5
$\begingroup$

Let $f$ be a function twice differentiable on $\mathbb{R}$. Suppose that $f(0)>0, \; f'(0)=0$ and $\; f''(x)<0$ for all $x>0$. Prove that $f(x)=0$ has exactly one positive root.

We have $f''(x)<0 \implies f'(x)$ is strictly decreasing on $(0,\infty)$. Then $f'(0)=0$, so $f'(x)<0$ on $(0,\infty)$. So $f$ has at most 1 positive root.

$f'(x)<0 \implies f(x)$ is strictly decreasing on $(0,\infty)$.

How do I make use of the fact that $f(0)>0$ to show $f(x)=0$ has at least 1 positive root?

$\endgroup$
1
  • 1
    $\begingroup$ You're going to have to use $f''(x) < 0$ more strongly, because merely being strictly decreasing is not enough: $f(0) > 0$ and $f'(0) = 0$ and "$f$ strictly decreasing" does not imply that $f$ has a positive root. $\endgroup$ – Patrick Stevens Nov 27 '18 at 6:47
4
$\begingroup$

The condition $f''<0$ implies that $f'$ is strictly decreasing. Since $f'(0)=0$, $f'(a)$ must be negative for some positive $a$ that is close to zero. But then, as $f'$ is strictly decreasing, $f'(x)<f'(a)$ for every $x\ge a$. By mean value theorem, when $x>a$, $f(x)-f(a)=f'(c)(x-a)$ for some $c\in(x,a)$. Hence $f(x)<f(a)+f'(a)(x-a)$ whenever $x>a$, meaning that $f(x)$ is eventually negative when $x\to+\infty$. Hence $f(b)<0$ for some $b>a$. Now $f(0)>0>f(b)$. By intermediate value theorem, $f(x)=0$ has a positive root. As $f$ is strictly decreasing, this root is unique.

$\endgroup$
2
  • $\begingroup$ I'm struggling with the part in the proof where you say $f(x)$ is eventually negative. Why? $\endgroup$ – Jan Nov 27 '18 at 7:34
  • $\begingroup$ @Jan $f'(a)$ is negative. When $x-a>\frac{f(a)}{-f'(a)}$, $f(a)+f'(a)(x-a)$ would be negative and hence $f(x)<f(a)+f'(a)(x-a)$ is negative too. $\endgroup$ – William McGonagall Nov 27 '18 at 7:38
3
$\begingroup$

Hint. Since $f$ is strictly concave in $(0,+\infty)$, we have that for $x_0,x>0$, $$f(x)\leq f'(x_0)(x-x_0)+ f(x_0).$$ that is the graph of $f$ stays under its tangent at $x_0$. Note that here $f'(x_0)<0$. Take the limit as $x\to +\infty$. What may we conclude?

$\endgroup$
0
$\begingroup$

You may show this also as follows:

  • Uniqueness: Assume $x_1,x_2 > 0$ with $f(x_1) = f(x_2) = 0$ and $x_1 \neq x_2$ $$\Rightarrow \exists \xi > 0: f'(\xi) = 0 \mbox{ contradiction to } f' \mbox{ strictly decreasing and } f'(0) = 0.$$
  • Existence: Consider $x \geq 1$ and use MVT for continuous functions: $$f(x) = f(1) + \int_{1}^x f'(t)\;dt \stackrel{f'(t) \leq f'(1) <0}{\leq} f(1) + (x-1)\cdot \underbrace{f'(1)}_{<0} \color{blue}{<0} \mbox{ for } x \mbox{ large enough.}$$ So, $f$ changes the sign on $[0,\infty)$.
$\endgroup$
0
$\begingroup$

Since the second derivative $f''$ is negative in $(0,\infty) $, the first derivative $f' $ is strictly decreasing in $[0,\infty)$. And we have $f'(0)=0$ so that $f'$ is negative in $(0,\infty)$. It thus follows that $f'(x) $ either tends to a negative limit or diverges to $-\infty $ as $x\to\infty$.

The fact that derivative $f'$ is negative leads us to the conclusion that $f$ is strictly decreasing on $[0,\infty)$ and hence $f(x) $ either tends to a limit or diverges to $-\infty $ as $x\to\infty $. If $f(x) \to L$ then by mean value theorem we have $$f(x+1)-f(x)=f'(\xi)$$ for some $\xi\in(x, x+1)$. Now as $x\to\infty$ the LHS of above equation tends to $L-L=0$ and the RHS remains strictly away from $0$ (see last paragraph) which is a contradiction. Thus $f(x) \to-\infty$ as $x\to\infty $. Since $f(0)>0$, it follows by intermediate value theorem that $f(x) $ vanishes at least once in $(0,\infty) $ and since $f$ is strictly monotone it vanishes only once. Thus $f$ has exactly one positive real root.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.