# A Real Analysis question on anti-derivatives

1. May 20, 2012

### mike1988

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.

2. May 20, 2012

### gopher_p

Edit:

I think you mean $\int_m^nf(x)dx\leq(n-m)^2$. Again, definite integrals are not the same as antiderivatives.

Last edited: May 20, 2012
3. May 20, 2012

### Fredrik

Staff Emeritus
I don't understand what you mean by "assume anti-derivative" or "f(x)dx from m to n≤ (n-m)^2".

4. May 20, 2012

### mike1988

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. May 20, 2012

### Ray Vickson

Hint: mean-value theorem.

RGV

6. May 20, 2012

### mike1988

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. May 20, 2012

### gopher_p

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. May 20, 2012

### mike1988

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. May 20, 2012

### gopher_p

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).