Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Proving Integral Inequality

Let $f$ be defined on $[0,1]$ with $f(0)=0$ and $0 < f'(x) \leq 1$. Prove that $$ \int_{0}^{1} f^3(x)\ dx \leq \bigg[ \int_{0}^{1} f(x)\ dx\bigg]^2$$

Now I have been able to show that $0 < f(x) \leq x$ from this it is simple to show that $f(x) \leq 1$ and hence $f^3(x) \leq f(x)$ thus $\int_{0}^{1} f^3(x)\ dx \leq \int_{0}^{1} f(x)\ dx$.

share|cite|improve this question

marked as duplicate by Marvis, Henry T. Horton, Douglas S. Stones, Eric Naslund Jan 18 '13 at 5:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Are you also given that $f(0) = 0$? Try to look at the function $g(t) = \left(\int_0^t f(x)dx\right)^2 - \int_0^t f^3(x)dx$ and study its derivatives. – user27126 Jan 18 '13 at 2:50
If $f(0) = 0$ then as you said, $0 \leq f(x) \leq x$, whence $\int f dx \leq 1/2$, and squaring we have $(\int f dx )^2 \leq 1/4$. From what you have so far, this completes the inequality. – A Blumenthal Jan 18 '13 at 2:52
How could $(\int f\ dx)^2 \leq 1/4$ complete the inequality? – Eric Jan 18 '13 at 3:36
up vote 1 down vote accepted

As stated, the problem is false. Likely you need another condition which guarantees that we cannot shift the function by a constant.

Let $f(x)=x+c$ where $c$ is some constant. Then $f^{'}(x)=1$, and so the condition is satisfied. Evaluating the left and right hand side, we find that $$\int_0^1 f(x)^3 dx =c^3+\frac{3}{2}c^2+c+\frac{1}{4},$$ and $$\left(\int_0^1 f(x)\right)^2 =c^2+c+\frac{1}{4}.$$ The inequality $$c^3+\frac{3}{2}c^2+c+\frac{1}{4}\leq c^2+c+\frac{1}{4}$$ is evidently false for large $c$. For example, $c=1$ fails. That is the function $f(x)=x+1$ satisfies $0< f^{'}(x)\leq 1.$

share|cite|improve this answer
I am so sorry I didn't realize that I forgot to put $f(0) = 0$. As someone above has pointed out this is a duplicate of another problem. Thank you so much for the effort though. I will accept your answer. – Eric Jan 18 '13 at 3:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.