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]

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]

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]

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

 

[PeroK]

Real,and both positive and negative valued

What about continuity?

 

[Apashanka]

What about continuity?

Its is well defined for every point in the volume.

 

[Mark44]

What about continuity?

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.