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Let $ E = l^{1} $, so that $ E^{*} = l^{\infty} $ Consider: $ N = c_0 = \{x= (x_k): \lim\limits_{k\to \infty}x_k = 0\} $ as a closed subspace of $ l^{\infty}$ . Determine:

$ N^\bot=\{x\in E : \langle f, x\rangle = 0 \quad \forall f\in N \} $

and

$ N^{\bot\bot}=\{f\in E^* : \langle f, x\rangle = 0 \quad \forall x\in N^\bot \}. $

Check that $ (N^\bot)^\bot \neq N $. Proof:

A Schauder Basis for $ E = l^{1} $ is $ (e_{k})_{k \in \mathbb{N}} = ( \delta_{i,k})_{k \in \mathbb{N}} $. If $ x \in l^{1} $ we have: $ x = \displaystyle\sum_{k=1}^{\infty} \alpha_k e_k$ . For any $ f \in (l^{1})^* = l^{\infty} $, we have:

$ f(x) = \displaystyle\sum_{k=1}^{\infty} \alpha_k f(e_k) = 0 $ . But $ (e_{k})_{k \in \mathbb{N}} $ is a Schauder basis, then:

$ \alpha_k = 0 \qquad \forall k \in \mathbb{N} $. Therefore:

$ N^\bot= \{x= (x_k): x_k = 0 \quad \forall k \in \mathbb{N}\}. $

It's correct?

How to determine $ N^{\bot\bot} $ ?

Gabriel Granda
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1 Answers1

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You found:

$$N^\bot= \{x= (x_k): x_k = 0 \quad \forall k \in \mathbb{N}\}.$$

This is correct, and is a very fancy way of writing $N^\bot = \{0\}$, i.e. it is the zero subspace.

So it follows that $N^{\bot\bot} = E^*$, since $\langle f, x \rangle = 0\; \forall x \in \{0\}$ is trivially true for every $f \in E^*$. Since $N \ne E^*$, as witnessed for example by the constant function $1 \in E^* \setminus N$, you are done.

Nate Eldredge
  • 97,710