An example of space $V$ such that $(V^{\perp})^{\perp} \neq V$ I know that if $W$ is a vector space of finite dimension then for any subspace $V$ ,$(V^{\perp})^{\perp} = V$. But I have heard that this is not true for infinite dimensional vector spaces. So I tried to construct a counter example but I could not get any. I tried the vector space formed by infinite tuples but whatever subspace I took it was satisfying $(V^{\perp})^{\perp} = V$. So if any one could give a counter example it would be great.Thanks.
 A: In a Hilbert space, $A^{\perp\perp}=\overline{L(A)}$ (= the smallest closed subspace containing $A$) for any set $A$. Thus the counterexamples are exactly the subspaces $V$ that are not closed.
For a concrete example, you can take $H=\ell^2$ and $V$ as the finitely supported sequences. Then it's easy to see directly that $V^{\perp}=\{0\}$, so $V^{\perp\perp}=\ell^2$.
A: The space of trigonometric polynomials is an example.  A function of the form
$$
x\mapsto\sum_{n=-N}^N c_n e^{inx}
$$
is a trigonometric polynomial.  With the inner product
$$
\langle f, g \rangle = \int_0^{2\pi} f(x) \overline{g(x)}\,dx,
$$
(where $\overline{g(x)}$ is the complex conjugate of $g(x)$) the dual of the dual of the space of trigonometric polynomials is the space of quadratically integrable functions, i.e. functions $f$ satisfying
$$
\int_0^{2\pi} f(x) \overline{f(x)}\,dx <\infty.
$$
The space contains every trigonometric polynomial but also contains every function that can be approximated arbitrarily closely by trigonometric polynomials, where closeness of approximation is measured by proximity in the metric defined by this inner product.
