I am working on exercise 1.2.1 of Chapter 1 of H. Brezis' Book "Functional Analysis, Sobolev Spaces, and Partial Differential Equations", but I am stuck in as I cannot find useful tool in the book to solve this problem.
Here stated the exercise:
Let $E$ be a vector space of dimension $n$ and let $(e_{i})_{1\leq i\leq n}$ be a basis of $E$. Given $x\in E$, write $x=\sum_{i=1}^{n}x_{i}e_{i}$ with $x_{i}\in\mathbb{R}$; given $f\in E^{*},$ set $f_{i}=<f, e_{i}>.$
Now, consider on $E$ the norm $$\|x\|_{1}=\sum_{i=1}^{n}|x_{i}|.$$
(a) Compute explicitly, in terms of the $f_{i}'s$, the dual norm $\|f\|_{E^{*}}$ of $f\in E^{*}$.
(b) Determine explicitly the duality map $F(x)$ for every $x\in E$.
Firstly, I am getting confused if $f_{i}'s$ means the dual basis of $E^{*}$? If so, how to check it is true? For instance, compute if the expression $$\delta_{i,j}=<e_{i},f_{j}>=\Big<e_{i},<f,e_{i}>\Big>$$ satisfies $$\delta_{i,j}=1,\ \text{if}\ i=j,$$ and $$\delta_{i,j}=0,\ \text{if}\ i\neq j?$$
If so, how to compute the inner product inside an inner product?
My attempt is rather pretty short here since I have no idea what should I use, so by definition, we have $$\|f\|_{E^{*}}:=\sup_{\substack{\|x\|\leq 1 \\ x\in E}}<f,x>=\sup_{\substack{\sum_{i=1}^{n}|x_{i}|\leq 1 \\ x\in E}}<f,x>.$$
Then, how could I connect this expression with $f_{i}'s$?
I believe if I can solve $(a)$ then $(b)$ would be straightforward. Since I provide few attempt here, I think any hint would be enough. What should I learn or to look up to solve this problem?
$f_i$'s just numbers. They are not elements of the dual space.
$f(x)=f( \sum\limits_{i=1}^{n} x_ie_i)= \sum\limits_{i=1}^{n} x_if(e_i)= \sum\limits_{i=1}^{n} x_if_i$ (where I have written $f(x)$ for $\langle f, x \rangle$).
Hence $|f(x)| \leq \{max_i |f_i|\} \sum\limits_{i=1}^{n} |x_i|=\{max_i |f_i|\} \|x\|_1$. This proves that $\|f\|_{E^{*}} \leq \{max_i |f_i|\}$. Also $\|f\|_{E^{*}} \geq |f(e_i)|=|f_i|$ for each $i$. [Here I have used the fact that $\|e_i\|=1$]. Conclusion: $\|f\|_{E^{*}}=\{max_i |f_i|\}$.
The definition of $F(x)$ is $F(x)(f)=f(x)$ so $F(x)(f)=\sum\limits_{i=1}^{n} x_if_i$.