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

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

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?

• Do I need to show that if we take any sequence which converges in l^1 then the operator norm will converge? Commented Aug 1, 2019 at 17:11
• You might note the edit I just made - it makes \N and \R do what you wanted... Commented Aug 1, 2019 at 17:22

## 1 Answer

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 $$\|f\|\le \|y\|_1.$$