Proving the inverse of an application is continuous. Exercise. Consider the application $T: (\Bbb{K}^n, \| \cdot \|_\infty) \rightarrow (X, \| \cdot \|),$ where $\Bbb{K}$ is a field of scalars (either $\Bbb{C}$ or $\Bbb{R}$) and the norm $\| \cdot \|_\infty$ is the usual infinite norm in these fields. On the other hand, $X$ is some normed space equipped with the norm $\| \cdot \|$ with finite dimension $n \in \Bbb{N}$. Let $\{e_1,\dots,e_n\}$ be a basis for the space $X$. The application is defined as follows:
$$ T(\alpha_1,\dots,\alpha_n) = \sum_{j=1}^n \alpha_j e_j$$
Show that $T$ is a linear and topological isomorphism.
My attempt. To show that $T$ is a linear and topological isomorphism, one should prove that $T$ is linear, bijective, continuous and that it admits an inverse ($T^{-1}$) which is also continuous. I was able to prove that $T$ is linear, bijective and continuous. I omit the proofs for linearity and bijective since they are obvious by simple definitions. Follows the proof for continuity:
Recall the usual delta-epsilon definition for continuity:
$$ \forall \epsilon > 0, \, \exists \delta > 0 \, : \, \| \alpha - \alpha_0 \|_{\infty} < \delta \Rightarrow \| T\alpha - T\alpha_0 \| < \epsilon $$
Let $\alpha_0$ be an arbitrary element of $\Bbb{K}^n$. Then, $\forall \alpha \in \Bbb K:$
$$ \| T\alpha - T\alpha_0 \| = \| \sum_{j=1}^n \alpha_j e_j - \sum_{j=1}^n \alpha_{0j} e_j \| \leqslant \sum_{j=1}^n \|(\alpha_j - \alpha_{0j})e_j\| = \sum_{j=1}^n |\alpha_j - \alpha_{0j}|\| e_j \| \leqslant \sum_{j=1}^n \|\alpha - \alpha_0 \|_{\infty}\|e_j\| $$
Since $\| \alpha - \alpha_0\|_{\infty}$ doesn't depend on $j$, we just remove it from the sum and we get:
$$ \|\alpha - \alpha_0\|_{\infty}\sum_{j=1}^n\|e_j\| < \delta \sum_{j=1}^n\|e_j\| < \epsilon \Rightarrow \delta < \frac{\epsilon}{\sum_{j=1}^n \|e_j\|}.$$
Thus, taking  $\delta < \frac{\epsilon}{\sum_{j=1}^n \|e_j\|}$ on the usual delta-epsilon definition of continuity is enough to show that $T$ is, indeed continuous. Here, note that $\sum_{j=1}^n \|e_j\| \neq 0$ (otherwise, the basis wouldn't be linearly independent).
Now, the only thing that's left is the proof that $T^{-1}$ is continuous. I am having some trouble doing this. Does anyone have a hint?
Thanks for any help in advance.
 A: @roro  I see your point (this was too long for a comment) $$\lvert \lvert T^{-1}(x)\rvert \rvert_{\infty} =\lvert \lvert T^{-1}(\sum_{i=1}^{n}x_{i}e_{i} \rvert \rvert_{\infty} =\lvert \lvert \sum_{i=1}^{n} (x_{i} T^{-1}e_{i} \rvert \rvert_{\infty} \leq  \sum_{i=1}^{n} \lvert \lvert   (x_{i} T^{-1}e_{i} \rvert \rvert_{\infty} \leq \max_{1\leq i\leq n}\lvert \lvert  T^{-1}e_{i} \rvert \rvert_{\infty}  \sum_{i=1}^{n}\lvert   x_{i}  \rvert $$,this proves that $T^{-1}$ is continuous (in fact this proves continuity at zero but for linear maps this is equivalent to continuity everywher) ,when $X$ is endowed with the norm $\lvert \lvert .\rvert \rvert_{1}: x\to \sum_{i=1}^{n}\lvert   (x_{i} \rvert $.Now all u need is to show that $\lvert \lvert .\rvert \rvert_{1}$ and $\lvert \lvert .\rvert \rvert$ are equivalent, because replacing a norm by an equivalent one does't alter continuity(In fact all norms are equivalent for finite dimensional normed spaces).another approach is to prove that $T$ is bounded below,$$\exists M>0,\forall a\in \mathbb{K}^{n} ;\lvert \lvert  a \rvert \rvert_{\infty}=1,\lvert \lvert  Ta \rvert \rvert \geq M$$.Suppose to the contrary that no such $M$ exists,the there is a sequence of unit vectors $a_{n}$ such that  $\lvert \lvert  Ta_{n}\rvert \rvert \to 0$,since the unit sphere of $\mathbb{K}^{n}$ is compact ,we can extract a convergent subsequence to a vector $a$( which I still denote $a_{n}$) ,so $$\lvert \lvert \lvert  Ta \rvert \rvert=\lim_{n\to \infty}\lvert \lvert  Ta_{n}\rvert \rvert = 0$$ ,this is impossible , since $T$ is injective and $\lvert \lvert  a \rvert \rvert=1$,hence $$\forall x\in X,\lvert \lvert  T^{-1}(x) \rvert \rvert_{\infty} \leq \frac{1}{M} \lvert \lvert x\lvert \rvert $$ which proves that $T^{-1}$ is continuous
A: All linear maps are continous provided the domain is a finite dimensioanl normed space,to see this u can apply the triangle inequality(which is exactly what u did in your proof), or remark that an injective linear map must be bounded below on the unit sphere(since it is compact in the finite dimensional case),the constant involved being the inverse of the norm of the inverse .To prove the continuity of the inverse ,simply u remark that the inverse of a linear map is linear, and your apply exactly the argument u used to prove contintuity of $T^{-1}$ instead of $T$.
