Consider the space $L^2[0,1]$ of complex valued square-integrable functions $f : [0,1] \to \mathbb{C}$. Let $\langle f, g \rangle = \int_0^1 f \bar{g}$ denote the standard $L^2$ inner product. For $M \subseteq L^2[0,1]$, define $M^\perp = \{ f \in L^2[0,1] : \langle m, f \rangle = 0, \text{ all } m \in M \}$ (note that $M$ does not need to be a subspace of $L^2[0,1]$).
I would like to show that $(\{1\}^\perp)^\perp =$ the set of constants functions on $[0,1]$.
One inclusion is easy: if $c$ is any constant and $\int_0^1 1 \cdot \bar{f} = 0$, then $\int_0^1 f \cdot \bar{c} = \overline{ c \int_0^1 1 \cdot \bar{f}} = 0$. This shows all constants functions are contained in $(\{1\}^\perp)^\perp$.
For the other inclusion, we suppose that $g \in (\{1\}^\perp)^\perp$, and my conjecture is that $g = \int_0^1 g$ almost everywhere on $[0,1]$. However, this is where I have gotten stuck. One thing I have noticed is that $h = g - \int_0^1 g \in \{1\}^\perp$, hence $\langle h ,g \rangle = 0$. Although I am not sure if this gets me anywhere.
Hints or solutions are greatly appreciated.
Easiest way might be to show: $\{c\}^{\perp}=\{1\}^{\perp}$, where $\{c\}$ denotes the constant functions.(This is essentially a tautology-and you have essentially already proved this). Then, the result follows as the constant functions form a closed subspace of $L^2[0,1]$.
EDIT: Your method also works. We will show $\langle h, h \rangle =0$.
Note that $\langle h, h \rangle=\langle h, g-c \rangle=\langle h, g \rangle- \langle h,c \rangle$, where $c= \int_{0}^{1}g$
$\langle h, g \rangle=0$.
$\langle h, c \rangle = \langle g-c, c \rangle =0$, by direct computation.