In which case these integrals are equal?

Question

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.

Answer 1

$\newcommand{\d}[1]{\, \mathrm{d}#1}$ 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.

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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 $f(x) \equiv 0$.

Answer 2

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 $0$.