A corollary of Hahn–Banach theorem states that

Let $E$ be a normed vector space, $M$ a proper closed subspace and $x \in E$. If $d(x,M) = \delta > 0$, so exists $f \in E'$ such that $\|f\|=1$, $f(x)=\delta$ and $f(m)=0$ $\forall m \in M $.

Consider $T: \ell_\infty \rightarrow \ell_\infty$ a bounded linear operator defined by $$T(x_1,x_2,x_3,\dots) = (x_2,x_3,\dots).$$

Let $M=\{ x-T(x) : x \in \ell_\infty\}$, so $M$ is a subspace of $\ell_\infty$. If $e=(1,1,1,\dots) \in \ell_\infty$, so $d(e,M)=1>0$. Then, applying the corollary above in $\overline{M} $, exists $f \in \ell_\infty'$ such that $\|f\|=1$, $f(e)=d(x, \overline{M}) = d(e, M) =1$ and $f(x)=0~\forall x \in M\subset \overline{M} $.

I was able to show that $$f(x_1,x_2,x_3,\dots) = f(x_2,x_3,\dots) ~~\forall (x_n) \in \ell_\infty.$$ Besides that, we have that $\forall x = (x_n) \in c = \{ (x_n) \in \ell_\infty : x_n \text{ is convergent} \}$ $$f(x) = \lim_{n \rightarrow \infty} x_n$$

Now, let $x=(x_n), y=(y_n) \in \ell_\infty$ such that $x_n \geq y_n$ $\forall n \in \mathbb{N}$. How can I show that $f(x) \geq f(y)$?

  • 4
    $\begingroup$ If the typographical difference between $||f||$ and $\|f\|$ is not conspicuous to you, consider the difference between $||f|| ||g||,$ coded as ||f|| ||g||, and $\|f\|\|g\|,$ coded as \|f\|\|g\|. I made that correction in the question and added proper use of \text{}. I also changed $l_\infty$ to $\ell_\infty,$ which seems to be somewhat standard and also strikes me as more readily legible. $\endgroup$ – Michael Hardy May 3 '18 at 1:24

I'm going to try to answer this from the real $\ell^\infty$ perspective. Say we take a positive sequence $(x_n)$. Then, consider the sequence $$y_n = e_n - \frac{x_n}{\|x_n\|},$$ where $e_n = 1$ for all $n$. Note that $0 \le y_n \le 1$ for all $n$, so by definition of the norm of $f$, $$f(y_n) \le 1 = f(e_n).$$ Rearranging, \begin{align*} &f\left(e_n - \frac{x_n}{\|x_n\|}\right) \le f(e_n) \\ \implies \, &f(e_n) - \frac{f(x_n)}{\|x_n\|} \le f(e_n) \\ \implies \, &\frac{f(x_n)}{\|x_n\|} \ge 0 \\ \implies \, &f(x_n) \ge 0. \end{align*} By linearity, this implies what you want shown.

  • $\begingroup$ Very nice. $\ $ $\endgroup$ – Martin Argerami May 3 '18 at 2:13

The fact that $\|f\|=f(e)$ implies positivity. The proof is not specific to $\ell^\infty$, it works on any C$^*$-algebra $A$.

So assume that $A$ is a C$^*$-algebra, and $f:A\to\mathbb C$ a bounded linear functional, with $\|f\|=f(e)$, where $e$ is the unit of the algebra.

First, let $a\in A$ be selfadjoint ($a^*=a$) with $\|a\|=1$. For any $n\in \mathbb N$, \begin{align}\tag1 |f(a)+in|&=|f(a+ine)|\leq\|a+ine\|=\|(a-ine)^*(a-ine)\|^{1/2}\\ \ \\ &=\|(a+ine)(a-ine)\|^{1/2}=\|a^2+n^2e\|^{1/2}\\ \ \\ &=(1+n^2)^{1/2}. \end{align} The last equality is due to $a=a^*$, which implies that $a^2\geq0$. In particular $$ |\operatorname{Im} f(a)+n|=|\operatorname{Im}(f(a)+in)|\leq|f(a)+in|\leq(1+n^2)^{1/2}. $$ So $$ n-(1+n^2)^{1/2}\leq\operatorname{Im}f(a)\leq (1+n^2)^{1/2}-n. $$ By taking $n$ arbitrarily large, we obtain $\operatorname{Im}f(a)=0$. If we revisit $(1)$ with this knowledge, we get $$ (f(a)^2+n^2)^{1/2}\leq(1+n^2)^{1/2}, $$ so $f(a)^2\leq1$. So $f(a)$ is real, and $-1\leq f(a)\leq 1$.

If now $b$ is positive with $0\leq b\leq e$, the elements $2b-e$ is selfadjoint and $-e\leq 2b-e\leq e$, so $\|2b-e\|\leq 1$. By the above $$-1\leq f(2b-e)=2f(b)-1\leq 1,$$ which we may rewrite as $$ 0\leq f(b)\leq 1. $$ In particular $f(b)\geq0$, so $f$ is positive.

  • $\begingroup$ Yes, thanks for noticing. $\endgroup$ – Martin Argerami May 3 '18 at 1:39
  • $\begingroup$ This answer is beyond my knowledge. I'm an undergraduate student. I cannot grasp your answer. But I would like to learn more about C*-algebras. Can you recommend a book on the topic? $\endgroup$ – H R May 3 '18 at 2:09
  • 1
    $\begingroup$ Most books do quite a bit of Banach algebra before getting into C$^*$-algebras. That's what happens in Conway's "A Course in Functional Analysis" and in Murphy's "C$^*$-algebras and Operator Theory". A book that goes directly into C$^*$-algebras is Davidson's "C$^*$-algebras by Example". $\endgroup$ – Martin Argerami May 3 '18 at 2:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.