Not complete ortonormal set on a Hilbert space

Question

For all $h \in \mathbb{N}$ write $$E_h=\left\{x \in \mathbb{R^2}: h \leq |x| < h+1\right\}.$$ Let $v_h=\chi_{E_h}$, where $\chi$ is the characteristic function of $E_h$. I have checked that $$M=\left\{\frac{v_h}{\|v_h\|_2}:h\in \mathbb{N}\right\}$$ is an ortonormal set in $L^2(\mathbb{R^2})$. How can I prove that it's not complete?

Answer 1

Just take some function supported on $B(0,2)$ whose integral on $B(0,2)$ is zero.

For example $f : \mathbb{R}^2 \to \mathbb{R}$ defined as:

$$f(x,y) = \begin{cases} 1, & \text{if $\sqrt{x^2+y^2} \in [1,2\rangle$ and $x \ge 0$} \\ -1, & \text{if $\sqrt{x^2+y^2} \in [1,2\rangle$ and $x < 0$}\\ 0, & \text{otherwise} \end{cases}$$

We have:

$$\left\langle f, \chi_{E_1}\right\rangle = \int_{\mathbb{R}^2} f(x)\cdot \chi_{E_1}\,dx = \int_{B(0,2)\setminus B(0,1)} f(x)\,dx = 0$$

and for all $n \ge 3$

$$\left\langle f, \chi_{E_n}\right\rangle = \int_{\mathbb{R}^2} f(x)\cdot \chi_{E_n}\,dx = \int_{B(0,n+1)\setminus B(0,n)} f(x)\,dx = 0$$

Therefore $f \perp M$ and $f \ne 0$ so $M$ is not complete.