Let $f:\mathbb R\rightarrow \mathbb R$ be a differentiable function, and suppose $f=f'$ and $f(0)=1$. Then prove $f(x)\neq 0$ for all $x\in \mathbb R$
The way I solve this is kind of strange.
I first suppose there is a closed interval $[0, a]$ on the real line. Since $f$ is differentiable, there exists a $x_0\in[0, a] $ such that $\frac{f(a)-f(0)}{a}=f'(x_0)=f(x_0)$. Thus when $f(a)=1$, $f(x_0)=0$. Then I just let $f(a)=1$ and try to find some contradictions. Since $f(a)=f(0)=1$, by Rolle thorem, there should exist a $x_1\in[0, a]$ such that $f'(x_1)=f(x_1)=0$ and this $f(x_1)=0$ is supposed to be the maximum or minimum on the interval $[0,a]$. Then apply MVT again on the interval $[0, x_0]\implies\frac{f(x_0)-f(0)}{x_0}=\frac{-1}{x_0}=f'(x_2)=f'(x_2)\implies-1=x_0f(x_2)\implies f(x_2)<0$ for a point $x_2\in[0, x_0]$. The existence of $x_2$ make sure that $f(x_1)=0$ is not the minimum on $[0, a]$ and since $0<f(0)=f(a)=1\implies f(x_1)$ is not the maximum on the $[0, a]$. Also if there exists other points $\beta$, for example, and $f(\beta)$ is the maximum of $[0,a]$ this implies that $f'(\beta)=0=f(\beta)$ the contradiction remains. Thus $f(a)\neq 1\implies f(x_0)\neq 0$. Since $a$ is an arbitrary real number, this means $f$ has no zero point on $[0,\infty]$. By the similar idea $f$ doesn't have zero point on $[-\infty, 0].
Is this a correct idea? And any shorter version of proof？ Thanks in advance!
Best Answer
Let $g(x) = f(x)e^{-x}$. Then, $g$ is a differentiable function on $\mathbb R$. By the product rule, $g'(x) = 0$. So, $g$ is constant. Since $f(0)=1$, we have $g\equiv 1$. This proves $$f(x)=e^x$$ so that $f$ has no zeros.