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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.

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  • $\begingroup$ Have you managed to show the orthogonality part? $\endgroup$
    – J.A.L
    Nov 12, 2014 at 13:02
  • $\begingroup$ 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. $\endgroup$
    – J.A.L
    Nov 12, 2014 at 13:16

1 Answer 1

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It sounds like the orthonormality is not the issue giving you trouble, only the incompleteness.

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.

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