# $f$ has a second derivative $f'' < 0$ $\implies$ $f$ has a decreasing first derivative $\implies$ $\frac{f(x)}{x}$ is decreasing for $x > 0$.

I am currently working on the following problem and had a few questions regarding my work:

if $$f$$ has a second derivative $$f'' < 0$$ $$\implies$$ $$f$$ has a decreasing first derivative. Show this implies that $$\frac{f(x)}{x}$$ is decreasing for $$x > 0$$.

My Work Thus Far:

Let $$g(x) = \frac{f(x)}{x}$$, for $$x >0$$. According to the question we have that $$f''<0$$ then $$f$$ has a decreasing first derivative, meaning $$f'$$ is decreasing. Now, taking the derivative of $$g$$ yields: $$g'(x) = \frac{xf'(x) - f(x)}{x^{2}}$$ $$=$$ $$\frac{f'(x)}{x} - \frac{f(x)}{x^{2}}$$, where $$x >0$$.

I know that $$- \frac{f(x)}{x^{2}}$$ is decreasing, but how do I know for sure that $$\frac{f'(x)}{x}$$ is decreasing so I can deduce that $$g'(x) <0$$, $$\forall x >0$$ and that $$g(x)$$ is a decreasing function?

Remark: I think what has really confused me is why can we say $$\frac{f'(x)}{x}$$ is decreasing? Is this simply because $$f'(x)$$ is decreaing? If this is the case, why? $$f'(x)$$ and $$\frac{f'(x)}{x}$$ are two different functions.

For the sake of context, this question was pulled from the problem: let $$f: [0, \infty) \to [0, \infty)$$ be increasing and satisfy $$f(0) = 0$$ and $$f(x) > 0$$ $$\forall x >0$$. If $$f$$ also satisfies $$f(x+y) \leq f(x) + f(y)$$ $$\forall x,y \geq 0$$, then $$f \circ d$$ is a metric whenever $$d$$ is metric. Show each of the following conditions is sufficient to ensure that $$f(x+y) \leq f(x) +f(y)$$ $$\forall x,y \geq 0$$:

$$a)$$ $$f$$ has a second derivative satisfying $$f'' \leq 0$$;

$$b)$$ $$f$$ has a decreasing first derivative.

$$c)$$ $$\frac{f(x)}{x}$$ is decreasing for $$x > 0$$.

To prove this claim, I figured out that it would be easier to show that $$a)$$ $$\implies$$ $$b)$$ $$\implies$$ $$c)$$; hence where my question arose.

• Counterexample: $f(x) = -(x-2)^2$. Then $f(1)/1 = -1$ and $f(2)/2=0$. Aug 23 '20 at 2:02
• @angryavian Indeed. I really dislike posts which ask one to prove something that is false. (+1) Aug 23 '20 at 2:05
• @MarkViola - See my most recent edit to my question. Aug 23 '20 at 2:23

Geometric intuition: $$f(x)/x$$ is the slope of the line connecting $$(x,f(x))$$ to the origin. This leads us to a simple counterexample for the original claim: with $$f(x) = -(x-2)^2$$ we have $$f(1)/1 = -1$$ and $$f(2)/2 = 0$$.

However, the claim is true with the added condition $$f(0)=0$$. The concavity of $$f$$ (given by $$f'' < 0$$) then implies that this slope decreases as $$x$$ increases.

Let $$0. If we show $$\frac{f(x)}{x} \ge \frac{f(y)-f(x)}{y-x},\tag{*}$$ then we have $$f(y) = f(x) + \frac{f(y)-f(x)}{y-x} (y-x) \le f(x) + \frac{f(x)}{x} (y-x) = y \frac{f(x)}{x}$$ which is what we want. To prove ($$*$$), note that the mean value theorem implies $$\frac{f(x)}{x} = \frac{f(x)-f(0)}{x-0} = f'(a)$$ for some $$0 \le a \le x$$ and $$\frac{f(y)-f(x)}{y-x} = f'(b)$$ for some $$x \le b \le y$$. Using the fact that $$f'$$ is decreasing, we have $$f'(a) \ge f'(x) \ge f'(b)$$ which proves ($$*$$).

• It should be sufficient to require that $f(0) \ge 0$. Aug 23 '20 at 2:39
• (+1) Nicely done Aug 23 '20 at 2:51

As mentioned above this is true for $$f(0)\ge 0$$ Assuming this condition:

now continuing from your attempt $$g'(x)=\frac{xf'(x)-f(x)}{x^2}$$

let us consider $$h(x)=xf'(x)-f(x)$$

$$h'(x)=xf''(x)<0$$ or $$h(x)$$ is decreasing for all positive x

Also $$h(0)<0$$ using $$f(0)\ge 0$$ or $$h(x)<0$$ for all positive x.

this implies $$g'(x)=\frac{h(x)}{x^2}<0$$ for all positive x

• Minor typo: $h'(x) = x f''(x)$ Aug 23 '20 at 18:02
• @angryavian thank you fixed it Aug 23 '20 at 18:04