Let $X$ be a Banach space. We have $\tau : X \rightarrow$ $X^{**}$ is the canonical map, isomorphic and isometric but suppose that X is not reflexive, so the map is not surjective. Now $\tau (X)$ is a closed subspace of $X^{**}$. Why can I say that $\tau(X)$ is closed? Is it something related to the map?
Image of the canonical map is closed in a non-reflexive space
functional-analysisisometryreflexive-space
Related Solutions
Making my comments into an answer: No there are no such Banach spaces.
Assume that every proper subspace of $X$ is reflexive. Take a non-zero continuous linear functional $\varphi: X \to \mathbb{R}$. Let $Y = \operatorname{Ker}{\varphi}$ and choose $x_0 \in X$ with $\varphi(x_0) = 1$. By continuity of $\varphi$ the space $Y$ is a closed subspace. The map $Y \oplus \mathbb{R} \to X$ given by $(y,t) \mapsto y + t x_0$ is continuous with continuous inverse $x \mapsto (x - \varphi(x)\cdot x_0,\, \varphi(x))$, hence $X \cong Y \oplus \mathbb{R}$. By hypothesis $Y$ is reflexive, hence so is $Y \oplus \mathbb{R}$ and thus $X$ is reflexive, too. Replacing $\mathbb{R}$ by $\mathbb{C}$ gives the same for complex Banach spaces.
Weakening the hypotheses as Robert suggested makes it a bit more subtle, but still manageable:
Every non-reflexive Banach space contains a non-reflexive closed subspace of infinite codimension.
(Bourbaki, Topological vector spaces 1, Exercise 12 to Chapter IV, §5, page IV.69.)
Passing to the contrapositive, if every closed subspace of infinite codimension is reflexive, then $X$ must be reflexive.
The following is a slightly expanded version of the hint given by Bourbaki:
By the Eberlein–Šmulian theorem, a non-reflexive Banach space $X$ contains a bounded sequence $(x_n)_{n=1}^{\infty}$ without weak accumulation point. Note that this implies that the $x_n$ must span an infinite-dimensional subspace of $X$. Using the Riesz-lemma on almost orthogonal vectors to closed subspaces, it is not difficult to extract a subsequence $(x_{n_k})_{k=1}^{\infty}$ and a topologically independent sequence $(y_k)_{k=1}^{\infty}$ such that $\|x_{n_k} - y_{k}\| \leq \frac{1}{k}$ [topologically independent means that no $y_k$ is in the closed linear span of $\{y_n\}_{n \neq k}$ ]. This yields that every weak accumulation point of the $y_k$ is also a weak accumulation point of the $x_n$. The closed subspace $Y$ generated by the $\{y_{2k}\}$ is thus a non-reflexive subspace of $X$ (by Eberlein–Šmulian again) and it has infinite codimension by construction.
To show that $j_Yy = y^{**}$, let $y^* \in Y'$ and $x^*\in X'$ a Hahn-Banach extension of $y^*$. We have \begin{align*} (j_Yy)(y^*) &= y^*(y)\\ &= x^*(y)\\ &= j_Xx(x^*)\\ &= \tilde y(x^*)\\ &= y^{**}(x^*|_Y)\\ &= y^{**}(y^*) \end{align*} Hence, $j_Yy = y^{**}$.
Best Answer
$\tau$ is an isometry so its range is complete. This implies that it is closed in $X^{**}$.
More simply argue as follows: Let $\tau(x_n) \to x^{**}$ in $X^{**}$. Then $\|x_n-x_m\| =\|\tau (x_n)-\tau (x_m)\|\to 0$. Since $X$ is complete there exists $x \in X$ such that $x_n \to x$. Since $\tau$ is continuous we get $\tau (x_n) \to \tau (x)$. Hence $x^{**}=\tau (x) \in \tau (X)$.