On existence of a bounded linear functional $f.$

Let $\mathcal H$ be a separable Hilbert space and let $\{e_n\}_{n \geq 1}$ be an orthonormal basis for $\mathcal H.$ Does there exist a bounded linear functional $f : \mathcal H \longrightarrow \mathbb C$ such that $f(e_n) = \frac {1} {\sqrt n}$ for all $n \geq 1\ $?

$\textbf {My Attempt} :$ I think there wouldn't be any such functional which is bounded. First of all such a functional would take the following form. Given $x \in \mathcal H$ we should have $$f(x) = \sum\limits_{n=1}^{\infty} \frac {1} {\sqrt n} \left \langle e_n, x \right \rangle.$$ Now in order to show $f$ is bounded we need to find a sequence of unit vectors $\{x_n\}_{n \geq 1}$ such that the absolute value of the image sequence $\left \{|f(x_n)| \right \}_{n \geq 1}$ is unbounded above. My initial attempt is to take $x_n = \frac {1} {\sqrt n} \sum\limits_{k = 1}^{n} e_k$ for all $n \geq 1$ but that won't help much since the image sequence $\left \{\frac {1} {\sqrt n} \sum\limits_{k = 1}^{n} \frac {1} {\sqrt k} \right \}_{n \geq 1}$ converges to $2$ and hence in particular it is bounded. Could anyone come up with some clever sequence of unit vectors which will do the trick? Any suggestion regarding this would be warmly appreciated.

Thanks a bunch.

Solution 1:

Take $y_n= \sum\limits_{k=2}^{n} \frac 1 {\sqrt k \ln k}e_k$ and $x_n=\frac {y_n} {\|y_n\|}$.

Note that $\|y_n\| \to \sum \frac 1 {k (\ln k)^{2}} <\infty$ and $f(y_n)= \sum_{k \leq n} \frac 1 {k \ln k} \to \infty$.