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.

  • $\begingroup$ Read the question and think about what it is asking? $\endgroup$
    – H. Gutsche
    Mar 16 '18 at 0:02
  • $\begingroup$ What is it that you mean? $\endgroup$ Mar 16 '18 at 0:36
  • $\begingroup$ He means the question is asking you to verify if $f(x)=0$ satisfies the conditions set out in the problem! $\endgroup$ Mar 16 '18 at 1:39
  • $\begingroup$ 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". $\endgroup$
    – Paramanand Singh
    Mar 16 '18 at 1:43
  • 2
    $\begingroup$ Consider the function $f(x) =x^4$ for $x<0$ and zero otherwise. $\endgroup$
    – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.