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While studying Fourier series, I encountered the fact that

The set of piecewise continuous functions along with the inner product $$\langle f,g \rangle=\int_0 ^{2\pi}f(x)g(x) \, \mathrm{d}x$$ form an inner product space.

However, when trying to verify this fact, I was unable to prove positive definiteness. I know that $\int_0 ^{2\pi}|f|^2 \, \mathrm{d}x=0\implies f=0$ if $f$ is continuous on the entire interval, but the allowance of a finite number of discontinuities complicates things. Is the result, as stated, incorrect or am i missing something? If it is incorrect, why can we find Fourier expansion of a function using inner product-space methods for functions which have a finite but non-zero number of discontinuities?

PrincessEev
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Vulgar Mechanick
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    If you identify functions that agree everywhere except at a finite number of points, then you get a positive definite inner product. If not, then you only get a pseudo inner product. – Danny Sep 05 '24 at 16:46

1 Answers1

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If $f$ is piecewise continuous on $[a,b]$, then you can write $$ [a,b] = \bigcup_{i=1}^n [x_i,x_{i+1}] $$ where $f$ is discontinuous at the points $x_1<x_2<x_3<\cdots<x_n$ in $[a,b]$. For notational pleasantry we'll also let $x_0 = a$ and $x_{n+1}=b$. Then $$ \int_a^b |f(x)|^2 \, \mathrm{d}x = \sum_{i=1}^n \int_{x_i}^{x_{i+1}} |f(x)|^2 \, \mathrm{d}x $$ so concluding from here is trivial.

Reply to a comment:

"How did you deduce continuity at the end points of the subintervals?"

It was an oversight; they should only be continuous on the open intervals, and discontinuous at the endpoints thereof. I rewrote the answer to the above form to be (hopefully) a little clearer.

The key point of note though is that the continuity at the endpoints of the induced subintervals (i.e. at the $x_i$ as written above) doesn't matter; you can show that if, say, $g$ is Riemann integrable on $[a,b]$ and $h$ is given by $$ h(x) := \begin{cases} g(x), & x \ne x_0 \\ A \ne g(x_0), & x=x_0 \end{cases} \text{ for some given } x_0 \in [a,b]$$ (i.e. $h$ agrees with $g$ except at one point - or even finitely many) that $\int_a^b h(x)\, \mathrm{d}x = \int_a^b g(x)\, \mathrm{d}x$, so the values at finitely many points won't affect the Riemann integral.

PrincessEev
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    How did you deduce continuity at the end points of the subintervals? Isn't the function which is 0 everywhere and 1 at the origin a piecewise continuous function? – Vulgar Mechanick Sep 05 '24 at 15:24
  • And you are not allowed to do the almost everywhere equivalence class caveat? – AHusain Sep 05 '24 at 15:38
  • @AHusain Im sorry but I don't follow. – Vulgar Mechanick Sep 05 '24 at 15:42
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    You can say two functions $f,g$ are equivalent if $\int |f-g|=0$. Two equivalent functions will have the same Fourier series. – Steen82 Sep 05 '24 at 16:33
  • "How did you deduce continuity at the end points of the subintervals? Isn't the function which is 0 everywhere and 1 at the origin a piecewise continuous function?" -- I replied to this with an edit to the post due to length. – PrincessEev Sep 05 '24 at 22:21
  • Just to clarify, to make this work, we need to define equality of the functions as mentioned by @Steen82, rather than the canonical definition (where both functions must agree everywhere)? – Vulgar Mechanick Sep 06 '24 at 05:00
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    I don't believe that would be necessary. It's largely an idea inherited from measure theory & Lebesgue integration that makes observations like these nigh-trivial though. – PrincessEev Sep 06 '24 at 05:09
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    If you do the change what is meant by equivalent and work with those equivalence classes, you are making this work correctly. The problem in @PrincessEev is that h and g are different without that caveat so $g-h$ is not zero but acts just like 0 for the integration pairing. That is the problem. – AHusain Sep 06 '24 at 17:16