Isn't $E\cong E^{**}\iff \dim(E)<\infty $ wrong?

Question

Let $E$ a $K$-vector space and denote $E^{**}$ its bidual. Let $$\Phi: E\longrightarrow E^{**}$$ defined by $$\Phi(x)=\left<f,x\right>,\quad f\in E^*.$$ A theorem in my course says

$$\Phi\text{ is an isomorphism}\iff E\text{ has finite dimension}.$$

The implication (i.e. $\Rightarrow$) is not correct, no ? For example, if $p\in (1,\infty )$ we have that $L^p(\mathbb R)$ is reflexive, which mean exactly that $\Phi$ is bijective, right ? So I have a truble with this theorem. Moreover, the proof looks correct :

The fact that $\Phi$ is injective is fine. We show the contrapositive, i.e. we suppose $E$ has infinite dimension and we prove that $\Phi$ is not surjective. Fo the surjectivity, let $F=\text{Span}(e_1^*,...,e_n^*)\neq E$. I know that there is a non zero linear form $f:E^*\to K$ s.t. $F\subset \ker(f)$. For all $x=(x_i)_{i\in I}\in E$, $$\Phi(x)(e_{i}^*)=x_i.$$ Therefore, if $\Phi (x)(F)=0$, then $x=0$ and thus $\Phi (x)=0$. Therefore, $f$ is not in the range of $\Phi$ and thus its not surjective.

What do you think ? By the way, I don't understand why $\Phi(x)(F)=0$ implies that $\Phi(x)=0$... Indeed, we can have $g\notin F$ s.t. $\Phi(x)(g)\neq 0$ no ?

Does $E^*$ denote the algebraic dual or the continuous dual? — Ben Grossmann, Jul 25 '18 at 12:21
Reflexivity is about the dual space of a Banach or Hilbert space comprising just the continuous linear functionals not the vector space comprising all linear functions. — Rob Arthan, Jul 25 '18 at 12:21
The problem here is that you are making confusion with the continuou dual and the alebraic dual. The algebraic dual of $L^p$ contains also unbounded linear functionals. — Crostul, Jul 25 '18 at 12:22
@Omnomnomnom If $E^*$ denotes the algebraic dual, how can we prove this result? — Math, May 31 '20 at 22:15

score 4 · Answer 1 · answered Jul 25 '18 at 14:09

I also don't understand why $\Phi(x)(F)=0$ implies $\Phi(x)=0$.

Consider this argument instead: let $(e_i)_{i\in I}$ be a basis for $E$ and let $(e_i^*)_{i\in I}$ be the dual functionals in $E^*$ (defined by $e_i^*(e_j) = \delta_{ij}$).

Let $f$ be a linear functional in $E^{**}$ such that $f(e_i^*) = 1, \forall i \in I$. To convince yourself that such $f$ exists, note that $(e_i^*)_{i\in I}$ is linearly independent in $E^*$ so it can be extended to a basis $B$ for $E^*$. Now define $f$ on $B$ as $f(e_i^*) = 1, \forall i \in I$ and something arbitrary on the rest of $B$.

Assume that there exists $x = \sum_{i\in I}x_ie_i \in E$ with only finitely many $x_i \ne 0$ such that $\Phi(x) = f$. Then in particular $$1 = f(e_i^*) = \Phi(x)(e_i^*) = e_i^*(x) = x_i$$ which is a contradiction because now $x_i \ne 0, \forall i\in I$, and $I$ is infinite.

Hence $f$ is not in the range of $\Phi$.

Isn't $E\cong E^{**}\iff \dim(E)<\infty $ wrong?

1 Answers1