Lema prior to completion of finite normed vectorial spaces. Let $(V, \mathbb{K},||.||)$, with $n=$dim($V)< \infty$ and $\beta= \lbrace u_1, u_2, ... u_n \rbrace$ a basis of $V$. I want to prove that $\forall x \in V$, if $x=\sum_{i=1}^{n} a_i u_i$ $\Rightarrow$ $\exists c>0$ $\backepsilon$ $c \cdot \sum_{i=1}^{n} |a_i | \leq ||x||= ||\sum_{i=1}^{n} a_i u_i||$...(1). I found this result is being used on a proof about completion of finite dimentional normed spaces.
My attempt so far: By induction on $n$.
The case for $n=1$ is trivial since $|a_1| || u_1 ||= ||a_1 u_1 || =||x||$ and $u_1 \neq 0_V$
Then I tried to use the property $| |x| - |y| |\leq |x-y|$ so I obtained something like this:
$\exists c_1 >0 \backepsilon$ $$c_1 \cdot \sum_{i=1}^{n} |a_i| - |a_{n+1}| ||u_{n+1} ||$$
$$\leq  || \sum_{i=1}^{n} a_i u_i|| - |a_{n+1}| ||u_{n+1} ||$$
$$\leq |\text{   } ||\sum_{i=1}^{n} a_i u_i|| - |a_{n+1}| ||u_{n+1} || \text{   } | $$
$$\leq ||\sum_{i=1}^{n} a_i u_i +a_{n+1} u_{n+1} || $$
$$= ||\sum_{i=1}^{n+1} a_i u_i || =x$$
But I can't go further. Any ideas are welcome. Thanks.
 A: As far as I can tell, the proof cannot work down the path you chose.
What you are trying to prove is the nontrivial part of the argument that all norms are equivalent on a finite-dimensional normed space.
The usual way is to consider the map $\Gamma:\mathbb K^n\to V$ given by
$$
\Gamma(a_1,\ldots,a_n)=\sum_{j=1}^n a_ju_j.
$$
This map is trivially bijective, and the reverse inequality of the one you want holds, so $\Gamma$ is bounded. The set $$B=\{(a_j):\ \sum_{j=1}^n|a_j|=1\}\subset\mathbb K^n$$ is compact, so the continuous function $$(a_j)\longmapsto\Big\|\sum_{j=1}^na_ju_j\Big\|$$ maps $B$ to a compact subset of $\mathbb K$. In particular there exists $d$ such that $\|\Gamma((a_j))\|≤d$ on $B$. This gives
$$
\sum_{j=1}^n|a_j|=1\}\leq d\,\Big\|\sum_{j=1}^na_ju_j\Big\|.$$
A: Let $s=\sum_{i=1}^{n} |a_i|$. If $s=0 \Rightarrow a_j=0$ $1\leq j \leq n$. Therefore (1) is true $\forall c \in \mathbb{K}$, in partiular $\forall c >0$.
Supose now that $s>0 \Rightarrow$
$||\sum_{i=1}^{n} a_j x_j || \geq c ( \sum_{i=1}^{n} |a_j|) =cs =c|s| \Leftrightarrow \dfrac{1}{|s|} ||\sum_{i=1}^{n} a_j x_j ||\geq c \Leftrightarrow ||\dfrac{1}{s} \sum_{i=1}^{n} a_j x_j ||\geq c $
$\Leftrightarrow ||  \sum_{i=1}^{n} (\dfrac{a_j}{s}) x_j || \geq c \Leftrightarrow || \sum_{i=1}^{n} \beta_j x_j || \geq c$.
Where $\beta_j = \dfrac{a_j}{s}$ and $\sum_{i=1}^{n} |\beta_j| =1$.
Thus, (1) is equivalent to prove that for a l.i. subset $\{x_1, x_2 , ... x_n \} \subset V$, $\exists c \in \mathbb{K} \backepsilon \forall \beta_1, \beta_2 , ... \beta_n  \in \mathbb{K} \backepsilon \sum_{i=1}^{n} |\beta_j| =1 \Rightarrow ||\sum_{i=1}^{n} \beta_j x_j || \geq c $ ... (2)
By contradiction, suppose that its false. (i.e. for a l.i. subset $\{x_1, x_2 , ... x_n \} \subset V$ $(\forall c \in \mathbb{K})$ ($\exists \beta_j \in \mathbb{K}, 1\leq j \leq n$) [($\sum_{j=1}^{n}|\beta_j|=1$) $\Rightarrow$ ($||\sum_{i=1}^{n} \beta_j x_j ||\leq c$)].
With this, we have in particular that $\forall m \in \mathbb{N}$ $\exists \{\beta_{1}^{(m)},\beta_{2}^{(m)},...,\beta_{n}^{(m)} \} \subset \mathbb{K}$ $\backepsilon $ $\sum_{j=1}^{n}|\beta_{j}^{(m)}|=1 \Rightarrow ||\sum_{i=1}^{n} \beta_j^{(m)} x_j ||\leq \dfrac{1}{m}$
$\Rightarrow \exists \{ y_m \} \subset V$, where $\forall m \in \mathbb{N}, y_m = \sum_{i=1}^{n} \beta_{j}^{(m)} x_j \backepsilon \text{lim}_{m \rightarrow \infty} ||y_m||=0$ because $ ||y_m || \leq 1/m  $.
Because $\sum_{i=1}^{n} | \beta_j^{(m)} | =1 \forall m \in \mathbb{N} \Rightarrow \forall j \in \{1,2,...n\}$,   $0 \leq |\beta_j^{m}| \leq \sum_{i=1}^{n} | \beta_j^{(m)} | =1$ $\Rightarrow 0\leq |\beta_j^{(m)}| \leq 1  \forall m \in \mathbb{N} , 1 \leq j \leq n$. That implies that the sequence $\{ \beta_j^{(m)}\}=\{ \beta_j^{(1)} , \beta_j^{(2)} , ...\}$ is bounded and by construction countable $\Rightarrow $ (Bolzano-Weierstrass Theorem) $\{ \beta_j^{(m)}\}´ \neq \emptyset$ ($1\leq j \leq n$). Let $\beta_j \in \{ \beta_j^{(m)}\}´$ ($1\leq j \leq n$)  $\Rightarrow \exists $ a subsequence $\{ \beta_j^{(k_m)}\} \subset \{ \beta_j^{(m)}\}-\{\beta_j \} \backepsilon $ $\text{lim}_{m \rightarrow \infty } \beta_{j}^{(k_m)} = \beta_j$ for ($1\leq j \leq n$).
Let $j=1$. taking the natural numbers that index the convergent subsequence $\{ \beta_1^{(k_m)}\}$, we generate  a corresponding subsequence of $ \{ y_m \} $, wich we denote by $\{ y_{1,m} \}$.
Let $j=2$, taking the natural numbers that index the convergent subsequence $\{ \beta_2^{(k_m)}\}$ and intersecting them with the subscripts of $\{ y_{1,m} \}$, we generate a corresponding subsequence of $\{ y_{1,m} \}$ wich we denote by $\{ y_{2,m} \}$.
Continuing with this way, afther $n$ steps, we obtain a subsequence $ \{ y_{n,m} \} = \{ y_{n,1}. y_{n,2} , .... \}$ of $ \{ y_m \} $ whose terms are of the form
\begin{align}
& y_{n,m} = \sum_{j=1}^{n} \gamma_{j}^{(m)}x_{j} &   &  \text{with} & ( \sum_{j=1}^{n} |\gamma_{j}^{(m)}| =1 ) \\
\end{align}
whose scalars $ \gamma_{j}^{(m)}$ satisfy $\text{lim}_{m \rightarrow \infty} \gamma_{j}^{(m)} = \beta_j$ for ($1\leq j \leq n$). Therefore
\begin{equation}
\text{lim}_{m \rightarrow \infty} y_{n,m} = \text{lim}_{m \rightarrow \infty} \sum_{j=1}^{n} \gamma_{j}^{(m)} x_j = \sum_{j=1}^{n} x_j \text{lim}_{m \rightarrow \infty} \gamma_{j}^{(m)} = \sum_{j=1}^{n}  x_j \beta_j = \sum_{j=1}^{n} \beta_{j} x_j := y \end{equation}.
Where also
\begin{equation}
\sum_{j=1}^{n} |\beta_j | = \sum_{j=1}^{n} | \text{lim}_{m \rightarrow \infty} \gamma_{j}^{(m)} | = \sum_{j=1}^{n} \text{lim}_{m \rightarrow \infty} | \gamma_{j}^{(m)} | = \text{lim}_{m \rightarrow \infty}  \sum_{j=1}^{n} | \gamma_{j}^{(m)} | = \text{lim}_{m \rightarrow \infty}  1 = 1
\end{equation}.
$\therefore \sum_{j=1}^{n} |\beta_j | = 1 \Rightarrow $ not all $ b_j $ are zero. Since $ \{ x_1, x_", ..., x_n\}$ is l.i. $\Rightarrow y \neq 0$, that can be easily seen by contradiction.
On the other side, $\text{lim}_{m \rightarrow \infty} y_n,m =y \Rightarrow \text{lim}_{m \rightarrow \infty} ||y_n,m || = ||y|| $ because the continuity of the norm. Remember for hypothesis that we had $\text{lim}_{m \rightarrow \infty} ||y_m || =0 $, and by the fact that $ \{ y_{n,m} \} $ is a subsequence of a convergent sequence, it must be that $\text{lim}_{m \rightarrow \infty} ||y_n,m ||=0$ $\Rightarrow $ (uniqueness) $||y||=0 \Leftrightarrow y=0$
$\therefore y=0 \wedge y \neq 0 \mathbb{!}$ So (2) must be true and with that it's also true (1). Q.E.D
This proof can be found on page 72 of the book written by Erwin Kreyszig "Introductory functional analysis with applications" (Wiley, 1978).
