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.
@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