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For two Riemann integrable, $2\pi$-periodic functions $u$ and $v$, we define

$$(u,v)=\frac{1}{2\pi}\int_{0}^{2\pi} u(x)\overline{v(x)} dx$$

where the complex number $(u,v)$ is called the scalar product of the functions $u$ and $v$.

Is is not at all clear to me why this definition works? What is the reason for multiplying one function/vector by the conjugate of another function/vector? And then why would we want to integrate the result of that multiplication?

Eiraus
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  • As you say, you are interested in Fourier Series. This is an inner product for which the Fourier basis functions $\exp(inx)$ are orthogonal. – Angina Seng Sep 24 '17 at 13:11
  • What do you mean by "why this definition works"? Do you want to know if the function is well-defined or do you want to know if the function is really an inner product? – edm Sep 24 '17 at 13:39
  • @edm If I compare with the linear algebra inner product (a,b)=|a||b|cos(x). It is easy to understand why this "works" i.e. helps us to find orthogonal vectors. I suppose I am looking for a similar way to understand the definition above helps us to find orthogonal functions. – Eiraus Sep 24 '17 at 13:55
  • Well, that's the thing: the relationship between scalar products and orthogonality goes the other way. Which functions are considered orthogonal depends on what scalar product you use. – Arthur Sep 24 '17 at 15:41

2 Answers2

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The definition works because it fulfills the criteria in the definition of scalar product / inner product. It also works because it's a very natural extension of the standard inner product on finite complex vector spaces.

We conjugate because we want the scalar product of a function with itself to be real (and positive, as long as the function isn't $0$). We integrate because that's one of the most natural ways to assign a single number to a function.

Arthur
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Multiplying by $\overline{v(x)}$ instead of just $v(x)$ ensures the positivity property of inner products:

$$ (u, u) = \frac{1}{2\pi} \int_{0}^{2\pi} u(x)\overline{u(x)}\:dx = \frac{1}{2\pi} \int_{0}^{2\pi} |u(x)|^2\:dx \ge 0. $$

Inner products must output scalar values, so integrating is a way getting this while maintaining the properties (positive-definiteness, additivity and homogeneity in the first slot and conjugate symmetry) and extracting useful information of the functions behavior on a certain interval. We could define an inner product which would be easier to calculate, such as

$$ (u, v) = u(a)\overline{v(a)} $$

for some fixed constant $a$. But this would only give us information about how $u$ and $v$ behave at the fixed point $a$ instead of a whole interval and wouldn't satisfy the definiteness property because we could have $u(a) = 0$ and then $(u, u) = 0$ with $u \neq 0$.

WeakestTopology
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  • Thankyou @cpassos this helps. I didn't realise that we were taking the inner product over an interval. May I ask a follow up question of why it is necessary to then multiply the integral by 1/2pi? – Eiraus Sep 24 '17 at 13:31
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    I don't know about Fourier series, but maybe $\frac{1}{2\pi}$ was chosen to normalize certain functions. $(1, 1) = 1$, for example. – WeakestTopology Sep 24 '17 at 13:41