Is f Uniformly 0 if Integral of f^2 on [a,b] equals 0?

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Discussion Overview

The discussion centers around the question of whether a continuous function \( f \) defined on the interval \([a,b]\) must be uniformly zero if the integral of \( f^2 \) over that interval equals zero. The scope includes theoretical reasoning and mathematical justification.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant, KRC, poses the initial question about the implications of the integral of \( f^2 \) being zero.
  • Another participant expresses confusion regarding the notation and suggests changing \( f \) to \( x \) to simplify the question, but does not clarify the implications of the integral being zero.
  • A later reply questions the assumption that \( f \) could be non-zero at some point, suggesting that if \( f \) is not everywhere zero, then the integral cannot be zero due to the continuity of \( f^2 \).
  • One participant agrees with the reasoning that if \( f \) is continuous and non-negative, then the integral must be positive if \( f \) is non-zero at any point within the interval.

Areas of Agreement / Disagreement

Participants generally agree on the reasoning that if \( f \) is continuous and non-zero at any point, the integral of \( f^2 \) cannot be zero. However, there is no consensus on the implications of the original question, as some participants express confusion about the notation and the conditions of the problem.

Contextual Notes

There are limitations in the discussion regarding the assumptions about the function \( f \) and its continuity, as well as the interpretation of the integral condition. The notation used by participants may also lead to misunderstandings.

krcmd1
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given f continuous on [a,b] with a not equal to b,
does
[tex]\int{a_b}(f^2) = 0 [\tex]<br /> <br /> imply that f must be uniformly 0?<br /> <br /> (still not able to get Latex right, either. sorry)<br /> <br /> thanks<br /> <br /> KRC[/tex]
 
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I don't know why they used 'f' since 'f' is generally used to be a function not describe them. If we change the f to x it makes this question easier.

I don't know what it is saying when it asks is x is uniformly zero.
If you find the definite integral of x^2 and make it equal to zero you get:
a^3 = b^3

Since a equals b there will never be any area under the curve. I'm sorry I don't know where to progress from here.
 
Evales said:
I don't know why they used 'f' since 'f' is generally used to be a function not describe them. If we change the f to x it makes this question easier.

I don't know what it is saying when it asks is x is uniformly zero.
If you find the definite integral of x^2 and make it equal to zero you get:
a^3 = b^3

Since a equals b there will never be any area under the curve. I'm sorry I don't know where to progress from here.

Umm... What are you talking about? It seems fairly evident that f is representing a function.

To krcmd1: What happens if f is not everywhere 0 on [a,b]?
 
If f is continuous then f2 is both continuous and non-negative. Suppose there were some point, say x0 at which f(x0[/sup]) is not 0. By continuity, you can find some small neighborhood of x0 in which f2 is positive. The integral of f2 over that neighborhood is positive. Since there are no negative values of f2 to cancel that, the integral of f2 [a, b] is larger than or equal to that positive number: it can't be 0.
 
Thank you Halls of Ivy. That is just how I reasoned it. Don't have enough experience to be confident in my reasoning yet.

Ken
 

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