Strictly decreasing and concave down takes on negative values

Question

Visually it seems that a strictly decreasing twice-differentiable map $f: [0, \infty) \to \mathbb{R}$ with $f''(x) < 0$ for all $x$ should be negative at some point, but I can't come up with a formal proof.

Any help?

Answer 1

As somewhat of an elementary proof, pick any point $x_0\in[0,\infty)$. Note that $f'(x_0)<0$. Apply a proof by contradiction with the standard Calc I MVT to conclude that for all $x>x_0$ we have $f'(x)<f'(x_0)$. Hence, $f\left(x_0-\frac{f(x_0)}{f'(x_0)}\right)$ must be negative (again by the MVT).

Answer 2

Since $f(x)$ is twice differentiable, $f'(x)$ is (at least) once differentiable, hence continuous.

Since $f(x)$ is strictly decreasing, we have

$\forall x_1, x_2 \in [0, \infty) \; x_2 > x_1 \Longrightarrow f(x_2) < f(x_1); \tag 1$

since $f'(x)$ is continuous, it is integrable, and thus for $x_1 < x_2$,

$f(x_2) - f(x_1) = \displaystyle \int_{x_1}^{x_2} f'(s) \; ds; \tag 2$

I claim

$\exists \xi \in (x_1, x_2), \; f'(\xi) < 0; \tag 3$

if not,

$\forall \xi \in (x_1, x_2), \; f'(\xi) \ge 0; \tag 4$

then using (2),

$f(x_2) - f(x_1) = \displaystyle \int_{x_1}^{x_2} f'(s) \; ds \ge 0, \tag 5$

contradicting (1); thus

$\exists \xi \in [0, \infty), \; f'(\xi) < 0; \tag 6$

we have

$\forall x > \xi, \; f'(x) < f'(\xi); \tag 7$

if not,

$\exists x > \xi, \; f'(x) \ge f'(\xi); \tag 8$

then by the mean value theorem,

$\exists \xi', \;\xi' \in (\xi, x), \; 0 \le f'(x) - f'(\xi) = f''(\xi')(x - \xi) \Longrightarrow f''(\xi') \ge 0, \tag 9$

contradicting our hypothesis that $f''(x) < 0$ everywhere; thus (7) binds. Therefore, with $x > \xi$ we have

$f(x) - f(\xi) = \displaystyle \int_\xi^x f'(s) \; ds < \int_\xi^x f'(\xi) \; ds = f'(\xi)(x - \xi), \tag{10}$

or

$f(x) < f(\xi) + f'(\xi)(x - \xi) < 0 \tag{11}$

for $x$ sufficiently large.

Answer 3

$f'$ is strictly decreasing because its derivative $f''$ is negative. So $f'(y)<f'(0)<0$ when $y>0.$ So for $x>0$ we have $$f(x)-f(0)=\int_0^x f'(y)dy<\int_0^xf'(0)dy=xf'(0).$$ So $x>\frac {|f(0)|}{|f'(0)|}\implies f(x)<0$.