# How to show $f$ is non-negative.

I am stuck in the following problem. Can someone show me how shall I finish the track ?

the problem is : if $f:\mathbb{R}\rightarrow \mathbb{R}$ twice differentiable function and $f(0)=f'(0)=1$ and $f''\geqslant f$. Then show that $f$ is non-negative.

Thank you in advance.

• do you have an intuitive idea of what the second derivative actually means? Mar 27, 2016 at 14:03
• As an example you have the function $f(x)=e^x$ which fulfills all the three conditions. Maybe you can figure out an explanation by playing with this very simple example Mar 27, 2016 at 14:05
• Use mean value theorem? Mar 27, 2016 at 14:06
• Hint: If both f' and f'' are > 0, then f(x) must be strictly increasing. Mar 27, 2016 at 14:11
• A start: if not, there is a smallest $c>0$ with $f(c)=0$. Mar 27, 2016 at 14:18

## 2 Answers

Let $$g(x)=(f'(x)-f(x))e^x,$$ then $g'(x)=(f''(x)-f(x))e^x\geq 0,$ and we know that $g(0)=(f'(0)-f(0))e^0=0,$ thus $g(x)\begin{cases} \leq 0,x\leq 0;\\ \geq 0,x>0. \end{cases}$

Let $$h(x)=f(x)e^{-x},$$ then $h'(x)=(f'(x)-f(x))e^{-x}=g(x)e^{-2x}.$ Thus $h(x)\geq h(0)=1,$and $f(x)=h(x)e^x\geq e^x\geq 0$.

EDIT

The answer below does not fully answer the question.

HINT

Let us first prove this for $x \ge 0$.

If there exists such $x \ge 0$ that $f(x)<0$, we get that there exists such a $c$ that $f(c)=0$.

Let $c_1$ be the smallest such $c$.

Then, note that from MVT there exists such $d \in (0,c_1)$ that $$f'(d)=\frac{f(c_1)-f(0)}{c_1}=-\frac{1}{c_1}<0$$

Thus, from MVT there exists such $e \in (0,d)$ that $$f''(e)=\frac{f'(d)-f(0)}{d}<-\frac{1}{d}<0$$

Thus $0>f''(e)>f(e)>0$. A contradiction!

POSTSCRIPT

Thanks to @DavidMitra for giving the idea.