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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.

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  • $\begingroup$ This has been answered here many times. $\endgroup$ May 18 at 8:36
  • $\begingroup$ and this time the OP did not show any work. $\endgroup$
    – GEdgar
    May 18 at 8:59

1 Answer 1

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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 $\{π‘₯_1,π‘₯_2,..,π‘₯_𝑛\}$, I mean the union of the hulls of $\{𝑠(1)π‘₯_1,𝑠(2)π‘₯_2,...,𝑠(𝑛)π‘₯_𝑛\}$ where $𝑠:\{1,2,..,𝑛\}β†’\{+1,βˆ’1\}$ are sign functions. This can easily seen to be closed and not containing 0

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