4
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Theo Bendit Sep 18 '19 at 21:18
  • $\begingroup$ @Theo It's the first question of an exercise given to us by our Real Analysis teacher. Thank you for the reference. $\endgroup$ – Sami Fersi Sep 18 '19 at 21:59
  • 1
    $\begingroup$ @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 :) $\endgroup$ – Sami Fersi Sep 18 '19 at 22:10
  • 1
    $\begingroup$ 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 $(a,x)$ $\endgroup$ – Nanayajitzuki Sep 19 '19 at 2:10
  • 1
    $\begingroup$ 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. $\endgroup$ – Geoffrey Trang Sep 21 '19 at 0:03
3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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! $\endgroup$ – Sami Fersi Sep 20 '19 at 22:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.