i was study functional analysis and i found a interesting problem. Let $X$ an infinite dimensional normed space. Construct an operator $T: X \rightarrow X$ such that $T$ is injective and onto.
Also, we can define a new norm on $X$ given by $\| x \|_{T} = \| T(x) \| $. Prove that $(X,\| \cdot\|)$, $(X,\| \cdot \|_{T})$ are isometric.
My attemp: I was trying to define the operator $T$ in the usual form of the books, take a Hamel Basis $(x_{i})_{i \in I}$,and WLOG suppose $\|x_{i}\|=1$ for any $i \in I$ then take countable subsequence $J = \{j_{1},j_{2},\dots\}$ of $I$. And define then as follows: $T(x_{ji}) = ix_{ji}$, and $T(x_{i})=0$ if $i \in I-J$. Then we can extend the function and we see that is unbouded. The problem is that, in this way $T$ is neither injective nor onto.
At this point i do not how to redefine the function to be bijective. Any help is welcome, thanks!
This answer assumes A: The axiom of choice, so all vector spaces have a Hamel basis and B: You are working in a field of Characteristic 0. EDIT: As per @SeverinSchraven 's comment, if working working in a field of characteristic $p$, make the labels of $J$ from $\mathbb{N}\setminus p \mathbb{N}$ instead of $\mathbb{N}$
I decided to delete and restart to answer the generic question. Setting things to 0 loses surjectivity, instead take your countable subset $J$ of your index set $I$, label the $J's$ in the natural way, and define $$T(x_j)=jx_j,j\in J$$, $$T(x_i)=x_i\in I \setminus J$$ and extend via linearity. This defines a linear, unbounded operator, no matter the norm.
1-1 and onto are both done by looking at your elements as a finite linear combination of bases elements. If $T(x_1)=T(x_2)$, then $T(x_1 -x_2)=0$. But by definition of $T$, any basis elements in $x_1-x_2$ that come from $J$ would output $j\cdot$ the field element in that spot, so that element is 0 (assuming our field is characteristic 0, anyways!) Not even sure this is true if you have another field characteristic, I haven't worked with vector spaces over such fields.
If the basis element of $x_1-x_2$ comes from $I\setminus J$, then the coefficient of $T(x_1-x_2)$ will just be the underlying field element, so once again it must be 0. Ergo it is injective.
For surjective, take an arbitrary $x\in X$, it can be written as a linear combination of $x_i$ and $x_j$, it has the obvious preimage, take $\frac 1 j x_j$ times the field element for any of those terms and just the field element for any $x_i$ terms
For isometry, we need a linear map that makes it so the action of $T$ doesn't change the norms, that's easy, just use $T^{-1}$