I'm trying to find a proof for the following:

Let {$u_{1},...,u_{n}$} be a linearly independent set of a normed space $X$. Then, there is a constant $c>0$ such that for every set of scalars $\{\alpha_{1},...,\alpha_{n}\}:$

$$\| \alpha_{1}u_{1}+...+\alpha_{n}u_{n} \| \ge c(|\alpha_{1}|+...|\alpha_{n}|)$$

I don't even know how to start.


1 Answer 1


Let $S$ be the set of all tuples of scalars $(\alpha_1,\dots,\alpha_n)\in K^n$ such that $|\alpha_1|+\cdots+|\alpha_n|=1$.

Then the map $f:S\rightarrow [0,\infty)$ given as

$$f(\alpha_1,\dots,\alpha_n)=\|\alpha_1 u_1+\cdots+\alpha_n u_n\|$$

is continuous. Because $S$ is compact, the continuous function $f$ attains its infimum on $S$ at some $x_0\in S$, i.e. $\inf f(S)=f(x_0)$. Since the $u_1,\dots,u_n$ are linearly independent, we have $c:=f(x_0)>0$.


$$\|\alpha_1 u_1+\cdots+\alpha_n u_n\|\ge c$$

for all $\alpha_1,\dots,\alpha_n$ with $|\alpha_1|+\cdots+|\alpha_n|=1$. For a general set of scalars $\alpha_1,\dots,\alpha_n$ (not all zero) we just divide them all by $|\alpha_1|+\cdots+|\alpha_n|$ to obtain a set of scalars in $S$. Then homogeneity of the norm yields

$$\|\alpha_1 u_1+\cdots+\alpha_n u_n\|\ge c\cdot (|\alpha_1|+\cdots+|\alpha_n|)$$

for all scalars $\alpha_1,\dots,\alpha_n$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .