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Thread: Finite dimensional normed spaces

  1. #1
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    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?
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    Re: Finite dimensional normed spaces

    First, what is this a "corollary" to? Second, which definition of "continuous" are you using?
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    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:
    \[ \varrho(x,a)< \delta \Rightarrow \varrho(f(x), f(a))< \epsilon\]
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    Re: Finite dimensional normed spaces

    Quote Originally Posted by TheMR View Post
    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.
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    Re: Finite dimensional normed spaces

    Quote Originally Posted by TheMR View Post
    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.
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