Enlarging the space $PC(a,b)$ to include functions with one or more infinite singularities

Question

I'm reading a Fourier analysis book and on the chapter about convergence and completeness of orthogonal sets of functions I have one part which I don't understand. I have uploaded the part as an image and highlighted the relevant parts in the text:

What I'm interested about is the purple part (the other colored boxes are there to just give more info). My question is:

Why can one enlarge the space $PC(a,b)$ to include functions as in the example in the image by simply allowing improper integrals in the definition of the inner product and the norm for functions in $PC(a,b)?$ This in unclear for me, can someone make this more specific? Example?

Thank you for any help! Please let me know if you need more information or if the image is unclear. I tried to give all the needed information in the image.

P.S.

To enlarge the image do: right click --> show image / open image in new tab.

Answer 1

The space $PC(a,b)$ of piecewise continuous functions (p.c. functions) on the interval $[a,b]$ is not complete; this is the content of the example in the left picture in the OP. The definition of p.c. functions is given on the top right picture in the OP: we note that an element of $PC(a,b)$ can have a finite number of "jumps" and all of them must be finite.

How to achieve completeness? The author tried to enlarge $PC(a,b)$ with functions $f$ admitting

1. non finite jumps (like $f(x)=x^{-\frac{1}{4}})$ for $x>0$ and $f(0)=0$ in the example)
2. absolute converging integral, i.e.

$$\int_a^b |f(x)|dx <\infty $$

Let us call this space $PC_\infty(a,b)$; it contains the original $PC(a,b)$. The enlarged space $PC_\infty(a,b)$ admits the same scalar product defined on $PC(a,b)$ (due to absolute convergence of elements of $PC_\infty(a,b)$ ); the integral can be improper though, as shown in

$$\langle f, 1 \rangle := \int_0^1\frac{1}{x^\frac{1}{4}}dx:=\lim_{t\rightarrow 0^+} \int_t^1\frac{1}{x^\frac{1}{4}}dx$$

where we considered $PC_\infty(0,1)$ and $f$ given as in the example in the OP. With $1$ we mean the constant function $g(x)=1$ on $[0,1]$. The result of the scalar product $\langle f, 1\rangle$ is finite!

The problem is that even $PC_\infty(a,b)$ is not large enough to be complete: this is the meaning of the paragraph in the violet box in the left picture of the OP.