Context.
I am attending a Functional Analysis class and we're currently going over normed spaces. In one of the classes, my teacher proved the boundedness of the operator below using a really strong auxiliary lemma that I will link in the next paragraph. After doing so, we proved equivalence of norms in any finite-dimensional normed space and many more results based off this.
So, I am looking for a proof that DOES NOT involve equivalence of norms in any finite-dimensional normed space (because we proved this equivalence based off this). Also, I'm looking for a straight-forward proof – not using this famous auxiliary lemma that is visible here.
Question. Let $X = (X,\|\cdot\|)$ be a finite-dimensional normed space such that $\dim(X)= n \in \Bbb N$. Consider the linear operator $T: (X,\|\cdot\|) \to (\Bbb K^n, \|\cdot\|_\infty)$ (here, note that $\|\cdot\|_\infty$ is the usual supremum norm in $\Bbb K^n$ and $\Bbb K = \Bbb R$ or $\Bbb K = \Bbb C)$ defined as
$$ T(x) = T\left(\sum_{i=1}^{n}\alpha_ie_i\right) = (\alpha_1,\dots,\alpha_n).$$
where $\{e_1,\dots,e_n\}$ is a basis of $X$ and $\alpha_j \in \Bbb K, \, \forall j \in \{1,\dots,n\}.$ Show that $T$ is bounded.
My attempt. Proving that $T$ is bounded is simply proving that:
$$ \forall x \in X, \quad \| T(x) \|_\infty \leqslant c\|x\|, \quad \text{for some $c>0$}.$$
Well,
$$\| T(x)\|_\infty = \|(\alpha_1,\dots,\alpha_n) \|_\infty = \max_{1\leqslant i \leqslant n} |\alpha_i| \leqslant \sum_{i=1}^n |\alpha_i|\left(\frac{\|e_i\|}{\|e_i\|}\right)$$
And from here, we can apply some norm properties,which follow below:
$$ \sum_{i=1}^n |\alpha_i|\left(\frac{\|e_i\|}{\|e_i\|}\right) = \sum_{i=1}^n \| \alpha_i e_i \| \frac{1}{\|e_i\|}$$
But I don't know what to do next (?)
If anyone can help with this one I would be really apreciatted.
NOTE. If it somehow helps, the inverse operator $T^{-1}$ exists and it is bounded (in fact, it is a well known operator).
Best Answer
It looks pretty much the same than the proof that all norms are equivalent.
Let $L=T^{-1} \colon (\Bbb K^n,\|\cdot\|_{\infty}) \to (X,\|\cdot\|)$. As you said, $L$ is bounded. Hence, it is continuous (it is a property of linear maps).
From Bolzano-Weierstrass, the unit sphere $S$ of $(\Bbb K^n,\|\cdot\|_{\infty})$ is compact. Consider $$ \begin{array}{r|ccc} f\colon & S & \longrightarrow & \Bbb R\\ &x & \longmapsto & \|L(x)\| \end{array}. $$ $f$ is continuous on $S$ compact, so that $c=\min f$ exists and is positive since $L$ is injective. By homogeneity, we have $$ \forall x \in \Bbb K^n,\quad c\|x\|_{\infty}\leqslant \|L(x)\|. $$ Taking $x=L^{-1}(y) = T(y)$ now yields $$ \forall y \in X,\quad \|T(y)\|_{\infty}\leqslant \frac{1}{c}\|y\|. $$