- Thread starter
- #1

- Feb 5, 2012

- 1,621

Here's a question with my answer, but I just want to confirm whether this is correct. The answer seems so obvious that I just thought that maybe this is not what the question asks for. Anyway, hope you can give some ideas on this one.

**Problem:**

Let \(X\) be a finite dimensional linear space. Let \(x_1,\,\cdots,\,x_n\) be a basis of \(X\). Define the norm,

\[\|x\|_p=\left\|\sum_{i=1}^{n}a_i x_i\right\|_p=\|(a_1,\,\cdots,\,a_n)\|_p\]

If \(f\) is a bounded linear functional on \(X\), find the norm \(\|f\|\).

**My Answer:**

We have been given the following theorem;

Let \((X,\,\|\cdot\|_X)\) and \((Y,\,\|\cdot\|_{Y})\) be two normed linear spaces over \(F\) and \(B(X,\,Y)\) denote the set of all bounded linear functions from \(X\) to \(Y\). Then the function \(\|\cdot\|:\,B(X,Y)\rightarrow\mathbb{R}\) defined by,

\[\|T\|=\sup_{x\in X,\, \|x\|_X\neq 0}\frac{\|T(x)\|_Y}{\|x\|_X}\]

for \(T\in B(X,\,Y)\) is a norm on \(B(X,\,Y)\).

From the above theorem we know that the set of all bounded linear functionals, \(B(X,\,\mathbb{R})\) has the norm,

\[\|f\|=\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\|x\|_p}=\sup_{x\in X,\, \|x\|_p\neq 0}\frac{|f(x)|}{\left(\sum_{i=1}^{n}|a_i|^p\right)^{1/p}}\]