How to prove the following assertion for bounded linear functionals?

Question

let $e_i$ be vector with $1$ at the $i^{th}$ pl ace and $0$ elsewhere for $i = 1, 2, \ldots$. Then how to prove that $\{f(e_i)\}$ converges for every bounded linear functional on $l^2$.

I start with $$|f(e_m)-f(e_n)| = |f(e_m-e_n)| \leq \|f\| \|e_n-e_m\|$$ Using $f$ bounded, we get $$|f(e_m)-f(e_n)| \leq K \sqrt{2}$$. But how to prove the convergence?

Answer 1

The space $l^2$ is a Hilbert space, so by the Riesz's representation theorem $f(x)=<\xi,x>$, for some $\xi=(\xi_i)_{i\geq 1} \in l^2$.

Therefore, $f(e_i)= \xi_i$, however since $\xi\in l^2$, $\xi_i\rightarrow 0$ as $i\rightarrow \infty$. $\blacksquare$

Answer 2

clark pointed out a solution with Riesz's representation theorem.

Here is an alternative way if you do not know it yet. After having looked at some examples, we can reasonably conjecture that $f\left(e_i\right)\to 0$.

Suppose that the sequence $\left(f\left(e_i\right)\right)_{i\geqslant 1}$ does not converge to zero. Then there exists a positive $\varepsilon$ and an increasing sequence $\left(i_l\right)_{i\geqslant 1}$ of integers such that $\left\lvert f\left(e_{i_l}\right)\right\rvert\gt\varepsilon$ for all $l$. Define $$v_N:=\sum_{l=1}^Nc_l\operatorname{sgn}\left(f\left(e_{i_l}\right) \right) \cdot e_{i_l},$$ where $\operatorname{sgn}(x)$ equals $1$ if $x$ is positive, $-1$ if it is negative, and $c_l\gt 0$ will be chosen later. We have $$\left\lVert f\right\rVert\cdot \left\lVert v_N\right\rVert\geqslant f\left(v_N\right)\geqslant \varepsilon\sum_{l=1}^Nc_l $$ hence choosing a sequence $\left(c_l\right)_{l\geqslant 1}$ such that $\sum_{l=1}^{+\infty}c_l=+\infty$ and $\sum_{l=1}^{+\infty}c_l^2\lt +\infty$ contradicts boundedness of $f$.