Does two functions have to have an inner product of zero for just one interval to be orthogonal?

Question

If I have a function space defined by the basis {$sin(x), cos(x)$}. And you have the vectors $v = (2,2)$ and $u = (-1,1)$.

So $v$ reperesent a point, on the plane define by the basis, the function $2sin(x) + 2cos(x)$ and $u$ represents $-sin(x) + cos(x)$.

The vectors $v$ and $u$ are orthogonal on the plane where first axis represents multiples of $sin(x)$ and the second multiplies of $cos(x)$. So since the vectors themselves are orthogonal, then it should be the case that the functions they represent need to be orthogonal. And if I take the inner product of the function on the interval $[-\pi, \pi]$, that is

$\int_{-\pi}^{\pi} (2sin(x) + 2cos(x)) * (-sin(x)+ cos(x)) dx = 0$

But this is not the case for another interval, forexample $[-3\pi/2, 3\pi/2]$.

So then does this mean the functions are only orthogonal for certain intervals only? Can they be considered orthogonal?

This is all of course if I havent misinterpreted the concept of inner product of functions, which I'm still struggling to understand.

Answer 1

Orthogonality of vectors (in this case functions residing in a vector space) requires knowledge of an inner product, $\langle\cdot ,\cdot\rangle$, and two vectors $u$ and $v$ are said to be orthogonal (with respect to our specific inner product) when it is true that $\langle u,v\rangle=0$

It is possible that the same vectors, and even the same choice of vector space, have multiple choices for inner products where although $u$ might be orthogonal to $v$ with respect to the first inner product, they need not be orthogonal with respect to a different inner product.

Your example is a perfectly good one and shows that $(2\sin(x)+2\cos(x))\perp (-\sin(x)+\cos(x))$ with respect to the inner product $\langle u,v\rangle_1=\int\limits_{-\pi}^\pi u\cdot v ~dx$, but not with respect to the inner product $\langle u,v\rangle_2=\int\limits_{-\frac{3\pi}{2}}^{\frac{3\pi}{2}}u\cdot v~dx$.

Context is very important when discussing things like orthogonality, and so whenever we say "two vectors are orthogonal", unless it is completely obvious what setting we are working in, you should always include a reminder as to what inner product you are using.

Answer 2

There's not just one inner product on any given space of functions. Multiplying them and integrating (with respect to Lebesgue measure) is only one of many possible ways to define an operation satisfying the definition of "inner product" (bilinear symmetric positive definite form). And as you saw, you can also choose what set to integrate over.

So there isn't an absolute notion of two functions being "orthogonal". Given two functions, they might be orthogonal with respect to some inner products and not others. So when you speak of functions being orthogonal, you have to say which inner product you are using.