# Twice differentiable function not equal to $0$

This is a question in the undergraduate-level textbook "Advanced Calculus" by Fitzpatrick.

Suppose that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is twice differentiable such that for $\forall x$, $f'(x)\leq f(x)$, and $f(0)=0$. Then is $f$ the zero function?

The answer to this is not true as I was able to find a counter-example $f^*(x)= 1- e^x$. However we have only just learned about differentiation, the mean-value theorem and how to find extremes using 1st and 2nd derivatives, and we have only seen derivatives of polynomials so far, but I don't know how to disprove the above statement by using these.

(EDIT) For $1−e^x$ to be a valid counter-example, I need to "officially know" that the exponential function's derivative is equal to itself. But exponential functions are in the next chapter. Therefore unless I want to "cheat", I need to think of another function.

• Read the question and think about what it is asking? – H. Gutsche Mar 16 '18 at 0:02
• What is it that you mean? – blabla zazoo Mar 16 '18 at 0:36
• He means the question is asking you to verify if $f(x)=0$ satisfies the conditions set out in the problem! – Laars Helenius Mar 16 '18 at 1:39
• Your counter example is correct. You just need to add one more sentence in your answer: "from the counter-example given above it is evident that $f$ is not necessarily the zero function". – Paramanand Singh Mar 16 '18 at 1:43
• Consider the function $f(x) =x^4$ for $x<0$ and zero otherwise. – Jose27 Mar 16 '18 at 5:58

Morally any function that is positive an decreasing on $x<0$ and $0$ otherwise will be a counterexample. To satisfy the smoothness assumption just pick a function that goes to $0$ fast enough at $x=0$; for example $f(x)=x^4$ for $x<0$ and $0$ otherwise will do the trick.