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Let $f\in C^1([0,\infty[$ such that $f(0)=0$ and $\forall x \in R^+_0$, $0\le f'(x)\le1$. Prove that $$\bigg(\int_0^x f(t) dt\bigg)^2\le\int_0^x f^3(t)dt$$ and find the cases that are equal.

I already prove inequality but I can't find the cases for the equality.

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  • $\begingroup$ Do you need all the cases? If you need just one we have $f(x) = 0$. $\endgroup$ Commented Jan 10, 2020 at 2:33
  • $\begingroup$ no, i need all cases when $\bigg(\int_0^x f(t) dt\bigg)^2=\int_0^x f^3(t) dt$ $\endgroup$ Commented Jan 10, 2020 at 2:37
  • $\begingroup$ Did you derive two times this equation of something like that ? I think fundamental theorem of calculus should leads to an ODE. $\endgroup$ Commented Jan 10, 2020 at 2:58

2 Answers 2

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$\newcommand{\d}[1]{\, \mathrm{d}#1}$ To solve for the following equality: $$ \left(\int_0^x f(t) \d{t}\right)^2 = \int_0^x f^3(t) \d{t} $$ We invoke Fundamental Theorem of Calculus and differentiate both sides: $$ 2f(x)\int_0^x f(t) \d{t} = f^3(x) $$ Let $F(x) = \int_0^x f(t) \d{t}$. Note that $F'(x) = f(x)$. If $f(x) \not\equiv 0$, this is equivalent to solving the differential equation $(F'(x))^2 = 2F(x)$. Differentiating again yields: \begin{align*} 2F'(x)F''(x) = 2F'(x) &\implies f'(x)f(x) = f(x) \\ &\;\color{red}{\implies f'(x) = 1} \\ &\implies f(x) = x + C \end{align*} Therefore, the only possible cases are $f(x) = 0$ or $f(x) = x + C$ for some $C \in \mathbb{R}$. Since $f(0) = 0$ we must have $f(x) = x$. I'll leave you to check that these two are indeed solutions to the equality.


EDIT: As @ΑΘΩ has pointed out, more justification is required to show that $f'(x) = 1$ for all $x \in [0, \infty)$, as my argument only proves that $f'(x) = 1$ whenever $f(x) \neq 0$ (the implication in red). This can be fixed by showing that if $f(x) = 0$ for some $x > 0$, then $f(x) \equiv 0$ on $[0, \infty)$. Note that $f(0) = 0$ and $f'(x) \geq 0$ $\forall x \geq 0$ implies $f(x) \geq 0$ $\forall x \geq 0$.

We prove this by contradiction. Suppose $f(x) = 0$ for some $x > 0$ but $f(x) \not\equiv 0$. Let: $$ x_0 = \sup\{x \in [0,\infty) \mid f(x) = 0\} $$ We consider two cases. If $x_0 = +\infty$, then there exists $x_1 < x_2$ such that $f(x_1) = y_1 > 0$ but $f(x_2) = 0$. By MVT, there exists $c \in (x_1, x_2)$ such that $f'(c) = \frac{y_1}{x_1 - x_2} < 0$, contradicting that $f'(x) \geq 0$ for all $x \in [0, \infty)$.

If $x_0 < +\infty$, then we have $f(x) \neq 0$ for $x > x_0$. This means that $f'(x) = 1$ for $x > x_0$, so $\lim_{x \to x_0^+} f'(x) = 1$. Since $f \in C^1[0,\infty) \implies f'$ is continuous, we have $f'(x_0) = 1$. If $f(x_0) = 0$, then there exists $x' < x_0$ sufficiently close to $x$ such that $f(x') < 0$, contradicting $f(x) \geq 0$ so we must have $f(x_0) = y_0 > 0$. However, by continuity of $f$, we have for some $\delta > 0$, $x \in (x_0 - \delta, x_0 + \delta) \implies f(x_0) > \frac{y_0}{2}$, this time contradicting the supremum property of $x_0$. This concludes that $f(x) \equiv 0$.

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  • $\begingroup$ oh amazing, thank you so much! $\endgroup$ Commented Jan 10, 2020 at 3:08
  • $\begingroup$ @Clement Yung: that still leaves one with the question of how to handle the zeros of $f$. $\endgroup$
    – ΑΘΩ
    Commented Jan 10, 2020 at 3:36
  • $\begingroup$ @AΘΩ what do you mean by "handle to zeroes of $f$"? $\endgroup$ Commented Jan 10, 2020 at 4:15
  • $\begingroup$ @Clement Yung: apologies for the delay in replying. By that phrase I am referring to the fact that you have produced an answer by making additional assumptions (the ''if $f(x) \neq 0$'' part), which do not in general occur as consequences of the original hypothesis of the problem. In other words, you have only solved the problem partially, only in the case that you have granted the fulfillment of the additional assumptions you are making (in your case, that of function $f$ having no zeros). This leaves the general situation -- the one in which $f$ does have zeros -- unanswered. $\endgroup$
    – ΑΘΩ
    Commented Jan 12, 2020 at 13:40
  • $\begingroup$ @ΑΘΩ I was assuming that $f$ is not identical to the zero function, not assuming that $f$ is nowhere zero. To make it clearer, I've edited my answer by changing $f(x) \neq 0$ to $f(x) \not\equiv 0$. $\endgroup$ Commented Jan 12, 2020 at 13:45
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Just for the sake of being fully rigorous, allow me to present one possible method of analysing what happens with the zeros of function $f$.

First of all, as by hypothesis the derivative of $f$ is positive (may I explicitly note that to me $x$ is positive means $x \geqslant 0$ whereas $x>0$ is to be read as $x$ is strictly positive), the Mean Value Theorem ensures the fact that $f$ is increasing.

Recall the notion of convex subset in general order theory: if $(A, R)$ is an arbitrary ordered set and $M \subseteq A$ an arbitrary subset, we say $M$ is convex (with respect to $R$) if

$$(\forall x, y)(x, y \in M \wedge x \leqslant_R y \Rightarrow [x, y]_R \subseteq M)$$

in other words when $M$ contains together with two comparable elements $x ,y$ the whole closed interval (with respect to $R$) determined by them.

It is clear that:

for any increasing map $f: A \to B$ between ordered sets $(A,R), (B,S)$ and any convex subset $N \subseteq B$ the inverse image $f^{-1}(N)$ is convex in $A$ (with respect to $R$).

Recall also that:

if $(A, R)$ is totally ordered by a conditionally complete order (which means that under this order any upper-bounded nonempty subset admits a supremum), then the convex subsets of $A$ are precisely the intervals with respect to $R$.

Since $f$ is increasing, the singleton $\{0\}$ is obviously convex and both $\mathbb{R}$ and $[0, \infty)$ are equipped with total, conditionally complete orders we gather on the one hand that the set of zeros $F:=f^{-1}(\{0\})$ is an interval and on the other hand that it is closed (by virtue of the continuity of $f$). Let us also not forget that $0 \in F$, by hypothesis.

There are only two types of closed intervals included in $[0, \infty)$ that contain $0$:

  • $[0, \infty)$ itself, which corresponds to the trivial case when $f$ is identically null
  • those of the form $[0, a]$ for a real number $a \geqslant 0$. In the context of the given problem, let us argue by contradiction that this case forces $a=0$. Indeed, assuming $a>0$ we gather that $f^{-1}(\mathbb{R}^*)=(a, \infty)$ and to this unique connected component of the set where $f$ does not vanish the reasoning that Clement Yung presented to us above does apply with the conclusion that $f(x)=x+b$ for a certain $b \in \mathbb{R}$ and every $x>a$. However, this means that $f$ fails to be derivable in $a$, since at this point the left derivative is $0$ (this is where the hypothesis $a>0$ comes into play, since for the interval $[0, \infty)$ $a$ is a point of accumulation on the left) whereas the right derivative is $1$.

Thus, it is indeed the case that under the given assumptions $f$ is either identically null or vanishes only at $0$.

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