Is it possible to have a norm in a vector space and a vector whose norm is smaller than the absolute value of one of its coordinates?

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In $\mathbb{R}^n$ the three norms $\|\cdot\|_1$,$\|\cdot\|_2$ and $\|\cdot\|_{\infty}$ verify that for any vector $v \in \mathbb{R}^n$ such that $v=\sum a_ie_i$, where the $e_i$'s are the standard basis vectors, it must be: $$|a_i|\leq\|v\|_j$$ where $i=1,\ldots ,n$ and $j=1,2,\infty$.
So, I wonder if it is the case that for any norm $\|\cdot\|$ in a (possibly finite dimensional) vector space $V$ it must hold that for any vector $v \in V$ s. t. $v=\sum a_ie_i$, where the $e_i$'s are the vectors of a normalized basis, the inequality above also holds.

I couldn't prove it by simply using the definition of the norm, so maybe there are more hypothesis needed to make the claim true. Any thoughts on how to do it?

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Bessel's inequality gives you something of that flavor. More or less it says that if you have an orthonormal basis (finite or countable) of a Hilbert space, then the sum of the squares of the moduli of the coefficients is less than or equal the square of the norm of that element, i.e.

$$\sum_{n=1}^{\infty} | \langle v, e_i \rangle |^2=\sum_{n=1}^{\infty} |a_i|^2 \leq \|v\|^2$$