Functional analysis: Inner Product and Haar, Rademacher, Walsh functions

Solution 1:

a) What you do is you construct $g$ so that it differs from $f$ on very small intervals around the discontinuity points. For an easy example, for $f=1_{[1/2,1]}$ you could take $$ g_n(x)=\begin{cases} 0,&\ 0\leq x< 1/2-1/n \\ n/2 (x-1/2+1/n),&\ 1/2-1/n<x<1/2+1/n \\ 1,&\ x\geq 1/2+1n\end{cases} $$

b) the Haar functions are in $PC^2[0,1]$ and are orthonormal and total in $L^2[0,1]$ which is bigger. So yes, they are an orthonormal basis. The thing is that considering an orthonormal basis of a non-complete inner product space leads to weird things: basically, not every sequence of coefficients is allowable, and some will produce functions which are not pointwise continuous.

c) Yes, that shows that it is orthonormal. You still need to show that it is not a basis: there exists a piecewise continuous function $f$ that is orthogonal to all Rademacher functions.

d) As in the previous cases (and all cases) the same proof will apply: you have your functions, you have your inner product, the proof won't change because you tell it "where to live". Now, it cannot be the case that the Walsh functions form an orthonormal basis for $PC^2[0,1]$; because, as you said yourself, $PC^2[0,1]$ is dense in $L^2[0,1]$.