Proving an Inequality - Functions

Question

Here is a question I came across:

Consider two functions $p(x)$ and $q(x)$ such that $p(x)$ is differentiable and $q(x)$ is twice differentiable, and $|p(x)| ≤ 1$, $q(x) = p'(x)$.

If $p²(0) + q²(0) = 9$, then show that there exists some $\alpha$ such that $q(\alpha)q''(\alpha) < 0$.

I could verify the result by assuming a function such as $p(x) = \sin3x$, but not in general. Could someone please help me with how to go about the general proof?

My thoughts:

We could assume four cases,

1. $q(\alpha) > 0$, $q''(\alpha) > 0$
2. $q(\alpha) < 0$, $q''(\alpha) < 0$
3. $q(\alpha) > 0$, $q''(\alpha) < 0$
4. $q(\alpha) < 0$, $q''(\alpha) > 0$

Now, 3 and 4 satisfy the required condition so let's not bother about them. Proving that 1 and 2 cannot happen for any given $\alpha$ is not what is asked but if we are able to show it by some sort of contradiction, then our job is done.

I could not proceed from here, for some general $p(x)$. As for selecting a particular function and verifying the result, I have done that.

Please help me with how to get forward with this method, or suggest a new method. A solution or explanation would be great, thanks!

Answer 1

$\def\d{\mathrm{d}}$To prove by contradiction, suppose $q(x) q''(x) \geqslant 0$ for all $x \in \mathbb{R}$, then$$ (q(x) q'(x))' = (q'(x))^2 + q(x) q''(x) \geqslant 0. \quad \forall x \in \mathbb{R} $$ First, if there exists $x_0 \in \mathbb{R}$ such that $q(x_0) q'(x_0) > 0$, without loss of generality, assume $q(x_0), q'(x_0) > 0$. Note that for any $x > x_0$,$$ q(x) q'(x) \geqslant q(x_0) q'(x_0) > 0. \tag{1} $$ If there exists $x_1 > x_0$ such that $q(x_1) \leqslant 0$, by continuity of $q$, there exists $x_0 < x_2 \leqslant x_1$ such that $q(x_2) = 0$, contradictory to (1). Thus for any $x > x_0$, there is $q(x) > 0$, and by (1) there is $q'(x) > 0$. Therefore, $q(x) \geqslant q(x_0) > 0$ for any $x \geqslant x_0$, and$$ p(x_0 + x) = p(x_0) + \int_{x_0}^{x_0 + x} q(t) \,\d t \geqslant p(0) + q(x_0) x, \quad \forall x \geqslant 0 $$ which is contradictory to $|p(x)| \leqslant 1$ ($\forall x \in \mathbb{R}$). Hence$$ q(x) q'(x) \leqslant 0. \quad \forall x \in \mathbb{R} $$

Next, if there exists $x_0 \in \mathbb{R}$ such that $q(x_0) q'(x_0) < 0$, without loss of generality, assume $q(x_0) > 0 > q'(x_0)$. Note that for any $x < x_0$,$$ q(x) q'(x) \leqslant q(x_0) q'(x_0) < 0. $$ Analogous to previous deduction, $q(x) > 0$ for any $x \leqslant x_0$, which implies $q'(x) < 0$ for any $x \leqslant x_0$, then $q(x) \geqslant q(x_0) > 0$ for any $x \leqslant x_0$ and$$ p(x_0 - x) = p(x_0) - \int_{x_0 - x}^{x_0} q(t) \,\d t \leqslant p(x_0) - q(x_0) x, \quad \forall x \geqslant 0 $$ which again is contradictory to $|p(x)| \leqslant 1$ ($\forall x \in \mathbb{R}$). Hence$$ q(x) q'(x) \geqslant 0, \quad \forall x \in \mathbb{R} $$ which implies $q(x) q'(x) \equiv 0$. Therefore, for any $x \in \mathbb{R}$,$$ (q(x))^2 = (q(0))^2 + 2 \int_0^x q(t) q'(t) \,\d t = (q(0))^2. $$ Because $q$ is differentiable and $|q(x)| \equiv |q(0)|$, then $q(x) \equiv q(0)$.

Now, because $1 + (q(0))^2 \geqslant (p(0))^2 + (q(0))^2 = 9$, then $q(0) \neq 0$. Thus$$ p(x) = p(0) + \int_0^x q(t) \,\d t = p(0) + q(0) x, \quad \forall x \in \mathbb{R} $$ which is contradictory to $|p(x)| \leqslant 1$ ($\forall x \in \mathbb{R}$).

Answer 2

Complements:

First, let us define some boundary $I = [a, b]$ with $a < b$, on which $p, q$ are defined, $0 \in I$. I'll assume that $p, q : I \to \mathbb{R}$.

Let us assume that $I \neq [0, 0] = \{ 0 \}$, otherwise the inequality cannot happen (take: $p : x \mapsto 1$ and $q : x \mapsto 2\sqrt{2}$ defined on $\{0\}$, there is no such $\alpha$.)

Now, because: $\forall x \in I, q(x) = p'(x)$, as $q$ is double differentiable, then $p'$ is double differentiable, then $p$ is triple differentiable.

So, at least: $p$ and $q$ are continuously differentiable.

If you were to proceed with contradiction, you have to assume:

(*) : $\forall \alpha \in I, q(\alpha)q''(\alpha) \geq 0$.

Let us rule out the case where: $\forall \alpha \in I, q(\alpha)q''(\alpha) = 0$.

That is, when $qq'' = 0$ over $I$, i.e. either $q = 0$ either $q'' = 0$ over $I$.

That is, either $q$ is the null function, either $q$ is a polynomial of degree $d \leq 1$.

If $q$ is a polynomial of degree $d \leq 1$ (the null function is a particular case of this type of function), i.e. $\forall x \in I, q(x) = \alpha x + \beta, (\alpha, \beta) \in \mathbb{R}^2$.

As we have the relation between $q$ and $p$, we deduce $p$ to be a polynomial of degree $d' \leq 2$: $p'(x) = \alpha x + \beta$ gives us: $p(x) = \dfrac{\alpha}{2} x^2 + \beta x + \gamma, \gamma \in \mathbb{R}$.

Then: $\lvert p(x) \rvert \leq 1$ gives us: $\Big \lvert \dfrac{\beta^2}{2\alpha} + \gamma \Big \rvert \leq 1$.

Also: $p(0)^2 + q(0)^2 = \gamma^2 + \beta^2 = 9$.

So: $\beta^2 = 9 - \gamma^2$.

Now, let us fix $\alpha$, we must solve: $\lvert 9 - \gamma^2 + 2\alpha \gamma \rvert \leq 1$ with respect to $\gamma$.

Again, that is: $\lvert 9 + \alpha^2 \rvert \leq 2\alpha$.

That is: $\alpha^2 - 2\alpha + 9 \leq 0$, which is impossible for all values of $\alpha \in \mathbb{R}$, because $\alpha^2 - 2\alpha + 9 = (\alpha - 1)^2 + 8 \geq 8 > 0$.

Such $\alpha$ does not exist, then such polynomial does not exist.

Now, we know that $(*)$ is false under our conditions, so there is at least $x_0 \in I$ such that $q(x_0)q''(x_0) \neq 0$ for all $q$ which verifies our hypotheses.

The rest of the proof is already given in another answer. (I can remove it, if it's not enough.)