# How to find if this function is always zero?

The question I faced was:

Let $f(x)$ be a non-negative continuous and bounded function for all $x \ge 0$. If
$$(\cos x)f'(x) \le (\sin x - \cos x)f(x), \; \forall \; x \ge 0$$ then which of the following is/are correct?

(A) $f(6) + f(5) > 0$

(B) $x^2 - 3x + 2 + f(9) = 0$ has two distinct solutions

(C) $f(5)f(7) - f(6)f(5) = 0$

(D) $\lim\limits_{x \to 4} \dfrac{f(x) - \sin(\pi x)}{x-4} = 1$

(B), (C)

By observation, $f(x)=0$ satisfies the given conditions. But is it the only solution? If so, how to prove it is?

After rearranging the terms and combining them, I converted the inequality to this form: $$\left( f(x)\,\cos x \right)' + f(x)\,\cos x \le 0$$ Despite its allure, this inequality isn't getting me anywhere! It doesn't seem to have any information about $f(x)$, since it is stuck with a "$\cos x$". Even then, I don't see where I can go with it.

So how to solve this problem? Thank you.

• Your condition is equivalent to $$\frac{\mathrm{d}}{\mathrm{d}x} \ f(x) \cos x \ e^x \le 0$$ – Crostul Apr 28 '17 at 10:56
• @BST Actually the discriminant is $(-3)^2-4(2+f(9)) = 1-4f(9)$ which may be negative – Crostul Apr 28 '17 at 11:00
• Sorry, you're right. My mistake. I'll remove the comment. I miscalculated the discriminant. – be5tan Apr 28 '17 at 11:02

We assume that $f(x)$ is bounded and non-negative, for $x \geqslant 0$. This means that there is a non-negative function $g(x)$ with the property that $f(x) = g(x)e^{-x}$ for $x \geqslant 0$.

Plugging this into the inequality you found gives, \begin{equation} 0 \geqslant (f(x)\cos (x))' + f(x) \cos (x) = e^{-x} (g(x) \cos(x))'. \end{equation} And since $e^{-x} > 0$, we have \begin{equation} 0 \geqslant (g(x) \cos(x))'. \end{equation} This means that the function $g(x) \cos(x)$ is weakly decreasing. Because any point $x \in \mathbb{R}$ is between two zeroes of $\cos(x)$, we have that $g(x) \cos(x) = 0$ for all $x$.

• Instead of integrating the inequality, it gives more information simply to note that the derivative of $g(x)\cos x$ can never be positive. Since every $x$ is between two zeroes of $g(x)\cos x$, the mean value theorem forces it to be $0$ everywhere, not just where the cosine is positive. – Henning Makholm Apr 28 '17 at 11:07
• Ah, I missed that! Thank you, will edit my answer. – Peter Apr 28 '17 at 11:08
• Also, I think it would be slicker simply to define $g(x)=f(x)e^x$ at the beginning. Then we don't need to appeal to "bounded and nonnegative", and the first inequality is still true. – Henning Makholm Apr 28 '17 at 11:22
• This is brilliant! How did you figure out the substitution $f=ge^{-x}$? – FreezingFire Apr 28 '17 at 11:28
• I asked myself what function $h(x)$ would satisfy the strict version of your bound, i.e.~$h'(x) + h(x) = 0$. Evidently the function $h(x) = e^{-x}$ solves this equation. The rest was the result of a bit of experimentation with this function. – Peter Apr 28 '17 at 11:56

If you set $g(x)=f(x)\cos x$, your rearrangement tells you that $$\tag{1} g'(x) \le -g(x)$$

Consider an interval $[2\pi k-\frac12\pi , 2\pi k +\frac12\pi]$ where $\cos x$ is $\ge 0$. Then you know that $g(x)\ge 0$ on this interval, and $g(x)=0$ at the start of the interval. Then, because of (1), $g(x)$ must be identically zero on that interval, and so must $f(x)$.

This settles at least (A) and (C), because $5$, $6$ and $7$ are all within $\pi/2$ of $2\pi$.

For (B) and (D) this argument doesn't tell you enough; I recommend Peter's slicker answer instead.

Notice that whenever some $f$ satisfies the hypothesis, then so does any positive multiple of $f$. In particular:

Note that (B) is equivalent to $$0<(-3)^2-4(2+f(9)) = 1-4f(9)$$ i.e. $f(9)<1/4$. Hence (B) can hold if and only if $f(9) =0$ for any $f$ satisfying the hypothesis.