Suppose $E$ is a normed linear space, $\{x_k\}_{k=1}^{\infty}$ is linearly independent and $\|x_k\|=1, k=1,2,\cdots$. Give a example to show that there not necessarily exist uniformly bounded linear funtionals $\{f_j\}$ such that $f_j(x_k)=\delta_{jk},\ j,k\geq 1$.
I know there are already many posts on MSE discussing about dual basis of infinite dimensional space and the finite case of my question. But I have not founded the answer of my question. Appreciate any help or hint!
Consider an infinite dimensional Hilbert space with orthonormal basis $\{e_n\}.$ Let $$ x_1=e_1,\qquad x_n=\cos (n^{-1})\,e_1+\sin(n^{-1})\,e_n, \quad n\ge 2$$ The elements $x_n$ are linearly idependent, $\|x_n\|=1$ and $x_n\to x_1.$ If $f_1$ existed then $$0=\lim_n f_1(x_n)=f_1(x_1)=1,$$ which would give a contradiction.
In the example the sequence $x_n$ does not satisfy $$\qquad \qquad\|x_n-x_m\|\ge \delta>0,\qquad n\neq m, \qquad\qquad\qquad(1) $$ However, it is possible to construct a counterexample satisfying $(1).$
Indeed, let $$x_1=\sum_{k=1}^\infty {1\over \sqrt{k(k+1)}}e_k,\qquad x_n=e_{n-1},\quad n\ge 2.$$ Then the elements $x_n$ are linearly independent, $\|x_n\|=1,$ $\|x_n-x_m\|^2=2$ for $n>m\ge 2.$ Moreover $$\|x_2-x_1\|^2=\left (1-{1\over \sqrt{2}}\right )^2+\sum_{k=2}^\infty{1\over k(k+1)}\ge {1\over 2},$$ $$\|x_n-x_1\|^2={1\over 2}+\sum_{k\ge 2,\ k\neq n}{1\over k(k+1)}\ge {1\over 2},\qquad n\ge 3$$ We have $$\sum_{k=2}^n{1\over \sqrt{(k-1)k}}x_k=\sum_{k=1}^{n-1}{1\over \sqrt{k(k+1)}}e_k \underset{n}{\longrightarrow} x_1.$$ The functional $f_1,$ if existed, would satisfy $$0=\sum_{k=2}^n{1\over \sqrt{(k-1)k}}f_1(x_k)\underset{n}{\longrightarrow} f_1(x_1)=1$$