Set Annihilator Why the annihilator of the set M in $L_3[0, 1]$ space: $$M = \{x \in L_3[-1, 1]: \int_0^1 x(t)dt = 0 \}$$
equal to $Lin(\theta(t))$, where $\theta(t)$ is the Heaviside function in the space $L_{3/2}[0, 1]$?
 A: From the definition of $M$, I'll assume that we are considering $L_3[-1,1]$, not   $L_3[0,1]$. By definition of annihilator, $M^\perp=\{f\in L_3^\ast[-1,1]:\forall x\in M\;f(x)=0\}$. Without loss of generality, we will assume that we are considering the real spaces. From the Riesz representation theorem follows that the general form of a linear bounded functional on $L_3[-1,1]$ is $f(x)=\displaystyle\int\limits_{-1}^1x(t)y(t)dt$ where $y\in L_{3/2}[-1,1]$. So, if $f\in M^\perp$, then $\forall x\in M$ $\displaystyle\int\limits_{-1}^1x(t)y(t)dt=0$. By taking $x(t)=0$ at $t\in[0,1]$ and $x(t)=\operatorname{sign} y(t)$ at $t\in[-1,0)$ (check that $x\in M$) we'll get $\displaystyle\int\limits_{-1}^0|y(t)|dt=0$, i.e. $y(t)=0$ on the $[-1,0)$ almost everywhere.
So, for our functional, $\displaystyle\int\limits_{0}^1x(t)y(t)dt=0$ for $x\in M$. Further, we note that $\forall n\in\mathbb{N}$ $\cos\pi nt\in M$, thus $\displaystyle\int\limits_{0}^1y(t)\cos\pi ntdt=0$.
If we consider $z(t)=y(t)-\displaystyle\int\limits_{0}^1y(t)dt$ and corresponding functional $\varphi(x)=\displaystyle\int\limits_{0}^1x(t)z(t)dt$, then it's easy to show that $\varphi(1)=0$ and $\varphi(\cos\pi nt)=0$. But $\{1,\cos\pi nt\}$ forms complete system in $L_3[0,1]$, therefore $\varphi=0$ which is equivalent to $z(t)=0$ a.e. on the $[0,1]$. So $y(t)=\operatorname{const}$ a.e. on the $[0,1]$.
