# Rademacher functions form an incomplete orthonormal system in the real Hilbert space

I need to show that the Rademacher functions: $$r_n(x) = \text{sign}(\sin(2^nπx)), x \in [0, 1], n = 0, 1, 2, . . .$$ form an incomplete orthonormal system in the real Hilbert space $L^2([0,1])$

I know that a system is orthonormal if in addition to being orthogonal, we have that it is also normalized but I don't really understand the completeness criterion and I don't know how to show that these function do not form a complete orthonormal system. Any help will be appreciated.

• Have you managed to show the orthogonality part? Nov 12, 2014 at 13:02
• A complete orthonormal system $x_n$ (apart from being orthonormal) means that $\operatorname{span}(x_n)$ is dense in $L^2$. Thus every function in $f\in L^2$ can be written as $\sum_n^\infty (f,x_n)x_n$. To show incompleteness it suffices to find nonzero $f\in L^2$ for which $(f,r_n)=0$ for all $n$, though I don't know if that is the best approach in your case. Nov 12, 2014 at 13:16

An orthonormal system $\{X_n\}$ is complete if and only if $\langle f,X_n\rangle=0 \ \forall n\implies f=0$.
Let $f(x)=r_1(x)r_2(x)$. Then $\langle f,r_n\rangle=0$ for all $n$, yet $f$ is not the zero function.
By the way, the Walsh functions are a complete superset of the Rademacher functions on $[0,1]$ with respect to the standard inner product.