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This should be a fairly trivial question, but I would like to support my current intuition.

Let's consider two orthogonal functions $f$ and $g$, i.e. two function whose inner product is equal to zero over a certain domain $D$ (the weight function is omitted for simplicity):$$\int_D f(x)g(x) dx= 0$$

Does the orthogonality of the two functions hold only if the functions have the same arguments? In other words, would the following be true? $$\int_D f(y)g(x) dx= 0\space(1)$$

Well the equation above does not make much sense, since the $y$ argument is not specified. Let's say that $y$ itself depends on $x$, i.e. it exists a function $t$ such that $y = t(x)$. My guess is that $(1)$ is not true. In fact, if we consider $k(x) = (f \circ t)(x) = f(t(x))$, then we get:

$$\int_D k(x)g(x) dx= 0$$

Which it does not necessary hold, since the orthogonality between $k$ and $g$ must be proven in its own rights.

To summarize:

Am I correct in saying that the following is not true in the general case (although it could be true for same special cases, es. when $t$ is the identity function)?

$$\int_D f(y)g(x) dx= 0 \land y = t(x) \land k(x) = (f \circ t)(x) = f(t(x)) \Rightarrow \int_D k(x)g(x) dx= 0$$

Finally, what if $f$ and $g$ were elements of an orthonormal basis function set, would this make any difference?

The actual problem:

I'm currently doing some work involving spherical harmonics. At one point I found myself faced with the following equation:

$$h = \int_S Y_l^m(\vec n_x)Y_{l'}^{m'}(\vec x)dx\space(2)$$

Where $S$ is the surface of a 3D model, $dx$ is the area element of this surface, $\vec x$ is the position of each area element and $\vec n_x$ is the model's normal at the $\vec x$ position (note: SH functions are usually defined over the two angles $\theta$ and $\phi$. Please assume that these angles are implicitly represented by the directions of $\vec x$ and $\vec n_x$).

Initially, I thought that the solution for $(2)$ was $h = 1$ if $m = m'\land l = l'$ and $h = 0$ otherwise, but now I'm convinced that given the different arguments, $\vec x$ and $\vec n_x$, this is not necessarily the case. Is my thinking right?

Thanks in advance!

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Yes, you're right that the arguments must be the same for orthogonality to hold, and your intuition is spot on about why this should be so. In your example,

$$\int_D f(t(x))g(x)$$

one way to see this is to take $t(x)$ something dumb, like a constant function. For example, let's have

$$f = \cos, g = \sin, t(x) = 0$$

so:

$$\int_0^\pi \cos(x)\sin(x) = 0$$

but

$$ \int_0^\pi \cos(t(x))\sin(x) = \int_0^\pi \cos(0)\sin(x) = \int_0^\pi 1\cdot\sin(x) = 2 $$

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  • $\begingroup$ Great! Feel good to be right :) I have still a doubt though. Orthogonality might still hold in same cases (es. when t is the identity function). Do you think that the case I mentioned in the end on my question is one of these special cases for which orthogonality continues to hold? I think not... $\endgroup$ – Geoffrey91 Feb 11 '16 at 17:48
  • $\begingroup$ Which question? Do you mean: "what if $f$ and $g$ are elements of an orthonormal basis function set?" $\endgroup$ – Eli Rose Feb 11 '16 at 18:37
  • $\begingroup$ Not only that (although having a question for that would be useful), but also regarding the "The actual problem" section of my question. $\endgroup$ – Geoffrey91 Feb 11 '16 at 18:41

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