Case:
which reminds me a lot about Bourbaki–Alaoglu theorem i.e. a unit case on a ball because of the last inequality and something with adjoint Hahn-Banach.
My proposal by thinking the case on a ball [depreciated]
In the unit ball $D(x) = \{ u : \|u \| \leq 1 \}$ is not a compact set. Consider any family of subspaces $E_{1} \subset E_{2} \subset ... \subset E_{n} \subset ...$ where $\text{dim } E_{n} = n$. They are closed subspaces (since they are of finite dimension). By Lemma of 5.2.1 of Eidelman et al, there exists a sequence $u_{i} \in E_{i}$ with $\| u_{i} \| = 1$ such that $\text{dist}(y_{i}, E_{i-1}) \geq 1/2$. Then, for any $i \neq j$, $\| y_{i} - y_{j} \| \geq 1/2$. Obviously, $\{ y_{i} \}$ is not a Cauchy sequence and there is no Cauchy subsequence of it. $\square$
Sources
- Functional Analysis An Introduction by Eidelman, Corollary 5.2.2 and lemma 5.2.1
Comments
- To take a Cauchy sequence in Hahn-Banach here does not feel good because this variant of Hahn-Banach may be some complex version of it. I think this can be proved without considering any sequences.
- Eigenvalues come to my mind also by the statement and the Identity operator: $I : X \mapsto X$ that is not compact, like is not the case here.
My proposal #2 after the first hint
Consider $\| \varphi \| < 1$ of the equation. There exists $\varphi \in X'$ such that $\varphi = Span(u,X')$ for all $u \in X'$. Then, $\varphi(u) < 1$ because $\| \varphi \| < 1$. If $\varphi(u) > 1$, then $\| \varphi(u) \| > 1$, which is a contradiction to the initial assumption. Then, $\| \varphi \| < 1$ of the equation. $\square$
Consider $\| \varphi \| = 1$ of the equation. There exists $\varphi \in X'$ such that $\varphi = Span(u,X')$ for all $u \in X'$. Then, $\varphi(u) = 1$ because $\| \varphi \| = 1$. If $\varphi(u) \neq 1$, then $\| \varphi \| \neq 1$, so a contradiction to the initial assumption. Then, $\| u \| = 1$ so $\varphi(u) = \| u \|$. $\square$
Then, \begin{equation*} \| u \| = \sup\{ \| \langle u, \varphi \rangle \| : \varphi \in X', \| \varphi \| \leq 1 \}. \, \square \end{equation*}
Comments
- I used the proof of contradiction to show the both sides. Is there any sense in my proof?
How can you prove the rigorously the case about the norm on Banach space?
some hints:
First step: prove the "$\geq$" part of the equality $\|u\| = \sup \{ | \varphi(u) |: \varphi \in X', \|\varphi\| \leq 1 \}$. For this you only need to show that $\|u\| \geq \varphi(u)$ if $\varphi$ has the conditions mentioned.
Second step: show that there is a $\varphi \in X'$ with $ \varphi(u) = \| u \|$ using Hahn-Banach (start by defining $\varphi$ on $\mathrm{span} \, u$). Hahn-Banach also tells you that $\| \varphi \| \leq 1$. This shows you that the equality can be achieved.