Linearly independent vectors on a normed space

Question

I'm studying functional analysis and I have a problem I don't know how to solve or even how to start. Can anyone help me? The problem is the following:

Let $X$ be a normed space and $x_1,...,x_n$ a set of linearly independent vectors. Show that there exists $\delta>0$ so that if $y_1,...,y_n \in X$ satisfy $\|y_i\|<\delta$ for all $i$ then $x_1+y_1,...,x_n+y_n$ is a set of linearly independent vectors.

Thanks in advance.

Answer 1

Suppose no such $\delta$ exists, then for each $k\in \mathbb N$, we can find a $n$-tuple $(y_1,..,y_n)$ such that $||y_i||<\frac{1}{k}$ for each $i$ and $x_1+y_1,...,x_n+y_n$ is linearly dependent. Thus for some scalars $c_1,c_2,..,c_n$ associated to $k$ (with atleast one of them non-zero), we have $$\sum_{i=1}^nc_ix_i=-\sum_{i=1}^nc_iy_i$$ Taking norm on the both sides gives $$||\sum_{i=1}^nc_ix_i||=||\sum_{i=1}^nc_iy_i||\leq \sum_{i=1}^n|c_i|||y_i||<\frac{1}{k}\sum_{i=1}^n|c_i|$$ If we set $a_i=\frac{c_i}{\sum_{i=1}^n|c_i|}$ and adjust the signs of $y_i$'s and $x_i$'s to make all $c_i$'s non-negative, then we see from above that for each $k\in \mathbb N$, we can find an associated element $X_k=\sum_{i=1}^na_ix_i$ in the symmetric convex hull of $\{ x_1,..,x_n\}$ with norm less than $\frac{1}{k}$. Thus $\{X_k\}\to 0$ and hence $0$ must be in the symmetric convex hull of $x_1,..,x_n$ which is contradiction as these are linearly independent elements.

Just to clarify that, by symmetric convex hull of $\{𝑥_1,𝑥_2,..,𝑥_𝑛\}$, I mean the union of the hulls of $\{𝑠(1)𝑥_1,𝑠(2)𝑥_2,...,𝑠(𝑛)𝑥_𝑛\}$ where $𝑠:\{1,2,..,𝑛\}→\{+1,−1\}$ are sign functions. This can easily seen to be closed and not containing 0