# Is it true that for every subspace $N$, we have $N^{{\perp}{\perp}}=N$?

Let $‎X=C[-1,1]‎$‎‎ be inner product space with definition $$‎\langle f,g‎‎‎\rangle =‎\int_{-1}^1 f‎‎ \overline{g}‎ ‎dt ‎‎.$$ Let $M$ be the subspace defined by ‎$$‎M= ‎‎\left\{f‎ \in ‎X\mid ‎f(t)=0 ,‎ ‎‎-1 \leq‎ t ‎‎\leq ‎0 \right\}. ‎$$ Notice that $M^{{\perp}{\perp}}=M$. Can it be concluded that for every closed subspace $N$, we have $N^{{\perp}{\perp}}=N$?

The subspace $V$ of the form $V=N^\perp$ for some other subspace $N$ is necessarily closed as the intersection of kernels of continuous linear functions. Therefore for $N=N^{\perp\perp}$ to happen, $N$ must itself be closed.

• @jykary: Can it be concluded that for every closed subspace $N$ , we have $N ^{{\perp}{\perp}} =N$ ? thanks
– nim
Commented Apr 5, 2013 at 7:02
• It is true for all closed subspaces $N$ for example when $X$ is a Hilbert space. Have you proved this in class? It is not trivial, but not very difficult either. In more general cases - it depends... Commented Apr 5, 2013 at 7:02
• @jykary: but space $X= C [-1,1]$ is not Hilbert space. Again it is true? thanks
– nim
Commented Apr 5, 2013 at 7:05

The answer is no, by the following statement.

Proposition. Let $H$ be a Hilbert space and $H_0$ a proper dense subspace. Let $y_0 \in H \setminus H_0$. Let $K = {y_0}^\perp = \ker(\phi)$, where $\phi(x) = \langle x, y_0 \rangle$ for $x \in H$, and let $K_0 = K \cap H_0$. Then $K_0^\perp \cap H_0 = (0)$ .

The proof was sketched in my answer to this question.

Concretely, taking $X = C[-1, 1]$ with the $2$-norm, which is dense in $H = L_2[-1, 1]$, let $N = \{f \in X : \int_{-1}^0 f(t) dt = 0 \}$. Then $N$ is a closed hyperplane in $X$ with $N^\perp = (0)$, and $N^{\perp \perp} = X$.