I am studying functional analysis and I was asked to prove that $c_0$ is not reflexive. The point is I have no idea how to prove this. I don't even know how to show that a space is reflexive. What must I do on the practice?

I think it is hard to solve this kind of problem just looking for the canonical embedding and trying to discover it is surjective since several times we don't have any idea of which space is the bidual to our space.

My question is, what are useful theorems and techniques in order to show a space is/or not is, reflexive?


  • 2
    $\begingroup$ $c_0$ is very simple space, so, you can evaluate $((c_0)^*)$ and $(c_0)^{**}$ directly. Try to understand what element of $(c_0)^*$ is. $\endgroup$ – kp9r4d Mar 30 '17 at 19:22
  • $\begingroup$ hey, I am trying to do this unsuccessfully, could you explain to me? I tried to imagine what kind of properties $f \in (c_0)^{\ast}$ must satisfy, but I couldnt conclude too much. $\endgroup$ – L.F. Cavenaghi Mar 30 '17 at 20:15
  • 1
    $\begingroup$ You can notice that $f$ uniquely determined by its value on basis vectors $f(e_n)=a_n$. So, try to think which property need have $a_n$ to satisfy next: for each $(x_n) \in c_0, (a_n x_n) \in c_0$. $\endgroup$ – kp9r4d Mar 30 '17 at 20:28

If $X$ is reflexive, then for every $f\in X^*$ there is $x\in X$ with $\|x\|_X\le1$ and $f(x)=\|f\|_{X^*}$. So if we could find a functional such that $|f(x)| < \|f\|$ for all $x$ with $\|x\|\le1$ then $X$ cannot be reflexive.

One such functional is $$ f(x):=\sum_{k=1}^\infty 2^{-k}x_k. $$ It holds $|f(x)|\le \|x\| $, hence $\|f\|\le 1$. To show that $\|f\|=1$, consider the sequence $x_n$ with $x_n=(1,\dots, 1,0\dots)$ with first $n$ elements one. Then it is clear $f(x_n)\to 1$, hence $\|f\|=1$.

It remains to show that the norm is not attained for some $x$. Take $x\in c_0$ with $\|x\|_\infty\le1$ and $\epsilon\in (0,\|x\|_\infty)$. Then there is $N$ such that $|x_n|<\epsilon$ for all $n>N$. This implies $$ |f(x)| \le \|x\|_\infty\sum_{k=1}^N 2^{-k} + \epsilon \sum_{k=n+1}^\infty 2^{-k} < \sum_{k=1}^\infty 2^{-k} = 1 = \|f\|. $$ And $c_0$ cannot be reflexive.

  • $\begingroup$ @KennyWong thanks! corrected :) $\endgroup$ – daw Mar 31 '17 at 5:58
  • $\begingroup$ I am sorry can I ask why if $X$ is reflexive, then for every $f\in X^*$ there is $x\in X$ with $\|x\|_X\le1$ and $f(x)=\|f\|_{X^*}$? $\endgroup$ – Answer Lee Mar 28 '18 at 16:47
  • $\begingroup$ @AnswerLee please ask a new question. also: Banach-Alaoglu and weak compactness $\endgroup$ – daw Mar 29 '18 at 10:41

Another way (which can be helpful if the bidual is not as easily computable as in the case of $c_0$) is via the Krein-Milman theorem: If a Banach space $X$ is reflexive then its unit ball $B_X$ is weakly compact and hence has extreme points. It is easy to see that $B_{c_0}$ does not have extreme points: As $x\in B_{c_0}$ tends to zero there is a component $|x_n|<1/2$ so that $y^\pm$ defined by $y_k^\pm=x_k$ for $k\neq n$ and $y_n^\pm =x_n\pm 1/2$ also belong to $B_{c_0}$ and satisfy $x=\frac 12 (y^++y^-)$. Hence $x$ isn't an extreme point.


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.