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Let $V$ be complex normed vector space infinite dimensional and let $\{v_1,...,v_n\}\subset V$ a linearly independent finite subset of $V$.

we know that exists $c>0$ such that $\lVert a_1 \cdot v_1+...+a_n \cdot v_n \rVert \ge c \cdot (\lvert a_1\rvert +...+ \lvert a_n\rvert) $ for any $a_1,...,a_n \in \Bbb C$

I would like to know if $c$ is determinable or if $c$ is bounded from any relation on $v_1,...,v_n$

2 Answers 2

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Consider the following operator

$$F:\mathbb{C}^n\to V$$ $$F(a_1,\ldots,a_n)=a_1\cdot v_1+\cdots +a_n\cdot v_n$$

We consider $\mathbb{C}^n$ with taxi cab norm. Now let

$$G:\mathbb{C}^n\to\mathbb{R}$$ $$G(v)=\lVert F(v)\lVert$$

Obviously $G$ is continous (because $F$ is as a finite dimensional operator). Let $\mathbb{S}^n=\{v\in\mathbb{C}^n\ |\ \lVert v\lVert=1\}$ be an $n$-dimensional sphere. Obviously $G$ achieves minimum on $\mathbb{S}^n$ (because $\mathbb{S}^n$ it is compact). Denote this minimum by $c$. Then you have

$$\lVert F(v)\lVert=\lvert v\lvert \cdot \bigg\lVert F\bigg(\frac{v}{\lvert v\lvert}\bigg) \bigg\lVert \geq \lvert v\lvert\cdot c$$

Let $v=(a_1,\ldots, a_n)$. Since we have taxi cab norm in $\mathbb{C}^n$ then

$$\lVert F(a_1,\ldots, a_n)\lVert\geq c\cdot\sum_{i=1}^n\lvert a_i\lvert$$

This is the inequality you've asked us to look at.

So in order to find $c$ you just have to find infimum of $G$ on $\mathbb{S}^n$. In other words

$$c=\inf\bigg\{\bigg\lVert \sum a_i\cdot v_i\bigg\lVert\mbox{, where}\sum\lvert a_i\lvert=1\bigg\}$$

So this is strict equality and there is no better choice of $c$.

Now by taking various points from the sphere (remember that there is the taxi cab norm on $\mathbb{C}^n$) you can find some interesting inequalities. For example for $(a_1,\ldots,a_n)=(0,\ldots,1,\ldots, 0)$ where $1$ is on $i$-th place and zeros otherwise you get

$$c\leq \lVert v_i\lVert$$ for all $i$.

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    thanks so much @freakish, is it possible to find a lower bounds of $c$ ?2017-01-13
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    @MateyMath The smallest possible value for $c$ is given by the $\inf$ formula.2017-01-13
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Perhaps this is too simple, but you can get an easy estimate for $c$ by taking $a_i =1$ and $a_j =0$ for $j \neq i$. In this case we get $$ c = c |a_i| \le \Vert a_i v_i \Vert = \Vert v_i \Vert. $$ Since this works for all $i$ we find that $$ c \le \min\{\Vert v_i \Vert : i=1,\dotsc,n\}. $$

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    ok thanks @Glitch2017-01-13