Question: Let $f:[0,1]\to\mathbb{R}$ be a continuous function with $\int_0^1f(t)dt=0$. Prove that there exists $c\in[0,1]$ such that $$\int_0^cf(t)dt=f(c)^3.$$

Solution: Let $g:[0,1]\to\mathbb{R}$ be such that$$g(x)=\int_0^xf(t)dt-f(x)^3, \forall x\in[0,1].$$

Now since $f$ is continuous $\forall x\in[0,1]$, thus, by the first fundamental theorem of calculus, we can conclude that $g$ is continuous $\forall x\in[0,1]$.

Thereafter, observe that $g(x)=0$ for some $x\in[0,1]\iff \int_0^xf(t)dt=f(x)^3$ for some $x\in[0,1]$. Hence, to prove the statement of the problem it is sufficient to show that $g(c)=0$ for some $c\in[0,1]$.

Now $g(0)=-f(0)^3$ and $g(1)=-f(1)^3$.

Observe that if $f(0)$ and $f(1)$ are of different signs, then $g(0)$ and $g(1)$ are also of different signs, in which case, by IVT we can conclude that $\exists c\in(0,1)\subset[0,1],$ such that $g(c)=0$. Hence, we are done in this case.

Again, if $f(0)=0$ or $f(1)=0$, then at least one of $g(0)$ and $g(1)=0$, in which case we are done.

Now, we are left with the case that both $f(0)$ and $f(1)$ are of the same sign. Thus, let us assume WLOG that $f(0)>0$ and $f(1)>0$. Hence, $g(0)<0$ and $g(1)<0$. Now since $\int_0^1f(t)dt=0$ and $f(0),f(1)>0$, implies that $\exists$ at least two points $a,b\in(0,1)$, such that $b>a$ satisfying $f(a)=f(b)=0$. Thus, we can conclude that $\exists c_1\in(0,1),$ such that $f(x)>0, \forall x\in[0,c_1)$ and $f(c_1)=0$. Hence, we have $$g(c_1)=\int_0^{c_1}f(t)dt-f(c_1)^3=\int_0^{c_1}f(t)dt>0.$$ Thus, we have $g(c_1)>0$ and $g(1)<0$, which implies that, by IVT we can conclude that $\exists c\in(c_1,1)\subset[0,1]$, such that $g(c)=0$. Hence, we are done in this case too.

Hence, we are done with all the cases and in each case we have shown that $\exists c\in[0,1]$ such that $g(c)=0$. Thus, we are done.

Is this solution correct and rigorous enough? If yes, is there any alternative solution?

  • $\begingroup$ Looks fine and clearly explained without omitting any details. +1 for your efforts. $\endgroup$
    – Paramanand Singh
    May 14 '20 at 9:32

Appears correct and mostly rigorous to me, also clear and not too long. A proof by IVT is a valid idea. Two points:

  1. Are you sure that you're applying the second FTOC, not the first FTOC?
  2. You say that there exists $0<c_1<1$ such that $f(x)>0$ for all $0\leq x < c_1$ and $f(c_1)=0$. How do you know the $f(x) > 0$ for all $0\leq x < c_1$ part holds?

For part 2., you need to essentially show that $f$ has a smallest positive zero (assuming, for instance, that $f(0)>0$). Can you do it?

As a final note, the proof you gave allows one to slightly generalise the result. Namely, you can use any continuous $h: [0, 1] \to \mathbb{R}$ that preserves sign at $f(0), f(1)$ with $h(0) = 0$ instead of the cube function, i.e. instead of $f(c)^3$ you could put $h(f(c))$ without any trouble. Here is a slightly altered proof which, similarly to your proof, works for this generalised case (with details to be filled in by reader). Sign-preserving of $h$ is irrelevant at $f(1)$ for this proof, hence may be omitted.

Proof. For concreteness, let $f(0) > 0$. Let $x_1$ be the smallest positive zero of $q(x) := \int\limits_{0}^{x}f(t)\,\mathrm{d}t$. We may assume $f(x_1) < 0$. Then by IVT at some point $z$ in $(0, x_1)$ it is the case that $f(z) = 0$ and $q(z) > 0$. Therefore, $g := q - h(f)$ will have changed sign in $(0, z)$, completing the proof.

  • 3
    $\begingroup$ Fantastic, +1, I always knew there was a generalization and an easier proof!! $\endgroup$ May 14 '20 at 16:44

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