# Show that $f(0)>0,\; f'(0)=0,\; f''(x)<0$ imply that $f(x)=0$ has exactly 1 positive root.

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?

• 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. – Patrick Stevens Nov 27 '18 at 6:47

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, $$f'(x) for every $$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) 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.

• I'm struggling with the part in the proof where you say $f(x)$ is eventually negative. Why? – Jan Nov 27 '18 at 7:34
• @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. – William McGonagall Nov 27 '18 at 7:38

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?

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 $$[0,\infty)$$.

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.