Learning how to work with the operator norm in functional analysis.

Question

$\newcommand\N{\Bbb N} \newcommand\R{\Bbb R}$

I apologize if this question has been asked before,

1.) For $1\leq p\leq \infty$, let $l^p=L^p(\N,\mu)$ where $\mu$ denotes the counting measure. Hence $l^P$ consists of sequences $(x_n)$ such that $||(x_n)||_p = (\sum_{k=1}^\infty |x_n|^P)^{1/p}<\infty$ in case $1\leq p < \infty$ and $||(x_n)||_\infty$=$\sup\{|x_n| \mid n\in \N\}<\infty$ in case $p=\infty$.

Let $c_0$ be the closed subspace of $l^\infty$ consisting of all sequences that converge to $0$.

(a.) Prove: If $(y_n)$ is a sequence in $l^1$ and $f$ is defined for $(x_n)$ in $c_0$ by $f((x_n))=\sum_1^\infty x_ky_k$ then $f$ is a bounded linear functional on $c_0$ and $||f||=||(y_n)||_1$.

This is what I have tried so far. I am not sure how to approach the reverse inequality.

(a.) Claim: $f((x_n))$ is linear. Proof: Let $a\in \R$ $(x_n), (z_n) \in c_0$ then $f((ax_n))=\sum_1^\infty ax_ky_k=a\sum_1^\infty x_ky_k=af((x_n))$ and $f((x_n)+(z_n))=\sum_1^\infty (x_k+z_k)y_k=\sum_1^\infty x_ky_k+z_ky_k=\sum_1^\infty x_ky_k+\sum_1^\infty z_ky_k=f((x_n))+f((z_n))$.

Claim: $f((x_n))$ is bounded. Proof: $|f((x_n))|=|\sum_1^\infty x_ky_k|\leq \sum_1^\infty| x_k||y_k|=||x_k||_\infty\sum_1^\infty|y_k|=||x_k||_\infty||y_k||_1<\infty$.

Claim: $||f||=||(y_n)||_1$. The operator norm of $f$, $||f||$, is the smallest $c$ such that $|f((x_n))|\leq c|(x_n)|$. Or equivalently $||f||=\sup_{||(x_n)||_\infty=1}||f((x_n))||$. Well from above we already know that $|f((x_n))|\leq||x_k||_{\infty}||y_k||_1$. If we assume $||(x_n))||_{\infty}=1$ then we can say $|f((x_n))|\leq||y_k||_1$. Now we need to show the reverse inequality, right?

Answer 1

For each $j\in \Bbb N$ define $x(j)=(x(j)_k)_{k\in \Bbb N}\in l^{\infty}$ as follows:

For $k\le j$ let $x(j)_k=1$ if $y_k\ge 0$ and let $x(j)_k=-1$ if $y_k<0.$

And let $x(j)_k=0$ for all $k>j.$

Then for every $j$ we have $\|x(j)\|_{\infty}=1$ and $f(x(j))=|f(x(j))|=\sum_{k= 1}^j|y_k|.$

Therefore $\|f\|\ge \sup \{\,|f(x(j))|:j\in \Bbb N\,\}=\|y\|_1.$

You have already shown that $\|f\|\le \|y\|_1.$