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Follow-up on my last question.

It is well known that two functions $f(x)$ and $g(x)$ are orthogonal, if

$$\int f(x)g(x)\text{dx}=0$$

and it is also known that $f(x)=\sin(x)$ and $g(x)=\cos(x)$ satisfy this condition. This confuses me, because $$\int \sin(x) \cos(x)\text{dx}=-0.5\cos^2(x) $$ which obviously is not always zero, but only if I choose the interval of integration correctly. This interval relates to their periodicity.

My question:

  • How do I define the interval of integration for non-periodic functions, whose inner product is non-periodic?
  • Or put differently: Can I, and if so how, define orthogonality of non-periodic functions?
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    This can't be true. The $x$ exists only inside the integral so it can't appear on the right. – 2017-02-02
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    And yes, you can but you only define it for those for which the scalar product with themselves (or equivalently, the norm) is finite: https://en.wikipedia.org/wiki/Square-integrable_function – 2017-02-02

2 Answers 2

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To state if two object $x,y$ in a space $X$ with a scalar product are orthogonal you have to test $=0$, where the scalar product is a function from $X\times X$ to $\mathbb R$ (or $\mathbb C$).

In your case, the indefinite integral is NOT a scalar product, since its image is still $X$, and not a scalar field.

In general, when dealing with functions, the most common scalar products are definite integrals.

In the case of periodic function, the range of the integral spans the repeated set. If you have non-periodic functions, you can use the integral from $-\infty$ to $\infty$, but you have to be sure such integral exist. In fact, usually the functions considered are an Hilbert subspace of the space of all functions.

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Exodd is right. To put what he is saying in the example you give, $\cos$ and $sin$ are orthogonal in the interval $(0, 2\pi)$ e.g. you have to use a definite integral. Their periodicity means that they are also orthogonal in every interval shifted by $\pi$.

There is no example in which indefinite integral is used for a scalar product - you always integrate over the entire space where you have defined your functions, but this s still a definite integral.