B Spatial Average of squared functions

  • Thread starter Apashanka
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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?
 
386
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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??
 
386
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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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mathman

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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.
 

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