B Spatial Average of squared functions

Apashanka

If averaging of a function over a volume is defined as $\frac{\int_v f(x,y,z,t) dv}{\int_v dv}$.
Now if the average $f^2(x,y,z,t)$ is given 0 over a volume,then $f(x,y,z,t)$ has to be necessarily 0 in the volume domain??

PeroK

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Can you list all the relevant assumptions on $f$ that would make that true?

Apashanka

Can you list all the relevant assumptions on $f$ that would make that true?
I am asking ,is it will be??if so then what $f$ had to satisfy??

PeroK

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I am asking ,is it will be??
Surely it's better if you learn to think for yourself.

Apashanka

Surely it's better if you learn to think for yourself.
Yes it is,but can't you give any hints or ideas that would help to think better....

PeroK

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Yes it is,but can't you give any hints or ideas that would help to think better....
Is $f$ real- or complex-valued?

Apashanka

Is $f$ real- or complex-valued?
Real,and both positive and negative valued

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Apashanka

Its is well defined for every point in the volume.

Mark44

Mentor
Its is well defined for every point in the volume.
@PeroK asked if the function was continuous. That's not the same as being well-defined.

Apashanka

@PeroK asked if the function was continuous. That's not the same as being well-defined.
Yes it is continous

mathman

This seems to be a long drawn out discussion of the obvious, that is: a real valued function which is non-negative (such as the square of a real valued function) and has a zero integral over a domain, is 0 almost everywhere. If continuous, then 0 everywhere in the domain.

• PeroK

"Spatial Average of squared functions"

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