# $\forall x\in \mathbb{R} \exists b\in (x, x+a) \frac{f'(x) } {f(x) }=e^{a{f'(b) }/{f(b) } }$

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.

• What is the context for this problem, i.e. where did you come across it? It might be helpful to read up on delay differential equations. – Theo Bendit Sep 18 '19 at 21:18
• @Theo It's the first question of an exercise given to us by our Real Analysis teacher. Thank you for the reference. – Sami Fersi Sep 18 '19 at 21:59
• @Theo I'm familiar with the mean and extreme value theorems but never heard of Fermat's Theorem about stationary points. Actually I welcome any solution you can come up with. If you have to use heavy machinery then let it be, don't worry about it :) – Sami Fersi Sep 18 '19 at 22:10
• if you write $\frac{f'(x)}{f(x)}=\frac{f(x+a)}{f(x)}$, your conclusion is just the intermediate value theorem of $\ln f(x)$ but on the interval $(x,x+a)$, not $(a,x)$ – Nanayajitzuki Sep 19 '19 at 2:10
• Of course, you could remove the words "such that" and replace the words "there exists" with the symbol $\exists$ in both the title and the question without changing the meaning. – Geoffrey Trang Sep 21 '19 at 0:03

Define a new function $$g(x)=\ln{\big(f(x)\big)}$$. We note that $$0 so $$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.

• I indeed made a mistake while writing the question, it should be $(x,a+x)$ instead of $(a,x)$ (I'm so sorry!). Thank you for the answer Omer! – Sami Fersi Sep 20 '19 at 22:19