Convergence of Integrals: Proving f = 0 Using Continuity and Power Functions

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In summary, the theorem states that for a bounded function f(x), there exists a function p(x) such that |p(x)-f(x)|<e for all x in [a,b]. This means that if f is continuous on [a,b], then f=0.
  • #1
Treadstone 71
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"Suppose f is continuous on [a,b] and for all k = 0,1,2,3,4,5,... [tex]\int_a^b x^kf(x) = 0[/tex]. Prove that f = 0."

What I know so far:

f(c)=0 for some c in [a,b]

[tex]\int_a^b f=0[/tex]

[tex]\int_a^b x^k \int_a^b f=0[/tex]

Any hint on how to proceed would be appreciated.
 
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  • #2
Use Weierstrass approximation, and consider [itex]\int f^2[/itex].
 
  • #3
I can't see the solution. Can you elaborate a little more?
 
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  • #4
First, do you know the Weierstrass approximation theorem? Next, if p is any polynomial, what is [itex]\int pf[/itex]? Putting those two facts together, what can you conclude about [itex]\int f^2[/itex]?
 
  • #5
Weierstrass approximation theorem tells me that f(x) can be uniformly approximated by polynomials. So give e>0, for some p(x), |p(x)-f(x)|<e for all x. x^k * f(x) is apprximately x^k * p(x). But I don't understand how this has to do with f^2.
 
  • #6
f is bounded, so there exists M such that M > |f(x)| for all x in [a,b]. For all e > 0, there exists a polynomial p(x) such that |p(x) - f(x)| < e/M for all x. Thus, for such a polynomial, |f(x)p(x) - f²(x)| < e

[tex]|\int f^2| = |\int f^2 + 0| = |\int f^2 - \int fp| = |\int f^2 - fp| \leq \int |fp - f^2| \leq (b-a)e[/tex]

Since we can make e arbitarily small, [itex]|\int f^2| = 0[/itex], but f² is non-negative continuous, so f² is zero, hence so is f.
 
  • #7
At no point did you use the hypothesis of 'x^k for all k=0,1,2,...' . This would imply that if f is continuous on [a,b] and [tex]\int_a^b f=0[/tex], then f=0.
 
  • #8
Treadstone 71 said:
At no point did you use the hypothesis of 'x^k for all k=0,1,2,...' . This would imply that if f is continuous on [a,b] and [tex]\int_a^b f=0[/tex], then f=0.

Try to justify each of AKG's steps carefully.
 
  • #9
I understand now. Thanks for the help.
 

1. What does it mean to prove that f = 0 in an integral?

Proving that f = 0 in an integral means to show that the function f has a definite integral value of 0 over a given interval. This means that the area under the curve of the function f is equal to 0.

2. Why is it important to prove that f = 0 in an integral?

Proving that f = 0 in an integral is important because it allows us to determine the behavior and properties of a function. It also helps us to solve problems in various fields such as physics, engineering, and economics.

3. How do you prove that f = 0 in an integral?

To prove that f = 0 in an integral, we can use various integration techniques such as substitution, integration by parts, or partial fractions. We can also use properties of integrals, such as linearity and the fundamental theorem of calculus, to simplify the integral and show that it equals 0.

4. What are some applications of proving that f = 0 in an integral?

Some applications of proving that f = 0 in an integral include finding the area under a curve, calculating work done by a force, finding the average value of a function, and determining the probability of an event in statistics.

5. Can f be any function when proving that f = 0 in an integral?

No, f cannot be any function when proving that f = 0 in an integral. The function must be integrable, meaning it must have a definite integral value over a given interval. This excludes functions that have infinite or undefined values, or functions that are not continuous over the interval.

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