Proving f=0 on [a,b] with Continuous Nonnegative f

  • Thread starter Thread starter CarmineCortez
  • Start date Start date
  • Tags Tags
    Continuous
Click For Summary
SUMMARY

To prove that a continuous nonnegative function \( f: [a,b] \to \mathbb{R} \) is zero on the interval if the integral \( \int_a^b f(x) \, dx = 0 \), one must leverage the properties of continuity and nonnegativity. The argument hinges on the contradiction that arises when assuming \( f > 0 \) at any point in the interval, leading to a positive lower sum, which contradicts the integral being zero. Thus, it is established that \( f(x) = 0 \) for all \( x \in [a,b] \).

PREREQUISITES
  • Understanding of integral calculus, specifically the properties of definite integrals.
  • Knowledge of continuous functions and their implications in analysis.
  • Familiarity with the concept of lower sums in the context of Riemann integration.
  • Basic understanding of nonnegative functions and their behavior.
NEXT STEPS
  • Study the properties of continuous functions and their implications on integrals.
  • Learn about Riemann sums and their relationship to definite integrals.
  • Explore examples of nonnegative functions and their integrals to solidify understanding.
  • Investigate the implications of discontinuous functions on integral values.
USEFUL FOR

Students of calculus, particularly those studying real analysis, as well as educators looking for clear explanations of integral properties and continuity in functions.

CarmineCortez
Messages
30
Reaction score
0

Homework Statement



Let f map [a,b]-->R be a continuous nonegative function. Suppose Integral f(x)dx from a to b = 0 show that f = 0 on [a,b]


The Attempt at a Solution



Just not sure if this is good or not..

so the lower sum <= 0 = integral f(x) dx

but the lower sum must be 0 since f is non negative.

now if f >0 then the lower sum >0 so this is a contradiction, f = 0.
 
Physics news on Phys.org
you need to work the fact that f is continuous in there.

If you consider a function i.e f(x)= 1 for rationals , 0 for irrationals; the integral of this is zero, the lower sum is zero, but f is in no way 0 throughout the entire interval.
 
Sort of. The idea is right. But you didn't use that f is continuous, and if f isn't continuous, it's not true. Show that if f is NOT equal to 0, then that implies the integral of f is greater than zero. Pick a point x where f(x)>0. Can you see how to use continuity at x to show you can put a positive area rectangle under f (like in a lower sum)?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
4
Views
1K
Replies
7
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K