Is it true in any normed vector space that the modulus of each coordinate of a given vector is less than the norm of the vector itself ? It seems clear for euclidean norms when using an orthonormal basis (even orthogonal), also true of the modulus on the complex plane, but whether true or not for any norm does anyone have a proof or counterexample ?

  • $\begingroup$ It is not always true. The norms satisfying this requirement are sometimes called monotone norms in numerical linear algebra. $\endgroup$ Sep 27, 2016 at 19:57

1 Answer 1


If I understand correctly, your question is the following: Let $X$ be a normed space and $B=\{b_i: i\in I\}$ be a (lets say normalized, to avoid trivial counterexamples) Hamel basis of it. Every $x\in X$ can be written as a linear combination $x=\sum_{i\in F_x} \lambda_i b_i$, where $F_x$ is a finite subset of $I$.

Is it true that $|\lambda_i|\leq \|x\|$ for every $x=\sum_{i\in F_x} \lambda_i b_i\in X$ and $i\in F_x$ ?

The answer depends on the underlying space, however any infinite dimensional Banach space will provide a counterexample. To see why this is the case, remember that for every element $b_i$ of the Hamel basis, we can associate a linear functional $b_i^{\#}$ with the property that $b_i^{\#}(b_j)=\delta_{ij}$, for every $j\in I$. These $b_i^{\#}$'s are called the coordinate functionals associated with the basis $\{b_i:i\in I\}$.

It is known (for example see Are the coordinate functions of a Hamel basis for an infinite dimensional Banach space discontinuous?) that in every Banach space, at most finitely many coordinate functionals are continuous.

To return to your question, the inequality $|\lambda_i|\leq \|x\|$, for every $x\in X$, implies that $b_i^{\#}$ is bounded with $\|b_i^{\#}\|\leq 1$. So, this can only happen for at most finitely many such $i's$.

The answer, of course, is quite different if by "coordinates" you mean coordinates associated with a Schauder basis instead of a Hamel basis. In this case they are always bounded and in most working examples they are uniformly bounded as well.


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