$A^* \subset B^* \to B\subset A$

Question

While working with the Hahn-Banach theorem, I came across this question:

Let $ A, B $ be two closed subspaces of the Banach space $ C $. If $ B^* \subset A^* $, do we have $ A \subset B $?

I would say simply yes. In fact, if they were not, we could consider $ f \in A^* $ and, if $b \in B\setminus A$, define $f(b)=\infty$, which makes it absurd . However this reasoning seems too trivial to me...

Answer 1

Hint: Recall what the elements of the dual spaces $A^*$ and $B^*$ actually are. Then take another look at what the inclusion $B^* \subseteq A^*$ actually means in practice.

Answer 2

For your question to make sense you really need to say what you mean by $B^*\subset A^*$. There is no canonical way of seeing a function $B\to\mathbb C$ as a function $A\to\mathbb C$ if you don't specify any relation between $A$ and $B$.

If you want to say that you see $B^*\subset A^*$ by restriction, you are implying that $A\subset B$ in the definition of your inclusion, so the answer is yes in a trivial sense.

For example, let $$C=\ell^1(\mathbb N)\oplus \ell^2(\mathbb N),\qquad A=\ell^1(\mathbb N)\oplus 0,\qquad B=0\oplus \ell^2(\mathbb N).$$ Then $A^*=\ell^\infty(\mathbb N)$ and $B^*=\ell^2(\mathbb N)$ canonically, so we have a natural inclusion $B^*\subset A^*$. But this tells us nothing about $A$ and $B$ (which have $A\cap B=\{0\}$ in the above example. Or we could have $C=\ell^\infty(\mathbb N)$, $A=\ell^1(\mathbb N)$, $B=\ell^2(\mathbb N)$, and now we have $A\subset B$ and $B^*\subset A^*$ canonically.