a $\delta >0$ such that:
$\varrho(x,a)< \delta \Rightarrow \varrho(f(x), f(a))< \epsilon$

4. ## Re: Finite dimensional normed spaces

Originally Posted by TheMR
Let X and Y be normed spaces. Suppose that $dimX<\infty$. Then every linear operator $T\colon X\rightarrow Y$ is continuous. How to prove the previous corollary?
The fact that $X$ is of finite dimensional means there is a finite basis which means we can bound the basis elements.
Here is a useful fact for you.
$\|\lambda x_0\|=|\lambda|\|x_0\|$ ,$\|x+v\|\le \|x\|+\|v\|$ thus $\left\| {\sum\limits_{k = 1}^n {{\lambda _k}{x_k}} } \right\| \leqslant \sum\limits_{k = 1}^n {\left| {{\lambda _k}} \right|\left\| {{x_k}} \right\|}$
This can be used to bound elements moreover $T\left( {\sum\limits_{k = 1}^n {{\lambda _k}{x_k}} } \right) = \sum\limits_{k = 1}^n {{\lambda _k}T\left( {{x_k}} \right)}$

For continuity at a point $x_0$ if $\varepsilon>0$ there exists $\delta>0$ such that if $\|x-x_0\|<\delta\|$ then $\|T(x)-T(x_0)\|<\varepsilon$

\begin{align*}\left\| {\lambda x - {\lambda _0}{x_0}} \right\| &= \left\| {{\lambda _0}(x - {x_0}) + (\lambda - {\lambda _0}){x_0} + (\lambda - {\lambda _0})(x - {x_0})} \right\|\\& \leqslant |{\lambda _0}|\left\| {(x - {x_0})} \right\| + |(\lambda - {\lambda _0})|\left\| {{x_0}} \right\| + |(\lambda - {\lambda _0})|\left\| {(x - {x_0})} \right\|\end{align*}

That last bit just shows you the sort of 'tricks'' one uses with these proofs.

5. ## Re: Finite dimensional normed spaces

Originally Posted by TheMR
Sorry, it is corollary to: Any two norms on a finite-dimensional linear space are equivalent.
I wish I had seen this reply before I posted what I did.
Well then, the norm can be redefined in such a way that $T$ is bounded.
Knowing that you have proven the norm equivalency theorem that makes a huge difference.
Here is a well worn theorem: A linear operator $T$ is continuous iff it is bounded.
Proof one way. If $\left\{x_n\right\}$ converges (in norm) to $0$ then $\|T(x_n)\|\le\|T\|~\|x_n\|\to 0.$

If you have access to a good mathematics library, I think highly of a text book by Robert B Ash.
Here are the Amazon reviews. Buy a used one it is a great reference for many courses.