Is my proof correct? Finite-dimensional normed vector spaces

Question

I'm trying to prove that every finite-dimensional normed space is topological isomorphic to $\mathbb{R}^n$. Let $(E,\|\cdot\|_E)$ such that $dimE=n$ and let $$ T:\mathbb{R}^n\to E\\ x\mapsto x_1\delta_1+\dots x_n\delta_n $$ where $B_F=\{\delta_1,\dots,\delta_n\}$ is an algebraic basis for $F$. It is easy to show that $T$ is a bijection. - $T$ is continuous I'm showing that $T$ is bounded.

For every $x\in\mathbb{R}^n$

$\|Tx\|_E=\|\sum x_i\delta_i\|_E\le M\|x\|_{\mathbb{R}^n}$.

In the last passage I used the fact that all norms are equivalent in $\mathbb{R}^n$, and so I used $\|x\|_{\mathbb{R}^n}:=\sum |x_i|$.

• $T^{-1}$ is continuous

Suppose by contradiction that $T^{-1}$ is not bounded. Then, there exists $$ (y_n)_n\subset E:\qquad \frac{\|T^{-1}y_n\|_{\mathbb{R}^n}}{\|y_n\|_{E}}>n\,\,\,\,\,\,\,\,\,\,\forall n\in\mathbb{N} $$ Consider $$ x_n:=\frac{T^{-1}y_n}{\|T^{-1}y_n\|_{E}},\quad (x_n)_n\subset\mathbb{R}^n $$ We have $$ \|x_n\|_{\mathbb{R}^n}=1\quad\forall n\in\mathbb{N} $$ but $$ \|Tx_n\|_{E}<\frac{1}{n}\quad\forall n $$ This is a contradiction since $T$ is injective.

Is it correct my proof?

Answer 1

You gave a nice proof but the contradiction isn't (at least for me) clear. To explain it we can do as this: since the sequence $(x_n)$ is bounded ($\|x_n\|=1$) in finite dimensional space (and here we used this assumption) then by the Weierstrass theorem there's a convergent sub-sequence to say $x$ and by the continuity of the norm we have $\|x\|=1$ but by the inequality $\|Tx_n\|<\frac1n$ we find that $Tx=0$ which contradicts the injectivity of $T$.

Answer 2

Here is a 'constructive' proof of boundedness of $T^{-1}$:

Define $\|x\|_* = \|Tx\|$. Then $\|\cdot\|_*$ is a norm on $\mathbb{R}^n$ and so is equivalent to, say, $\|\cdot\|_1$.

In particular, for some $c>0$, you have $\|x\|_* \ge c \|x\|_1$, or in other words, $\|Tx\| \ge c \|x\|_1$. Now let $v=Tx$, then you have ${1 \over c} |v\| \ge \|T^{-1} v\|_1$, so $\|T^{-1}\|$ is bounded by ${1 \over c}$.