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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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    $\|(x,y)\| = |x|+|y|$ on $R^2$? – 2017-02-20
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    @Behnam sorry, I don't understand the question – 2017-02-20
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    http://math.stackexchange.com/questions/2017878/exists-norms-such-that-x-ell-but-x-k-ell-for-some-k-1-ldots-n – 2017-02-23

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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.

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

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    So, does it mean the claim is true for any Hilbert space? – 2017-02-20
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    I think it works for any Hilbert space with a countable or finite orthonormal basis. – 2017-02-20
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    If by "standard basis" you mean a Hamel basis, then it works for finite dimensional Hilbert spaces, that is for $\mathbb{R}^n$ with the Euclidean norm. It's no longer true for infinite dimensional Banach spaces and, in particular, it can't hold for infinite dimensional Hilbert spaces either. But topological bases are much different than Hamel bases. So, if by "standard basis" you actually mean a Schauder basis (ex. an orthonormal basis), then it holds if the basis constant is less or equal than one. – 2017-02-23