Today i am going to be posting on something related to my current area of research, i am currently

researching on the C*-algebras and i will be posting on a very important theorem of finite-dimensional

normed linear spaces since they tend to have lots of relationship with the inner product spaces and this is

related to the Hilbert spaces, now since the Hilbert spaces are so diverse and wide, it is possible to define a

bounded linear operator on the Hilbert spaces and we can also define many other operators on the Hilbert

spaces, one of the operators i love to study is the adjoint operator defined on an inner product or Hilbert

space. I will be treating this topics in my later discussions. But for now i want to discuss on a very important

theorem of the finite-dimensional normed linear spaces.

Below is a thorem and proof.

 Therorem: Let $(X,||.||)$ be a finite dimensional normed linear space with

basis $\{x_1,x_2,\cdots,x_n\}$, then there is a constant $m>0$ such that for every choice of scalers

$\alpha_1,\alpha_2,\cdots,\alpha_n$ we have

$m\sum^n_{j=1}|\alpha_j|\leq{||\sum^n_{j=1}\alpha_{j}x_{j}||}$


Proof:  If $\sum^n_{j=1}|\alpha_j|=0$, then $\alpha_j=0$ for all $j=1,2,\cdots,{n}$ 

then the inequality holds for any $m>0$.

Assume that $\sum^n_{j=1}|\alpha_j|\neq{0}$ we shall 

proof the result for a set of scalers $\{\alpha_1,\alpha_2,\cdots,\alpha_n\}$ 

that satisfy the condition $\sum^n_{j=1}|\alpha_j|=1$. 

Let

$A=\{(\alpha_1,\alpha_2,\cdots,\alpha_n)\}\in\mathbb{F}^n|\sum^n_{j=1}|\alpha_j|=1$

Now since $A$ is a closed and bounded subset of $\mathbb{F}^n$, it is compact. Define 

$f:A\rightarrow\mathbb{R}$ by 

$f(\alpha_1,\alpha_2,\cdots,\alpha_n)=||\sum_{j=1}^n\alpha_jx_j||$

Since for any $(\alpha_1,\alpha_2,\cdots,\alpha_n)$ and $(\beta_1,\beta_2,\cdots,\beta_n)$ in $A$

$|f(\alpha_1,\alpha_2,\cdots,\alpha_n)-f(\beta_1,\beta_2,\cdots,\beta_n)|$

$=|\|\sum_{j=1}^n\alpha_jx_j\|-\|\sum_{j=1}^n\beta_jx_j\||$

$\leq\|\sum_{j=1}^n\alpha_jx_j-\sum_{j=1}^n\beta_jx_j\|$

$=\|\sum^n_{j=1}(\alpha_j-\beta_j)x_j\|\leq\sum^n_{j=1}|\alpha_j-\beta_j|\|x_j\|$

$\leq\max_{1\leq{j}\leq{n}}\|x_j\|\sum^n_{j=1}|\alpha_j-\beta_j|$

$f$ is continuous on $A$. Since $f$ is a continuous function on a compact set $A$, 

it attains its minimum on $A$, i.e there is an element $(\mu_1,\mu_2,\cdots,\mu_n)\in{A}$ such that

$f(\mu_1,\mu_2,\cdots,\mu_n)$=$\inf\{f(\alpha_1,\alpha_2,\cdots,\alpha_n)|(\alpha_1,\alpha_2,\cdots,\alpha_n)\in{A}\}$

let $m=f(\mu_1,\mu_2,\cdots,\mu_n)$. 

Since $f\geq{0}$, it follows that $m\geq{0}$. If $m=0$, then

$\|\sum^n_{j=1}\mu_jx_j\|=0\Rightarrow\sum^n_{j=1}\mu_jx_j=0$

Since the set $\{x_1,x_2,\cdots,x_n\}$ is linearly independent, 

$\mu_j=0$ for all $j=1,2,\cdots{n}$.

This is a contradiction since $(\mu_1,\mu_2,\cdots,\mu_n)\in{A}$

$0<m\leq{f(\alpha_1,\alpha_2,\cdots,\alpha_n)}$

$\Longleftrightarrow{m}\sum^n_{j=1}|\alpha_j|\leq\|\sum^n_{j=1}\alpha_jx_j\|$
 
Now, let $\{\alpha_1,\alpha_2,\cdots,\alpha_n\}$ be any collection of scalers and 

set $\beta=\sum^n_{j=1}|\alpha_j|$. If $\beta=0$, then the inequality holds vacuously. 

If $\beta>0$, then $(\frac{\alpha_1}{\beta},\frac{\alpha_2}{\beta},\cdots,\frac{\alpha_n}{\beta})\in{A}$ 

and consequently

$\|\sum^n_{j=1}\alpha_jx_j\|$

 $=\|\sum^n_{j=1}\frac{\alpha_j}{\beta}x_j\|$

$=(\frac{\alpha_1}{\beta},\frac{\alpha_2}{\beta},\cdots,\frac{\alpha_n}{\beta})\beta\geq{m}\beta$

$=m\sum^n_{j=1}|\alpha_j|$

That is $m\sum^n_{j=1}|\alpha_j|\leq{||\sum^n_{j=1}\alpha_{j}x_{j}||}$