$f(x) = \sum_{k=1}^{+\infty} 2^{-k+1} x_k$ is linear bounded for $x=(x_k) \in c_0$ and find its norm.

Question

The question is as follows:

Prove that the functional $f(x) = \sum_{k=1}^{+\infty} 2^{-k+1} x_k$ is linear bounded for $x=(x_1, x_2, \ldots) \in c_0$ and find its norm.

$\textbf{Some effort:}$

First we note that $f:c_0 \to \mathbb{R}$.

For to show that it is bounded, we have \begin{align} |f(x)| &= \left|\sum_{k=1}^{+\infty} 2^{-k+1} x_k \right|\\ &\le \sum_{k=1}^{+\infty} \left| 2^{-k+1} x_k \right|\\&\le \sum_{k=1}^{+\infty} \left| 2^{-k+1}\right|\left| x_k \right|\\&\le \left\|\sum_{k=1}^{+\infty}2^{-k+1} \right\|_{1} \left\| x_k \right\|_{\infty}\\& = 2 \sum_{k=1}^{+\infty} 2^{-k} \left\| x_k \right\|_{\infty}\\& = 2 \left\| x_k \right\|_{\infty} \end{align} So $f$ is bounded and $\|f\| \leq 2 $.

for to show that its norm is equal to 2, we have to show that $\|f\| \geq 2$. For to show this we have to find a sequence $x=(x_k) \in c_0$ such that by substituting it in $f(x)$ instead of $x$ and also by finding its norm $\|x\|_{\infty} = \sup_{k\in\mathbb{N}}\{|x_k|\}$ and by using $|fx|\leq \|f\|_1 \|x_k\|_{\infty}$, i.e. by moving $\|x_k\|_{\infty}$ from right side to the left side of the inequality, to find 2 out of this process.

But I couldn't find such sequence. Can someone please help me to find its norm?

And also please let me know if my calculation for boundedness is not correct?

Thanks!

Answer 1

Your solution for boundedness is fine.

Now take a sequence $y_n$ of sequences in $c_0$ namely $y_n=(1, 1,...,1,0,0,...)$ so $1$'s until the $n$-th number, after that everything is zero. You can check that $\Vert y_n \Vert=1$ and $|f(y_n) |\to 2$. That gives $\Vert f\Vert =2$ since you found a sequence converging to the (least) upper bound.

Answer 2

We know that $(c_0)^* = \ell^1$. Notice that your $f \in (c_0)^*$ is represented by the sequence $$(a_n)_{n=1}^\infty = \left(\frac{1}{2^{n-1}}\right)_{n=1}^\infty = \left(1, \frac12, \frac14, \frac18, \ldots\right) \in \ell^1$$

because $f(x) = \displaystyle\sum_{n=1}^\infty a_nx_n = \sum_{n=1}^\infty \frac{x_n}{2^{n-1}} $ for all $x \in c_0$.

Now $\|f\|$ is simply the $\ell^1$-norm of the sequence represnting it:

$$\|f\| = \left\|(a_n)_{n=1}^\infty\right\|_1 = \sum_{n=1}^\infty |a_n| = \sum_{n=1}^\infty \frac{1}{2^{n-1}} = 2$$