Assuming that $f$ is a bounded linear function on $X$, compute $\|f\|$.

Question

Let $X$ be a finite dimensional linear space and let $\{e_i\}^{n}_{i=1}$ be a basis. Hence, there exists unique scalars $\{\alpha_i\}^{n}_{i=1}$ such that \begin{align} x=\sum^{n}_{i=1}\alpha_i e_i.\end{align} Now, we define $\|\cdot\|$ by \begin{align} \|x\|=\left(\sum^{n}_{i=1}|\alpha_i |^2\right)^{1/2}.\end{align} Assuming that $f$ is a bounded linear function on $X$, I want compute $\|f\|$.

MY TRIAL

Since $f$ is a bounded linear function on $X$, then there exists $K\geq 0$ such that \begin{align} |f(x)|\leq K\|x\|,\;\;\forall\;x\in X.\end{align} So, taking $\sup$ over $\|x\|\leq 1,$ we get \begin{align} \|f\|=\sup\limits_{\|x\|\leq 1}|f(x)|\leq K,\;\;\forall\;x\in X.\end{align} I might be wrong. Kindly check if I'm wrong or right. If I'm wrong, can you give me an idea of what to do?

Answer 1

$\|f\|=\sup \{\frac {|f(\sum a_ie_i)|} {\|\sum a_ie_i\|}\}\leq\sqrt {\sum \|f(e_i)\|^{2}}$ by C-S inequality. The exact value of $\|f\|$ is not easy to write and it is not $\sqrt {\sum \|f(e_i)\|^{2}}$ in general. For example, User Damien has given example in a comment in which the norm is not equal to $\sqrt {\sum \|f(e_i)\|^{2}}$

Answer 2

I guess it should be: \begin{align} \|f\|&=\sup\limits_{x\neq 0}\dfrac{|f(x)|}{\|x\|}\\&=\sup\limits_{\alpha_i\neq 0}\dfrac{\left|f(\sum^{n}_{i=1}\alpha_i e_i)\right|}{\left(\sum^{n}_{i=1}|\alpha_i |^2\right)^{1/2}}\\&=\sup\limits_{\alpha_i\neq 0}\dfrac{\left|\sum^{n}_{i=1}\alpha_i f(e_i)\right|}{\left(\sum^{n}_{i=1}|\alpha_i |^2\right)^{1/2}}.\end{align}