$X'$ finite-dimensional implies $X$ finite-dimensional

Question

How would one prove, for any normed space $X$ that if $X'$ is finite dimensional, then $X$ is finite-dimensional? Here $X'$ denotes the space of all bounded functionals $f: X \to\mathbb F$

If anyone would give me a hint, no need for full answer. I really don't know where to start.

Appreciate all the answers.

Answer 1

Since $X'$ is finite dimensional we can choose a basis $x_1', \cdots, x_n'$ of $X'$. The vectors $(x_i'')_{i \in \{1, \cdots, n \}}$ defined by $x_i''(x_j') = \delta_{ij}$ are a basis of $X''$. This means $X''$ is finite dimensional too. The canonical inclusion $J: X \to X''$ defined by $$Jx: X' \to \mathbb{K}, \ (Jx)(x') := x'(x)$$ is injective we get $\dim(X) \leq \dim(X'') < \infty.$

Answer 2

It strikes me that the answer provided by our colleague Eddie is quite on the mark; the solution to this problem seems to center on two essential facts: first, that for finite dimensional $Y$, $Y'$ is also finite dimensional; second, the the natural mapping from $Y$ to its double dual $Y''$ given by evaluation on elements of $Y'$ is injective; that is, the functional $\hat y \in Y''$ defined by

$\hat y(\phi) = \phi(y), \; \forall \phi \in Y', \tag 0$

is determined solely by $y$; there is no other $z \in Y$ such that

$\hat z(\phi) = \phi(y), \; \forall \phi \in Y'; \tag{0.1}$

these two principles combine to yield the desired result, as Eddie has indicated, and as I discuss at some greater length below.

What I have done in preparing this answer is to try and expand upon these two principles and work out how they operate jointly and severally in some detail, as much for the sake of my own recollection as anything. So this answer may be a tad on the long-winded side; I thank you all in advance for your patience.

Now let' see . . .

For any finite dimensional normed space $Y$, $Y'$ is also finite dimensional, with

$\dim Y' = \dim Y; \tag 1$

for let

$y_1, y_2, \ldots, y_n \in Y \tag 2$

be a basis; then set

$\phi_i \in Y', \; \phi_i(y_j) = \delta_{ij}, \; 1 \le i \le n; \tag 3$

for $y \in Y$ we may write

$y = \displaystyle \sum_1^n \alpha_i y_i; \tag 4$

thus for $\phi \in Y'$,

$\phi(y) = \displaystyle \sum_1^n \alpha_i \phi(y_i); \tag 5$

I claim

$\phi = \displaystyle \sum_1^n \phi(y_j) \phi_j; \tag 6$

for with $y$ as in (4),

$\left (\displaystyle \sum_1^n \phi(y_j) \phi_j \right )(y) = \left (\displaystyle \sum_1^n \phi(y_j) \phi_j \right ) \left ( \displaystyle \sum_1^n \alpha_i y_i \right ) = \displaystyle \sum_1^n \phi(y_j) \phi_j \left ( \displaystyle \sum_1^n \alpha_i y_i \right )$ $= \displaystyle \sum_{i, j = 1}^n \phi(y_j) \alpha_i \phi_j(y_i) = \sum_{i, j = 1}^n \phi(y_j) \alpha_i \delta_{ij} = \sum_1^n \alpha_i \phi(y_i) = \phi(y), \tag 7$

showing the $\phi_i$ span $Y'$; the $\phi_i$ are also linearly independent, for if

$\displaystyle \sum_1^n \beta_j \phi_j = 0, \tag 8$

with say $\beta_k \ne 0$ we have

$\beta_k \phi_k = -\displaystyle \sum_{i = 1, i \ne k}^n \beta_i \phi_i; \tag 9$

now if we apply this to $y_k$ we find

$\beta_k = 0, \tag{10}$

a contradiction. So the $\phi_i$ are indeed linearly independent, and as such form a basis for $Y'$; thus, (1) holds.

Next, replacing $Y$ with $Y'$ we see that

$\dim Y'' = \dim Y'; \tag{11}$

now consider the map

$Y \to Y'', \; y \to \hat y, \; \hat y(\phi) = \phi(y), \; \forall \phi \in Y'; \tag{12}$

I claim $y \to \hat y$ is injective; if

$\exists y \in Y, \; \hat y(\phi) = 0, \; \forall \phi \in Y', \tag{13}$

then

$\phi(y) = \hat y(\phi) = 0, \forall \phi \in Y'; \tag{14}$

so we need to show that

$\phi(y) = 0, \; \forall \phi \in Y' \Longrightarrow y = 0; \tag{15}$

this can be accomplished via the Hahn-Banach theorem; if $y \ne 0$ we my consider the subspace

$\{0\} \ne \{\alpha y \} \subset Y; \tag{16}$

define $\phi$ on this subspace by

$\phi(\alpha y) = \alpha \vert y \vert; \tag{17}$

then

$\vert \phi(\alpha y) \vert = \vert \alpha \vert y \vert \vert = \vert \alpha \vert \vert y \vert, \tag{18}$

which shows that $\phi$ is bounded on $\{\alpha y \}$ with bound $\vert y \vert$ (note $\vert \phi y \vert = \vert y \vert$); thus $\phi$ may be extended to $Y$ with the same bound; but then we have $\phi \in Y'$ with $\phi(y) \ne 0$; this contradiction proves (15). Thus, $y \to \hat y$ is injective.

Therefore,

$\dim Y \le \dim Y'' = \dim Y'. \tag{19}$

Answer 3

Since you asked for a hint: aiming for a contradiction, assume that $X$ was finite-dimensional. Let $B = (b_n)_{n\in\mathbb{N}}$ be a countable collection of linearly independent elements of $X$ with $\|b_n\| = 1$ for $n \in \mathbb{N}$. Now construct a countable family of linearly independent functions $f_n \in X'$ by prescribing their values on $B$.

Answer 4

Here's a sketch:

Suppose $X$ has an infinite basis, say of cardinality $\alpha$. This choice of basis now yields a homomorphism $X\to X'$. The image under the map $X\to X'$ of this basis is still linearly independent. (You need to show that this map is well-defined and injective, in the finite dimensional case this yields the dual basis.) Thus every basis of $X'$ has at least cardinality $\alpha$, which contradicts the fact that $X'$ is finite-dimensional.