# Problem on continuously differentiable function on (0, ∞)

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.

• Looks good to me. It is essentially math.stackexchange.com/q/81003/42969, applied to the function $g$. May 4, 2022 at 9:10
• 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 May 5, 2022 at 16:46

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.