Question about bounded functions

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    Bounded Functions
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Spherical harmonics form a complete orthonormal basis for representing bounded single-valued functions on a sphere, as unbounded functions would require an infinite number of basis functions. This limitation arises from the triangle inequality, which restricts the representation to bounded functions. Higher harmonic orders correspond to increased angular frequency, meaning that more complex patterns require higher orders for accurate representation. However, capturing fast-changing functions is ultimately about scaling rather than simply increasing harmonic order. Understanding these concepts is crucial for effectively using spherical harmonics in mathematical applications.
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Hello,

Just reading an essay about spherical harmonics and it says that spherical harmonic form a complete orthonormal basis set of functions over the sphere and can be used to represent any bounded single-valued function over a sphere.

I am not sure I understand why we can only represent bounded functions by spherical harmonics. Is it because otherwise we would need an infinite number of the spherical basis functions?

EDIT: One more question. It says about Spherical harmonics that the angular frequency increases with harmonic order n. Does this mean that to capture fast changing functions, one would need higher harmonic orders?

Thanks,

Luca
 
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pamparana said:
Hello,

Just reading an essay about spherical harmonics and it says that spherical harmonic form a complete orthonormal basis set of functions over the sphere and can be used to represent any bounded single-valued function over a sphere.

I am not sure I understand why we can only represent bounded functions by spherical harmonics. Is it because otherwise we would need an infinite number of the spherical basis functions?
You cannot get unbounded functions from bounded ones because of the triangle inequality.
EDIT: One more question. It says about Spherical harmonics that the angular frequency increases with harmonic order n. Does this mean that to capture fast changing functions, one would need higher harmonic orders?
No, whatever capture means. In the end it is a matter of scaling. It is: the more complicated the pattern the more high order functions will be needed.
Thanks,

Luca
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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