In a book "Completeness and Basis Properties of Sets of Special Functions" by J.R.Higgins there is a proof of completeness of the Walsh system and I don't comprehend one point in it.
To satisfy the definition of completeness, let $f\in L^2(0,1)$ and set $F(x)=\int_0^x f(t)dt$, then $F(0) = 0$.
Suppose $\int_0^1 f(x)w_{k+1}(x)dx = 0 \; (k=0,1,..)$ then $$ \int_0^1 f(x)w_1(x)dx = 0 = F(1)\\ \int_0^1 f(x)w_2(x)dx = 0 = 2\left[F(1/4)+F(3/4)\right] = 0\\ \int_0^1 f(x)w_3(x)dx = 0 = 2\left[F(1/4)-F(3/4)\right] = 0 $$ whence $F(1/4)=F(3/4)=0$. Continuing in this way for $w_4, w_5, w_6, w_7$ yields $$ F(1/8) = F(3/8) = F(5/8) = F(7/8) $$ In general we see that $F=0$ on a dense subset of $(0,1)$. By continuity $F=0, f(x)=0$ a.e. and the proof is complete.
QUESTION:
We assume that all projections of $f(x)$ are zero, for example $\int_0^1 f(x)w_2(x)dx = 0$. How do we get $2\left[F(1/4)+F(3/4)\right] = 0$ since $F(x)=\int_0^x f(t)dt$?
USEFUL LINKS:
Integral vanishes on all intervals implies the function is a.e. zero
