Theorem Reminder: Integral of $\phi$ Over [0,1]

  • Thread starter Thread starter Bachelier
  • Start date Start date
  • Tags Tags
    Theorem
Bachelier
Messages
375
Reaction score
0
Could someone remind me what theorem is this:

##let \ \phi: [0,1] \rightarrow \mathbb{R} \ , \ \phi \geqslant 0 \ , \ \phi \in \mathbb{R} \\[20pt]

if \ \ \int_0^1 \mathrm{\phi(t)} \ \mathrm{d}t = 0 \ \Rightarrow \ \phi \equiv 0##

Thanks
 
Last edited:
Physics news on Phys.org
I assume you mean ##\phi \geq 0##, not ##\phi > 0## (else the conclusion ##\phi = 0## would be impossible). It's false without further assumptions. Let ##\phi## be the indicator function for any finite subset of ##[0,1]## for a counterexample. Do you mean to assume ##\phi## is continuous?
 
jbunniii said:
I assume you mean ##\phi \geq 0##, not ##\phi > 0## (else the conclusion ##\phi = 0## would be impossible). It's false without further assumptions. Let ##\phi## be the indicator function for any finite subset of ##[0,1]## for a counterexample. Do you mean to assume ##\phi## is continuous?

[strike]He's written ##\phi \equiv 0## so he probably means a.e.[/strike]

Maybe not. But yeah equality is only up to almost everywhere.
 
pwsnafu said:
[strike]He's written ##\phi \equiv 0## so he probably means a.e.[/strike]

Maybe not. But yeah equality is only up to almost everywhere.
Good point. OP, can you give us more context? Is it a Riemann integral or a Lebesgue integral?
 
jbunniii said:
I assume you mean ##\phi \geq 0##, not ##\phi > 0## (else the conclusion ##\phi = 0## would be impossible). It's false without further assumptions. Let ##\phi## be the indicator function for any finite subset of ##[0,1]## for a counterexample. Do you mean to assume ##\phi## is continuous?

fixed. ##\phi \geqslant 0, \ \phi## is continuous.
 
jbunniii said:
Good point. OP, can you give us more context? Is it a Riemann integral or a Lebesgue integral?

Riemann integral
 
Bachelier said:
fixed. ##\phi \geqslant 0, \ \phi## is continuous.
OK, I don't know if the theorem has a name, but it's easy to prove. If ##\phi## is not identically zero, take a point ##x## where ##\phi(x) > 0##. Then there is a neighborhood of positive radius around ##x## where ##\phi(y) > \phi(x)/2## for all ##y## in the neighborhood. So the integral is at least ##\phi(x)/2## times the width of the neighborhood.
 
jbunniii said:
Then there is a neighborhood of positive radius around ##x## where ##\phi(y) > \phi(x)/2## for all ##y## in the neighborhood.

Do we get this because of the continuity of ##\phi##?

So the integral is at least ##\phi(x)/2## times the width of the neighborhood.

did you mean then that the integral is at least ##\epsilon \cdot \frac{\phi(x)}{2}## Hence ##\equiv## to 0?
 
Bachelier said:
Do we get this because of the continuity of ##\phi##?
Yes.
did you mean then that the integral is at least ##\epsilon \cdot \frac{\phi(x)}{2}## Hence ##\equiv## to 0?
It's at least ##\epsilon \cdot \frac{\phi(x)}{2}## (assuming ##\epsilon## is the width of the neighborhood described earlier), so the integral is NOT zero.

To summarize, we proved that if ##\phi## is continuous and ##\phi(x) > 0## for some ##x##, then ##\int \phi > 0##. Equivalently, if ##\int \phi = 0##, then there cannot be any ##x## for which ##\phi(x) > 0##, i.e., ##\phi## must be identically zero.
 
Back
Top