Functional analysis: Inner Product and Haar, Rademacher, Walsh functions I am attempting an exercise in functional analysis: Let $PC_2[0,1]$ denote the vector space of piecewise continuous functions $f$ on $[0,1]$ such that $\int_0^1|f(t)|^2dt < \infty $. Then I can show that indeed $<f, g> =\int^1_0 f(t)\overline{g(t)} dt$ defines an inner product on $PC^2[0, 1]$. Now my difficulty is I want to:
a) Show the set of continuous functions in $PC^2[0, 1]$ is dense in $PC^2[0, 1]$.  So given any piecewise continuous $f$, there's a continuous $g$ so that $\int (f - g)^2 < \epsilon$. Given such an $f$, there are at most a finite number $x_1 < ... <x_N$ points of discontinuity.  So by set $g = f$ for the most of the intervals? I am not sure of the details hmm also, how to prove the same holds for any subset of step functions?
b) Show that 1 and the Haar functions form an othonormal basis for $PC^2[0, 1]$.
From Haar system forms an orthonormal system in $L_2[0,1]$, we can show inductively that
$$\chi_{[k2^{-n},(k+1)2^{-n}]}$$
lies in the span for all $0≤k<2^n$. This implies $\chi_{[a,b]}$ lies in the span (as an element of $L^2$, ie differing from that function on a null-set) whenever $a,b$ are of the form $2^n k$ for some $n,k$. That implies (since such points are dense in $[0,1]$) that the characteristic function of any sub-interval lies in the closure of the span, but the closure of that is $L^2[0,1]$ so the span of the wavelets was dense.
Start with $n=1$,
$$\psi_{0,0}+1=2\chi_{[0,1/2]}\qquad 1-\psi_{0,0}=2\chi_{[1/2,1]}$$
Now $\psi_{n,k}=\chi_{[k2^{-n},(k+1/2)2^{-n}]}-\chi_{[(k+1/2)2^{-n},(k+1)2^{-n}]}$. From induction we are supposing $\chi_{[k2^{-n},(k+1)2^{-n}]}$ is in the span, so it follows that
$$\chi_{[k2^{-n},(k+1/2)2^{-n}]}-\chi_{[(k+1/2)2^{-n},(k+1)2^{-n}]}+\chi_{[k2^{-n},(k+1)2^{-n}]}=2\chi_{[k2^{-n},(k+1/2)2^{-n}]}=2\chi_{[(2k)2^{-(n+1)},(2k+1)2^{-(n+1)}]}$$
Lies in the closure of the span for any $0≤2k<2^{n+1}$. The odd case follows by considering
$$\chi_{[k2^{-n},(k+1)2^{-n}]}-(\chi_{[k2^{-n},(k+1/2)2^{-n}]}-\chi_{[(k+1/2)2^{-n},(k+1)2^{-n}]})=2\chi_{[(k+1/2)2^{-n},(k+1)2^{-n}]}=2\chi_{[(2k+1)2^{-(n+1)},(2k+2)2^{-(n+1)}]}$$. Does this solution work for this case?
c) Show the Rademacher functions form an orthonormal sequence in $PC^2[0, 1]$. From Rademacher functions form an orthonormal system but not an orthonormal basis, I see that Rademacher functions form an orthonormal system for $L_2[0,1]$: ie To compute $\langle r_n, r_m\rangle$ with $n<m$, say, observe that 
$$\langle r_n, r_m\rangle=\int_0^1 \operatorname{sgn}(\sin(2^n\pi t))\operatorname{sgn}(\sin(2^m\pi t))\,dt\\=\sum_{k=0}^{2^n-1}\int_{k2^{-n}}^{(k+1)2^{-n}} \operatorname{sgn}(\sin(2^n\pi t))\operatorname{sgn}(\sin(2^m\pi t))\,dt$$
In each summand, $\operatorname{sgn}(\sin(2^n\pi t))$ is constant and $\operatorname{sgn}(\sin(2^m\pi t))$ runs over $2^{m-n}$ full periods.
Thus each summand is zero. Is this same proof correct for this case?
d) Show the Walsh function form an orthonormal basis for $PC^2[0, 1]$.
From prove Walsh functions form a closed orthonormal system, there is a proof showing Walsh functions are orthogonal with respect to the $L_2$ inner product but can this be used to show  orthonormality for $PC^2[0, 1]$?
Thank You for any help :)
 A: 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 $L^2[0,1]$. 
