Exists norms such that $\|x\|=\ell$ but $|x_k|>\ell$ for some $k=\{1,\ldots,n\}$? Let $E$ a finite dimensional normed vector space with dimension $n$. If the set $\{e_k:1\le k\le n\}$ is a basis then we can write any vector of $E$ in the form:
$$x=\sum_{k=1}^n x_k\cdot e_k$$
where the $x_k$ are elements of the field. Then the question is, exists some norm such that $\|x\|=\ell$ but for some coordinate we have that $|x_k|>\ell$? I tried to prove that this is not possible but I failed.
I know that this is not true for any $p$-norm, but Im unable to conclude that this is false for any norm. It seems that the triangle inequality is enough for this proof but I dont know how to write it.
 A: Strangely, this is possible. A trivial example is to define  a norm $\|\cdot \|$ in $\mathbb{R}^n$ with $\|x\|=\tfrac{1}{2}\|x\|_\infty$, for every $x\in \mathbb{R}^n$. Then each basis alement $e_i$, has $\|e_i\|=\tfrac{1}{2}$, but its $i$-th coordinate is equal to $e_i(i)=1>\tfrac{1}{2}$. 
If, like me, you think this example is too trivial and this norm rather useless, then you can construct some more interesting examples. Here is a rule
(lets work in $\mathbb{R}^2$ for simplicity):
Pick your favourite closed, convex set $K$ which is symmetric around $0$,  has nonempty interior and is not a subset of the unit square $S$.  Let me pick for example the rectangle $K=\{(x,y):|x|\leq 2, |y|\leq \tfrac{1}{2}\}$. The Minkowski functional $p_k(x)=\inf \{\lambda>0: x\in \lambda K\}$ defines a norm, the unit ball of which is exactly $K$. Then any element of $K\setminus S$ has $p_k$ norm less or equal than one, but supremum norm strictly greater than one. 
The same construction works in every $\mathbb{R}^n$. Additionally,  since there is a one to one correspondence between norms and closed, symmetric, convex subsets of $\mathbb{R}^n$ with non empty interior, you have actually found all the norms which satisfy this property: They are the norms the unit balls of which are not included in the unit ball of $(\mathbb{R}^n, \|\cdot\|_\infty)$. 
In an infinite dimensional setting, things are even worse as every norm has this property. This is an easy consequence of the  property that  there exist coordinate functionals which are not bounded: 
Let $(X, \|\cdot\|)$ be an infinite dimensional Banach space and $\{b_i:i\in I\}$   a Hamel basis of it. Then there exist $i\in I$ and $x\in X$ with the property that $|x_i|>\|x\|$, where $x_i$ denotes the $i$-th coordinate of $x$ with respect to the Hamel basis. 
