How does one prove that if a $X$ is a Banach space and $x^*$, a continuous linear functional from $X$ to the underlying field, then $x^*$ attains its norm for some $x$ in $X$ and $\Vert x\Vert = 1$?

My teacher gave us a hint that we should use the statement that if $X$ is a reflexive Banach space, the unit ball is weak sequentially compact, but I am not sure as to how to construct a sequence in this ball which does not converge.

Thank you.

  • $\begingroup$ Why do you want to find a subsequence that doesn't converge? And if you say "doesn't converge" doesn't converge in what sense? $\endgroup$ – t.b. Jun 3 '12 at 21:45
  • $\begingroup$ I was trying to prove that if the unit ball has a subsequence that is not convergent, then it is not sequentially compact and hence, X is not reflexive, leading to a contradiction. Is the implication valid both ways, though? $\endgroup$ – nada Jun 3 '12 at 22:17
  • $\begingroup$ Yes, the implication works in both ways: James's theorem (mentioned in Davide's answer) says that a Banach space is reflexive if and only if every functional attains its maximum on the unit ball. The easier part of this rather deep theorem (reflexive $\Rightarrow$ every functional attains its maximum) can be proved directly, so it's better practice to phrase it this way. I outlined the main points relying on your teacher's hint in my answer and Davide gave an alternative approach that leads to the goal as well. $\endgroup$ – t.b. Jun 3 '12 at 22:22
  • $\begingroup$ @dini Btw, do you know how to accept answers? $\endgroup$ – Rudy the Reindeer Jun 3 '12 at 23:00

Suppose that $\varphi \neq 0$. Let $x_n$ be a sequence in the unit sphere such that $|\varphi(x_n)| \geq \lVert \varphi\| - \frac{1}{n}$ which exists by the definition of the operator norm. Choose a subsequence such that $x_{n_k} \to x$ weakly with $\|x\| \leq 1$ by weak sequential compactness of the unit ball. Deduce that $|\varphi(x)| = \|\varphi\|$. Then argue that $\|x\| = 1$.

  • $\begingroup$ The notation changed from $x'$ to $x^\ast$ and I'll stick to $\varphi = x' = x^\ast$. $\endgroup$ – t.b. Jun 3 '12 at 21:53
  • $\begingroup$ I have a question. How does one argue that norm(x) = 1. And even so, how does that help to prove the result? $\endgroup$ – nada Jun 3 '12 at 22:26
  • $\begingroup$ I mean, how does one use the fact that norm(x) = 1 to help prove the result? $\endgroup$ – nada Jun 3 '12 at 22:32
  • $\begingroup$ Okay, I understood how to do that. $\endgroup$ – nada Jun 3 '12 at 22:33
  • $\begingroup$ Great, glad to help :) $\endgroup$ – t.b. Jun 3 '12 at 22:34

We can use a corollary of Hahn-Banach theorem, applied to the dual space $X'$ of $X$. We have $$\lVert x'\rVert=\max_{y\in X'',\lVert y\rVert=1}|y(x')|$$ (the maximum is reached for a $y_0$ that can be constructed thanks to Hahn-Banach theorem). For this $y_0\in X''$, since $X$ is reflexive we can find $u\in X$ such that $J(u)=y_0$, where $J\colon X\to X''$ is the canonical embedding. Hence by definition $J(u)(x')=x'(u)=y(x')$ and $u$ (or $-u$) is such that $|x(u)|=\sup_{\lVert v\rVert=1}|x(v)|$.

Note that it doesn't follow the hint given by your teacher. Note that the converse is true (if each linear continuous functional attains its norm, the Banach space is reflexive). It's a difficult result, from James I think.

  • 1
    $\begingroup$ +1: That's the argument I'd have used if there hadn't been the hint. Yes, the converse is James's theorem. Links to the original articles are on the Wikipedia page. $\endgroup$ – t.b. Jun 3 '12 at 21:56
  • $\begingroup$ I also agree that the argument with compactness, albeit correct, is very unnatural. $\endgroup$ – wroobell Dec 19 '15 at 12:58
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    $\begingroup$ which corollary do you use as you wrote " We can use a corollary of Hahn-Banach theorem, applied to the dual space X′ of X " $\endgroup$ – manhattan Mar 9 '17 at 18:31
  • $\begingroup$ and in the end how does one conclude the " u or -u with $|x(u)|=\sup_{\lVert v\rVert=1}|x(v)|$ " $\endgroup$ – manhattan Mar 9 '17 at 19:28

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