How to show that $f(x)-x \times df(x)/dx ≥ 0$ when f(x) is concave?

Question

How do you intuitively (perhaps even graphically) show that $f(x)-x {{df(x)}\over{dx}} ≥ 0$ when the function $f(x)$ is concave, e.g. when ${{df(x)}\over{dx}}>0$ and ${{d^2f(x)}\over{dx^2}} < 0$?

Under the conditions that $x ≥ 0$.

Answer 1

Following up on another answer from Did, the inequality is true for $x \geq 0$ if you additionally assume $f(0) \geq 0$. And you really do need that or else the inequality fails when you plug in $x=0$ because for $x=0$ the inequality reduces to $f(0) \geq 0$.

As for the intuition, there is an intuitive explanation at least under the assumption that $f(0) \geq 0$, per the above. A concave function has tangent lines that are always above the graph of the function. So if you take the tangent line with slope $df/dx$ that passes through the point $(x,f(x))$, then $f(0)$ must be below the intercept of the line. This says that in fact $f(x) - x\frac{df}{dx} \geq f(0)$, (which is always true for concave $f$ even if $f$ is sometimes negative), so if $f(0) \geq 0$ then you get your inequality.

Here is a visual example of the argument. Its made in OSX Grapher, and using the concave function $f(x)=ln(x+1)$. The OSX Grapher file can be found here .

Answer 2

This cannot be true. Take a concave function $g$ and consider $f=g-c$ for some large constant $c$. Then $f(x)-xf'(x)=g(x)-xg'(x)-c$ is negative for at least some values of $x$, if $c$ is large enough. What is true however is that if $f(x)-xf'(x)\geqslant0$ for some $x\geqslant0$ and if $f''\lt0$, then $(f(t)-tf'(t))'=-tf''(t)\gt0$ for every $t\gt x$ hence $f(t)-tf'(t)\gt 0$ for every $t$ in $(x,\infty)$.