The existence of bounded linear functional on a complex Hilbert space

Question

Let $H$ be a complex Hilbert space and $\{e_n\}$ be an orthonormal basis.

1. There exists a bounded linear functional $f:H \to \mathbb C$ such that $f(e_n)=\frac{1}{n}$.

2. There exists a bounded linear functional $f:H \to \mathbb C$ such that $f(e_n)=\frac{1}{\sqrt {n}}$.

$\textbf{My attempt:}$

Let us take an element $x=(1,1,1,\cdots, 1 )\in H$.

For (1), if there exists such $f$ then, $f(x)=f(\sum_{k=1}^n1\times e_k)=\sum_{k=1}^nf(e_k)=\sum_{k=1}^n\frac{1}{k}$ [Since $f$ is a linear functional]

Now taking $n\to \infty$ we get $f(x)=\sum_{k=1}^{\infty}\frac{1}{k}$, the Right Hand Side is divergent since it is Harmonic series. So such $f$ does not exist.

(2) By the similar approach of (1) we can say such bounded linear functional does not exist.

$\textbf{But the answer is given that (1) correct and (2) is not correct}$. But I am not getting why (1) is true. Can anyone give me any hints that how to proceed?

Thanks in advance.

Answer 1

In both cases a linear functional with the given values on the set of the ${e_n}$ will (due to linearity) necessarily satisfy $$f\left(\sum_{i=1}^N \lambda_i e_i\right) = \sum_{i=1}^N \lambda_i f(e_i)$$ and if $f$ is bounded (which implies continuity) the only way to define it for infinite sums looks the same with $N$ replaced by $\infty$.

In the second example, if you want to show that this cannot be a bounded functional, it is sufficient to find an example for which the sums $f\left(\sum_{i=1}^N \lambda_i e_i\right) $ cannot be bounded - for that you were already on track with the example you mention.

In case $f(e_k) = \frac{1}{k}$ note that $$ \left|\sum_{n=1}^N \lambda_n f(e_n) \right|= \left|\sum_{n=1}^N \frac{\lambda_n}{n} \right| \le \sum_{n=1}^N \left|\frac{\lambda_n}{n}\right| \le \sqrt{\sum_{n=1}^N |\lambda_n|^2 }\sqrt{\sum_{n=1}^N\frac{1}{n^2}} $$ This should be more than enough of a hint to complete the exercise.