Ngô Quốc Anh

February 23, 2017

In a normed space, finite linearly independent systems are stable under small perturbations

Filed under: Giải tích 8 (MA5206) — Ngô Quốc Anh @ 23:21

In this topic, we show that in a normed space, any finite linearly independent system is stable under small perturbations. To be exact, here is the statement.

Suppose $(X, \|\cdot\|)$ is a normed space and $\{x_1,...,x_n\}$ is a set of linearly independent elements in $X$. Then $\{x_1,...,x_n\}$ is stable under a small perturbation in the sense that there exists some small number $\varepsilon>0$ such that for any $\|y_i\| < \varepsilon$ with $1 \leqslant i \leqslant n$, the all elements of $\{x_1+y_1,...,x_n+y_n\}$ are also linearly independent.

We prove this result by way of contradiction. Indeed, for any $\varepsilon>0$, there exist $n$ elements $y_i \in X$ with $\|y_i\| < \varepsilon$ such that all elements of $\{x_1+y_1,...,x_n+y_n\}$ are linearly dependent, that is, there exist real numbers $\alpha_i$ with $1 \leqslant i \leqslant n$ such that

$\displaystyle \alpha_1 (x_1+y_1) + \cdots + \alpha_n (x_n+y_n) =0$

with

$\displaystyle |\alpha_1| + \cdots + |\alpha_n| >0.$

Clearly,

$\displaystyle \alpha_1 y_1 + \cdots + \alpha_n y_n =- (\alpha_1 x_1 + \cdots + \alpha_n x_n)$

Observe that

$\displaystyle \|\alpha_1 x_1 + \cdots + \alpha_n x_n \| \leqslant |\alpha_1| \|y_1\| + \cdots + |\alpha_n |\|y_n \| <\varepsilon (|\alpha_1| + \cdots + |\alpha_n| ).$

Therefore, if we denote $c_i = \alpha_i/(|\alpha_1| + \cdots + |\alpha_n| )$, then

$\displaystyle |c_1|+\cdots +|c_n| = 1, \quad \|c_1 x_1 + \cdots + c_n x_n \| < \varepsilon.$

Hence, we have just shown that for any $\varepsilon>0$, there exist $n$ numbers $c_i$ with

$\displaystyle |c_1|+\cdots +|c_n| = 1, \quad \|c_1 x_1 + \cdots + c_n x_n \| < \varepsilon.$

Set

$\displaystyle E = \{ (c_1,\dots, c_n) : |c_1|+\cdots +|c_n| = 1 \}.$

Since $E$ is closed and bounded, it is compact. We also set

$\displaystyle f : E \to [0, +\infty), \quad f(c_1,\dots,c_n) = \|c_1 x_1 + \cdots + c_n x_n \|.$

Clearly, $f$ is continuous. We now consecutively choose $\varepsilon = 1/k$ with $k \geqslant 1$. Then there exists a sequence $\{c^k\}_k \subset E$ such that $f(c^k) < 1/k$. Hence

$\inf_E f(c) =0.$

By the Weierstrass extreme value theorem, there exists some $c \in E$ such that $f(c)=0$. However, this is impossible because $\{x_1,...,x_n\}$ is the set of linearly independent elements. The proof is complete.

1. Define $f:S^{n-1} \to [0,\infty)$ by $f(c) = \|c_1x_1 + \cdots +c_nx_n\|$ with $c =(c_1,\cdots,c_n)$. $f$ is continous. Since $x_1, \ldots, x_n$ is linearly independent then $f(c) >0$ for any $c\in S^{n-1}$. Hence $b = \inf_{c\in S^{n-1}} f(c) > 0$. By triangle inequality, we have for any $c \in S^{n-1}$

$\displaystyle \|\sum c_i(x_i + y_i)\| \geq \|\sum c_i x_i\| -\sum |c_i| \|y_i\| \geq b - \max_i \|y_i\|\sum |c_i| \geq b -\sqrt{n} \max_i \|y_i\|.$

Choose $\epsilon < b/\sqrt{n}$, then for any $y_1, \ldots, y_n \in X$ with $\|y_i\| \ne 0$ which implies $x_1 +y_1, \ldots, x_n +y_n$ is linearly independent. 🙂

Comment by Van Hoang Nguyen — February 25, 2017 @ 5:44

• typos: Choose $latex \epsilon 0$ which then implies $x_1 +y_1,\ldots,x_n +y_n$ is linearly independent.

Comment by Van Hoang Nguyen — February 25, 2017 @ 5:52

• Cool!

Comment by Ngô Quốc Anh — February 25, 2017 @ 13:11

2. This result can be seen as a consequence of a perturbation theorem from theory of metric regularity: roughly speaking, the metric regularity of a set-valued mapping is stable under a small linear perturbation (Since $\{x_1,...,x_n\}$ is linearly independent, the mapping $F(x*):=(x*(x_1),...,x*(x_n))$. is metrically regular. So, for $y_1,...,y_n$ in X near 0, the linearly perturbed mapping $F_y(x*):= (x*(x_1+y_1),...,x*(x_n+y_n))$ is also metrically regular. The latter entails the linear independence of $\{x_1+y_1,...,x_n+y_n\})$.

Comment by Nguyen Huy Chieu — March 6, 2017 @ 14:29