Let X be an inner product space and Y a closed subspace. Give an example of Y such that Y is properly contained in $Y^{\perp \perp}$.
I have no idea how to construct an example. I checked this: Orthogonal complement of orthogonal complement, but realized that the subspace in this question is not closed.
Let $X = c_{00}$, the space of finitely supported sequences, with the inner product inherited from the standard inner product on $\ell^2$. Then $Y = \{x \in X: \sum_{n=1}^\infty \frac{1}{n}x_n = 0\}$ is a closed subspace of $X$ - this is because $(\frac{1}{n}) \in \ell^2$, so $x \mapsto \sum_{n=1}^\infty \frac{1}{n}x_n$ is a bounded linear functional on $X$, from which it follows that $Y$ is closed. But $Y^\perp = \{0\}$ - indeed, if $x \in Y^\perp$, as $e_1 - ne_n \in Y$ for any $n \geq 2$, we have $x_1 - nx_n = 0$, so $x_n = \frac{1}{n}x_1$. But $x \in c_{00}$ is finitely supported, so we must have $x_1 = 0$ and thus $x = 0$, i.e., $Y^\perp = \{0\}$, as claimed. But then $Y^{\perp\perp} = X$ strictly contains $Y$ - say, $e_1 \in X$ but $e_1 \notin Y$.
More generally it can be shown that if $X$ is an inner product space which is not complete then there exists a closed subspace $Y$ of $X$ such that $Y \neq Y^{\perp\perp}$.
Let $X$ be an inner product space which is not complete. Let $(u_{n})_{n\in\mathbb{N}}$ be a Cauchy sequence in $X$ which does not converge. A simple application of the Cauchy-Schwarz inequality shows that \begin{equation} \lim_{n\to\infty} (x, u_{n})_{X} \tag{1} \end{equation} exists for each $x\in X$. Furthermore, there is no $u\in X$ such that $(x, u)_{X}$ is equal to (1) for each $x\in X$. To see why, suppose for a contradiction that there is such a $u\in X$. Let $\varepsilon > 0$. Since $(u_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence in $X$, there is $N\in\mathbb{N}$ such that $m,n\in\mathbb{N}$ with $m\geq N$ and $n\geq N$ implies $\|u_{m} - u_{n}\|_{X} < \varepsilon$. Let $n\in\mathbb{N}$ with $n\geq N$. Choose $y\in X$ with $\|y\|_{X} = 1$ such that $(y, u_{n} - u)_{X} = \|u_{n} - u\|_{X}$. Then \begin{equation} \|u_{n} - u\|_{X} = |(y, u_{n} - u)_{X}| \leq \limsup_{m\to\infty}|(y, u_{n} - u_{m})_{X}| + \limsup_{m\to\infty}|(y, u_{m} - u)_{X}| \leq \varepsilon \end{equation} which follows by applying (1) and using that $(u_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence in $X$. Therefore $(u_{n})_{n\in\mathbb{N}}$ converges to $u$ in $X$ which is a contradiction.
Define $\phi : X\to\mathbb{C}$ by $\phi (x) := \lim_{n\to\infty} (x, u_{n})_{X}$. Note that $\phi$ is well-defined by (1). Moreover, $\phi$ is a bounded linear functional because $(u_{n})_{n\in\mathbb{N}}$ is a bounded sequence in $X$ and is even a non-trivial linear functional, as otherwise (1) would be equal to $(x,0)_{X}$ for each $x\in X$ and it has already been shown that this is impossible.
Next, define $Y := {\rm ker} \, \phi$. Then $Y$ is a closed $1$-codimensional subspace of $X$. Let $u\in Y^{\perp}$. Choose $x_{0}\in X$ such that $\phi (x_{0}) = 1$. Note that ${\rm span} \, (Y \cup \{x_{0}\}) = X$. Suppose for a contradiction that $(x_{0}, u)_{X} \neq 0$. Define $u_{0} := (u, x_{0})_{X}^{-1} y$. Then $\phi (x_{0}) = 1 = (x_{0}, u_{0})_{X}$. But it then follows from linearity that \begin{equation} \lim_{n\to\infty} (x, u_{n})_{X} = \phi (x) = (x, u_{0})_{X} \end{equation} for all $x\in {\rm span} \, (Y \cup \{x_{0}\}) = X$ which is a contradiction. Hence $(x_{0}, u)_{X} = 0$. It follows that $u\in (Y\cup \{x_{0}\})^{\perp} = ({\rm span} \, (Y\cup \{x_{0}\} )^{\perp} = X^{\perp} = \{0\}$. Therefore $Y^{\perp} = \{0\}$ and consequently $Y^{\perp\perp} = X \neq Y$. So $Y$ has the desired properties.
