Let \(X\) be any normed linear space and \(X'\) be the dual space of \(X\). Which of the following statements is/are TRUE?
Show Hint
Use the theorem that a separable dual forces the space to be separable, plus the fact that reflexivity of a space and its dual always go together, then argue by contrapositive.
If \(X\) is separable and \(X'\) is non-separable, then \(X\) must be reflexive.
If \(X\) is separable and \(X'\) is non-separable, then \(X\) cannot be reflexive.
If \(X\) is reflexive, then \(X'\) is reflexive.
If \(X'\) is separable, then \(X\) is separable.
Show Solution
The Correct Option isB, C, D
Solution and Explanation
Step 1: State the two building-block theorems needed here.
Theorem 1: for a normed linear space $Y$, if $Y'$ (its dual) is separable then $Y$ is separable. This is a one-way implication, the converse can fail.
Theorem 2: $X$ is reflexive if and only if $X'$ is reflexive, so reflexivity of a space and reflexivity of its dual always go together.
Step 2: Apply Theorem 1 directly to option (D).
Option (D) says exactly "if $X'$ is separable then $X$ is separable", which is Theorem 1 word for word, so option (D) is TRUE.
Step 3: Apply Theorem 2 directly to option (C).
Option (C) says "if $X$ is reflexive then $X'$ is reflexive", which is one direction of Theorem 2, so option (C) is TRUE.
Step 4: Work out options (A) and (B) using reflexivity and separability together.
Assume $X$ is reflexive. Then the canonical embedding $J: X \to X''$ is onto, so $X'' \cong X$.
If, in addition, $X$ is separable, then $X''$ is separable too (because $X'' \cong X$).
Now apply Theorem 1 with $Y = X'$: since $Y' = X''$ is separable, $Y = X'$ must be separable.
So reflexive and separable $X$ forces $X'$ to be separable as well.
Contrapositive: if $X$ is separable but $X'$ is NOT separable, $X$ cannot be reflexive, this is exactly option (B), so (B) is TRUE, and option (A), which claims the reverse conclusion, is FALSE.
The space $\ell^1$ illustrates this concretely: $\ell^1$ is separable, its dual $\ell^\infty$ is not separable, and $\ell^1$ is indeed a well known example of a non-reflexive space, matching (B) and contradicting (A).
Final Answer:
$$\boxed{\text{Statements (B), (C), (D) are TRUE; statement (A) is FALSE}}$$
