# Normed linear space question

Suppose $x_1, x_2, \ldots, x_n$ are linearly independent elements of a normed linear space $X$. Show that there is a constant $c>0$ with the property that for every choice of scalars $\alpha_1, \ldots, \alpha_n$ we have $$\|\alpha_1x_1+\cdots+\alpha_nx_n\|\geq c(|\alpha_1|+\cdots+|\alpha_n|)$$

I tried doing this by contradiction but I am stuck.

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Obviously it suffices to show this for $|a_1|+\dots+|a_n|\le 1$. The set of such $n$-tuples is a compact subset in $\mathbb R^n$. So it would suffice to show that $(a_1,\dots,a_n)\mapsto \lVert a_1x_1+\dots+a_nx_n\rVert$ is continuous. – Martin Sleziak Jun 30 '12 at 20:35

Here's an outline:

1) Use a "normalization argument" to show that it suffices to prove your result for $|\alpha_1|+\cdots+|\alpha_n|=1$.

2) Define a map from the unit sphere of $\ell_1^n$ to $\Bbb R$ via $(\alpha_1,\ldots,\alpha_n)\mapsto\Vert\alpha_1 x_1+\cdots+\alpha_n x_n \Vert$.

3) Using the axioms of norm, show that this function is continuous.

4) Using the fact that the unit sphere of $\ell_1^n$ is compact, show that this function attains a minimum value.

5) Using the independence of the $x_i$, show that this minimum value is positive.

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what exactly is the normalization argument. I thought about it and I got nowhere. – john Jul 1 '12 at 19:55
@john For $0\ne y=\alpha_1 x_1+\cdots+\alpha_n x_n$, let $\beta=|\alpha_1|+\cdots+|\alpha_n|$. Assuming your result holds for $|\alpha_1 |+\cdots+|\alpha_n |=1$, you can apply it to $y/\beta$ to obtain the general result. – David Mitra Jul 1 '12 at 20:00

The span $W$ of the vectors $x_k$ is a finite dimensional space. The mapping $$\sum_{k=1}^n \alpha_k x_k \mapsto \sum_{k=1}^n |\alpha_k|$$ is a norm. It is well-defined since the vectors $x_k$ form a basis for $W$. All norms on $W$ are equivalent, so the result follows.

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Try to solve the problem in case where $X$ is an inner product space.

Notice that we can assume without loss of generality that $x_i$ span $X$, and that all finite-dimensional normed spaces look a lot like inner product spaces (since all norms on a finite-dimensional space are equivalent).

In fact, you can probably choose an inner product where all the vectors in question are orthogonal.

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Take the functin $F: W \rightarrow \mathbb{R}$ and $W \subset [-1,1] \mathrm{x} [-1,1] \mathrm{x} \cdots [-1,1] \,n$ times $W=\{(\lambda _1,\lambda _2, \cdots , \lambda _n) | |\lambda _1 |+ |\lambda _2 |+ \cdots |\lambda _n |=1 \}$ $W$ is closed hence compact times such that $F(y)= \|\lambda_1x_1+\cdots+\lambda_nx_n\|$ now$F$ is continuous $F(y) >0$ and for every choise of $y \in W$ and since is compact it takes a minimum that is your $c$

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why can you assume the scalars add up to 1? – john Jun 30 '12 at 21:33
why can you assume that? – john Jun 30 '12 at 21:38
I should have taken absolute values.. It is the normalization argument of David mitra – clark Jun 30 '12 at 21:43
I think it is ok now right? – clark Jun 30 '12 at 21:46

If there exist scalars $\alpha_1, \ldots, \alpha_n$ not all zero such that for all $c > 0$ we have $$\left\|\sum_{i=1}^n\alpha_i x_i\right\| < c\sum_{i=1}^n\lvert \alpha_i\rvert,$$ then $\alpha_1 x_1 + \cdots + \alpha_n x_n = 0$, which means the $x_i$ are not linearly independent.

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