Spatial Average of squared functions

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Discussion Overview

The discussion revolves around the implications of averaging a function over a volume, specifically focusing on whether a zero average of the squared function implies that the function itself must be zero throughout the volume. The scope includes theoretical considerations and assumptions regarding the properties of the function.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether a zero average of ##f^2(x,y,z,t)## necessarily implies that ##f(x,y,z,t)## is zero throughout the volume.
  • There are requests for clarification on the relevant assumptions about ##f## that would support or refute this implication.
  • Participants discuss the nature of the function, noting it is real-valued and can take both positive and negative values.
  • Continuity of the function is brought into question, with some asserting it is well-defined at every point in the volume.
  • One participant asserts that a real-valued function that is non-negative and has a zero integral over a domain is zero almost everywhere, and if continuous, then zero everywhere in the domain.

Areas of Agreement / Disagreement

There is no consensus on whether the zero average of the squared function implies that the function itself must be zero. Multiple competing views and assumptions about the properties of the function remain in discussion.

Contextual Notes

Participants have not fully defined the assumptions necessary for the implications discussed, and there are unresolved questions regarding the continuity and nature of the function.

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??
 
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Can you list all the relevant assumptions on ##f## that would make that true?
 
PeroK said:
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??
 
Apashanka said:
I am asking ,is it will be??

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

Is ##f## real- or complex-valued?
 
PeroK said:
Is ##f## real- or complex-valued?
Real,and both positive and negative valued
 
Apashanka said:
Real,and both positive and negative valued

What about continuity?
 
PeroK said:
What about continuity?
Its is well defined for every point in the volume.
 
  • #10
PeroK said:
What about continuity?

Apashanka said:
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.
 
  • #11
Mark44 said:
@PeroK asked if the function was continuous. That's not the same as being well-defined.
Yes it is continous
 
  • #12
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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