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

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?

• I don't use the same definition of a bounded function... Are you sure of yours? Dec 27 '18 at 8:22
• @Damien: Yes, you can also check en.wikipedia.org/wiki/Bounded_operator, to verify! Dec 27 '18 at 8:24
• @Damien: Which do you use? Dec 27 '18 at 8:24
• It appears that the definition of a bounded linear function is different from the definition of a bounded function. I did not know. Yo are right. Sorry Dec 27 '18 at 8:35
• @Damien: No problems! We all learn! Dec 27 '18 at 8:51

$$\|f\|=\sup \{\frac {|f(\sum a_ie_i)|} {\|\sum a_ie_i\|}\}\leq\sqrt {\sum \|f(e_i)\|^{2}}$$ 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 $$\sqrt {\sum \|f(e_i)\|^{2}}$$
• Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$ Dec 27 '18 at 8:52
• @Damien I thought $f$ was a continuous linear functional but I now realize that it is a linear map of $X$ into $X$. The exact value of $\|f\|$ is hard to write. From what the OP has written it is not clear if he is trying to write an explicitly formula for $\|f\|$. Thanks for your comment. Dec 27 '18 at 9:09
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}