$f'(x) \geq x$, then $f(x) \leq f(0) + \frac12 x^2 ,$ when $x \leq 0$.

Let $f : \Bbb R \to \Bbb R$ is differentiable and $f'(x) \geq x$, then $f(x) \leq f(0) + \frac12 x^2 ,$ when $x \leq 0$.

we have $f(x) = f(0) + f'(0)x + \frac12f''(0)x^2+ \dots < f(0) + \frac12f''(0)x^2$, we can write this since $x \leq 0$.

Also $f'(x) \geq x \implies f''(x) \leq 1$. Hence $f''(0) \leq 1$.

And we have $f(x) \leq f(0) + \frac12 x^2$.

Is the solution correct? If not please correct me.

• "we can write this since" - no, because the signs of derivatives can be arbitrary May 7, 2018 at 11:06
• can you give some hints how to proceed? @Logic_Problem_42 May 7, 2018 at 11:07
• Integrate your inequality May 7, 2018 at 11:16

2 Answers

If the derivative is integrable, just integrate.

$$f(x) = f(0) + \int_0^x f'(t) \, dt \leq f(0) + \int_0^x t \, dt = f(0) + \frac{1}{2}x^2$$

• What if $f'$ is not continuous ? May 7, 2018 at 11:16
• I think there is a problem, but we really only need $f'$ to be integrable.
– user211599
May 7, 2018 at 11:25

Here's a way without any integrability or continuity assumption on $f'$.

Since $f'(x)-x\geq 0$, the derivative of $x\mapsto f(x)-\frac{x^2}2$ is $\geq 0$, so this function increases.

Thus, for $x\leq 0$, $f(x)-\frac{x^2}2 \leq f(0)-0$, that is $f(x)\leq f(0)+\frac{x^2}2$.