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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 $\|f\|$.

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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?

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  • $\begingroup$ I don't use the same definition of a bounded function... Are you sure of yours? $\endgroup$
    – Damien
    Dec 27 '18 at 8:22
  • $\begingroup$ @Damien: Yes, you can also check en.wikipedia.org/wiki/Bounded_operator, to verify! $\endgroup$ Dec 27 '18 at 8:24
  • $\begingroup$ @Damien: Which do you use? $\endgroup$ Dec 27 '18 at 8:24
  • $\begingroup$ 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 $\endgroup$
    – Damien
    Dec 27 '18 at 8:35
  • $\begingroup$ @Damien: No problems! We all learn! $\endgroup$ Dec 27 '18 at 8:51
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$\|f\|=\sup \{\frac {|f(\sum a_ie_i)|} {\|\sum a_ie_i\|}\}\leq\sqrt {\sum \|f(e_i)\|^{2}}$ 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 $\sqrt {\sum \|f(e_i)\|^{2}}$

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  • $\begingroup$ Sorry, I find this result very strange. For example, consider $n=2$, $f(e_1)=e_1$, $f(e_2)=2e_2$ $\endgroup$
    – Damien
    Dec 27 '18 at 8:52
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    $\begingroup$ @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. $\endgroup$ Dec 27 '18 at 9:09
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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}

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