A Real Analysis question on anti-derivatives

In summary: Once you have this, it is straightforward to show that the function cannot be continuous at the point where this occurs, and so it must be zero.
  • #1
mike1988
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0
Let f : R to R be a continuous function, and assume anti-derivative of f(x)dx from m to n≤ (n-m)^2 for every closed bounded interval [m,n] in R. Prove that f(x) = 0 for all x in R.

I tried using fundamental theorem of calculus but got stuck.

Any help/suggestion would be appreciated.
 
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  • #2
mike1988 said:
assume anti-derivative of f(x)dx from m to n≤ (n-m)^2 for every closed bounded interval [m,n] in R.

Edit:

I think you mean ##\int_m^nf(x)dx\leq(n-m)^2##. Again, definite integrals are not the same as antiderivatives.
 
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  • #3
I don't understand what you mean by "assume anti-derivative" or "f(x)dx from m to n≤ (n-m)^2".
 
  • #4
gopher_p said:
This makes no sense. Maybe copy the problem exactly as stated.

Also, definite integrals are not antiderivatives. Definite integrals are numbers, antiderivatives are functions (more precisely families of functions).

I am sorry. Here is the exact question:

Let f : R to R be a continuous function, and suppose that definite integral from m to n |∫(m to n)f(x)dx|≤(n-m)^2 for every closed bounded interval [m, n] in R. Then is it the case that f(x) = 0 for all x in R?
 
  • #5
mike1988 said:
I am sorry. Here is the exact question:

Let f : R to R be a continuous function, and suppose that definite integral from m to n |∫(m to n)f(x)dx|≤(n-m)^2 for every closed bounded interval [m, n] in R. Then is it the case that f(x) = 0 for all x in R?

Hint: mean-value theorem.

RGV
 
  • #6
Ray Vickson said:
Hint: mean-value theorem.

RGV

So after using Mean Value theorem and FUndamental theorem of Calc, I got to a point where F'(c)=f(c) ≤ m - n. BUt this still doesn't seem to work enough.
Is there anything crucial that I am missing?

Thanks!
 
  • #7
mike1988 said:
So after using Mean Value theorem and FUndamental theorem of Calc, I got to a point where F'(c)=f(c) ≤ m - n. BUt this still doesn't seem to work enough.
Is there anything crucial that I am missing?

Thanks!

I think he meant Mean Value Theorem for integrals:

If ##f## is continuous on ##[a,b]##, then there is ##c\in(a,b)## with ##\int_a^bf(x)dx=f(c)(b-a)##.

Given the FTC, it's just a rewrite of the "regular" mean value theorem.The key here is to apply this to your problem, with the assumptions given, and notice what the implications are when ##n-m## is very small.
 
  • #8
gopher_p said:
I think he meant Mean Value Theorem for integrals:

If ##f## is continuous on ##[a,b]##, then there is ##c\in(a,b)## with ##\int_a^bf(x)dx=f(c)(b-a)##.

Given the FTC, it's just a rewrite of the "regular" mean value theorem.


The key here is to apply this to your problem, with the assumptions given, and notice what the implications are when ##n-m## is very small.

Do you think I should go by contradiction instead? coz I am not really figuring out how to use this assumption here.
Some elaboration would really help!
 
  • #9
mike1988 said:
Do you think I should go by contradiction instead? coz I am not really figuring out how to use this assumption here.
Some elaboration would really help!

Yes, I think it would be easiest to assume that ##f\neq0## and show that this gives a contradiction.

Supposing that ##f\neq0## gives that there exists ##x_0## such that ...

Since ##f## is continuous, there is ##\delta>0## such that ... on ##(x_0-\delta,x_0+\delta)##.

Then there is ##c\in(x_0-\delta,x_0+\delta)## such that ##\left|\int_{x_0-\delta}^{x_0+\delta}f(x)dx\right|=\left|f(c)\right|\cdot 2\delta>## ..., contradicting the assumption that ##\left|\int_{x_0-\delta}^{x_0+\delta}f(x)dx\right|\leq##... .

Now you may need to fidget a bit with the ##\delta## to get the inequalities that you need, but the direction that you need to fidget is a "good" one.

There is a fairly straightforward way to get around using the Mean Value Theorem. However it requires you to identify that you are in a situation where ##\left|\int_{a}^{b}f(x)dx\right|=\int_{a}^{b} \left| f(x) \right|dx## (i.e. the triangle inequality is actually an equality).
 

FAQ: A Real Analysis question on anti-derivatives

What is an anti-derivative in real analysis?

An anti-derivative in real analysis is a function that, when differentiated, gives the original function. It is the inverse operation of differentiation and is represented using the integral symbol.

How do you find the anti-derivative of a function?

To find the anti-derivative of a function, you can use the fundamental theorem of calculus or apply known integration rules, such as the power rule, exponential rule, and trigonometric rule. It is also important to remember to add a constant term (C) to account for all possible solutions.

What is the relationship between anti-derivatives and indefinite integrals?

An indefinite integral is the general form of an anti-derivative. It represents the set of all possible anti-derivatives of a given function. In other words, an indefinite integral is the family of curves that can be obtained by adding a constant term (C) to the anti-derivative of a function.

Can all functions have an anti-derivative?

No, not all functions have an anti-derivative. For a function to have an anti-derivative, it must be continuous on its entire domain. Additionally, there are some special functions, such as the Dirichlet function, that do not have an anti-derivative.

What are the practical applications of anti-derivatives?

Anti-derivatives have many practical applications in fields such as physics, engineering, and economics. They are used to find the total change in a quantity over a given interval, which can be useful in calculating displacement, velocity, and acceleration in physics. In engineering, they are used in the design and analysis of circuits, while in economics, they are used to calculate total revenue and profit.

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