Q. Let $f$ be continuously differentiable on $(0, \infty)$ and let $f(0)=1$. Show that if $|f(x)| \leq e^{-x}$ for $x \geq 0$, then there is $x_{0}>0$ such that $f^{\prime}\left(x_{0}\right)=-e^{-x_{0}}$.

I think

Setting $g(x)=f(x)-e^{-x}, x \geq 0$. Then $g(0)=0, g(x) \leq 0$ and $\lim _{x \rightarrow \infty} g(x)=0$. If $g(x) \equiv 0$, then $f^{\prime}(x)=-e^{-x}$ for $x \in(0, \infty)$. So, suppose that there is $a>0$ such that $g(a)<0$. Then for sufficiently large $x$, say $x>M$, we have $g(x)>\frac{1}{2} g(a)$. Consequently, $g$ attains its minimum value at some $x_{0}$ in $(0, M )$. Thus $g^{\prime}\left(x_{0}\right)=0$.

Is this okay? Also give me another approach.

  • $\begingroup$ Looks good to me. It is essentially math.stackexchange.com/q/81003/42969, applied to the function $g$. $\endgroup$
    – Martin R
    May 4, 2022 at 9:10
  • $\begingroup$ Haven't thought this through, but since you want other approaches as well, maybe you could treat it as a fixed point problem. So $f'(x_0)=-e^{-x_0}$ is equivalent to $-\ln(-f'(x_0))=x_0$, and so define $g:(0,\infty)\to\mathbb{R}$ by $g(x)=-\ln(-f'(x))$ and show that $g$ has a fixed point somehow. No idea if it works, but if you want to you could give it a try $\endgroup$
    – Lorago
    May 5, 2022 at 16:46

1 Answer 1


Let $$f(x) = \begin{cases} 1, &x=0 \\ 0, &x>0 \end{cases}$$ then

  • $f$ is continuous and differentiable on $(0,\infty)$, and
  • $f(0)=1$, and
  • $|f(x)| \leqslant e^{-x}$ for all $x\geqslant 0$.

But $f'(x) = 0 \neq -e^{-x}$ for any $x>0$, thus there exists no $x_0$ as claimed.


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