$f(x) = \sum_{k=1}^{+\infty} (1 - \frac{1}{k}) x_k$ is bounded for $x=(x_k) \in \ell_1$ and find its norm.

Question

The question is as follows:

Prove that the functional $f(x) = \sum_{k=1}^{+\infty} (1 - \frac{1}{k}) x_k$ is linear bounded for $x=(x_1, x_2, \ldots) \in \ell_1$ and find its norm.

$\textbf{Some effort:}$

First we note that $f:\ell_1 \to \mathbb{R}$. So we can use Cauchy-Schwarz inequality and so we have

\begin{align} |f(x)| &= \left|\sum_{k=1}^{+\infty} \left(1 - \frac{1}{k}\right) x_k\right|\\ &\le \sum_{k=1}^{+\infty} \left|\left(1 - \frac{1}{k}\right)x_k\right|\\ &\le \sum_{k=1}^{+\infty} \left|1 - \frac{1}{k}\right| |x_k|\\ &\le \left\|\sum_{k=1}^{+\infty} \left(1 - \frac{1}{k}\right)\right\|_{\infty} \|x\|_1\\ &= \sup_{n \in \mathbb{N}} \left\|\sum_{k=1}^{n} \left(1 - \frac{1}{k}\right)\right\|_{\infty}\|x\|_1 \end{align}

So $f$ is bounded and $\|f\| \leq \sup_{n\in\mathbb{N}} \left| \sum_{k=1}^{n} \left(1 - \frac{1}{k}\right)\right| $

Please let me know if my calculation is not correct?

And please let me know that how can I find its norm?

Thanks!

Answer 1

For $x = (x_1, x_2, \ldots ) \in \ell^1$ we have:

$$|f(x)| = \left|\sum_{k=1}^\infty \frac{k-1}{k} x_k\right| \le \sum_{k=1}^\infty \underbrace{\frac{k-1}{k}}_{\le 1} |x_k| \le \sum_{k=1}^\infty |x_k| = \|x\|_1$$

so $f$ is bounded and $\|f\| \le 1$.

Now consider $e_n \in \ell^1$, the $n$-th canonical vector. We have:

$$\|f\| \ge \frac{|f(e_n)|}{\|e_n\|_1} = \frac{\frac{n-1}{n}}{1} = \frac{n-1}{n} \xrightarrow{n\to\infty} 1$$

So $\|f\| \ge 1$. We conclude $\|f\| = 1$.