Let $a\ge 0$ and $f:\mathbb{R}\rightarrow \mathbb{R}$ be a differentiable positive function such that $f'(x) =f(x+a) \forall x\in \mathbb{R}$. How can I prove that $\forall x\in \mathbb{R} \exists b\in (x, a+x) \frac{f'(x) } {f(x) }=e^{a\frac{f'(b) }{f(b) } }$. I tried the intermediate value Theorem but couldn't prove it. Thank you in advance for your help.
$\forall x\in \mathbb{R} \exists b\in (x, x+a) \frac{f'(x) } {f(x) }=e^{a{f'(b) }/{f(b) } }$
continuityderivativesreal-analysis
Best Answer
Define a new function $g(x)=\ln{\big(f(x)\big)}$. We note that $0<f(x)$ so $g$ is well defined as a composition. Furthermore, $g$ is continuous on any interval of the form $[x,x+a]$ and differentiable on any interval of the form $(x,x+a)$, and so $g$ satisfies the conditions for the Mean Value Theorem, that is: There exist $b\in(x,x+a)$ that satisfies: $$g'(b )=\frac{g(x+a)-g(x)}{(x+a)-x}=\frac{g(x+a)-g(x)}{a}\Longrightarrow(*)\space\space\space ag'(b)=g(x+a)-g(x)$$
We now take care of the LHS of $(*)$: $$ag'(b)=a\cdot\frac{d}{dx}\ln{f(x)}\bigg|_{x=b}=a\frac{f'(b)}{f(b)}$$
And now the RHS of $(*)$: $$g(x+a)-g(x)=\ln{f(x+a)}-\ln{f(x)}=\ln{\frac{f(x+a)}{f(x)}}=\ln{\frac{f'(x)}{f(x)}}$$
The second equality is from $\ln$ properties and the last equality is from the given relation $f'(x)=f(x+a)$. So we now substitute what we found back to $(*)$ to get:
$$a\frac{f'(b)}{f(b)}=\ln{\frac{f'(x)}{f(x)}}\Longrightarrow e^{a{f'(b)}/{f(b)}}=\frac{f'(x)}{f(x)}$$
And that settles the proof.