# 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.

"Spatial Average of squared functions"

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