Existence of $1/(e-1)$ value of a continous and differentiable function $f:[0,1]\to \mathbb R$ given $f(1)=1, f(0)=0$

Question

Let $f:[0,1]\to \mathbb R$ be a continous function on $[0,1]$ and differentiable on $(0,1)$, $f(0)=0, f(1)=1$. Prove that there exists $c\in (0,1)$ so that $f(c)+\frac{1}{e-1}=f'(c)$ where $e$ is the Euler number.

I tried to subs $f(x)=e^{x}g(x)$ so that $f'(x)-f(x)=e^{x}g'(x)$ and define $e^{x}g'(x)=w(x)$. Then I tried to use the mean value theorem (to ensure the existence of $r \in (0,1)$ s.t. $f'(r)-f(r)=e^{r-1}$) however then I have no idea how to continue. My guess is we use intermediate value theorem on $w(x)$ however I don't have any idea on how to do it. Is there any construction to solve this?

Answer 1

Let $$F(x)=e^{-x}\left(f(x)+\frac{1}{e-1}\right),\quad x\in[0,1],$$ then $F$ is continous on $[0,1]$ and differentiable on $(0,1)$, $$F(0)=F(1)=\frac{1}{e-1},\qquad F'(x)=e^{-x}\left(f'(x)-f(x)-\frac{1}{e-1}\right).$$ So MVT implies the result!

Answer 2

While the problem was solved above, here is a method that uses a stronger result (Darboux's theorem) but shows how to arrive systematically at the function used in the MVT.

Suppose that there doesn't exist $c \in (0,1)$ such that $f'(c) = f(c) + \frac 1{e-1}$. Then, note that $f'(c) - f(c)$ is a function that satisfies the intermediate value property on $(0,1)$ [where we use Darboux's theorem], therefore $f'(x) - f(x)$ is either entirely above or entirely below $\frac 1{e-1}$ on $(0,1)$.

Without loss of generality assume that $f'(x) - f(x) < \frac 1{e-1}$ (the opposite case is addressed by changing $f$ to $-f$) for all $x \in (0,1)$. We execute the following series of steps, aiming to create a functional inequality using the combination of both sides. : \begin{align} &f'(x) - f(x) < \frac 1{e-1} \\ \implies &e^{-x}f'(x) - e^{-x}f(x) < \frac {e^{-x}}{e-1} \\ \implies &(e^{-x}f(x))' < \frac{e^{-x}}{e-1} \\ \implies &e^{-b}f(b) - e^{-a}f(a) < \frac{e^{-a}- e^{-b}}{e-1} \quad \forall a<b \in (0,1), \end{align} where we obtained the last inequality by integrating the previous one from $a$ to $b$. By rearrangement, we conclude that $$ e^{-x}f(x) + \frac{e^{-x}}{e-1} \text{ is strictly decreasing on } (0,1). $$

Now, because $f(0)=0$ and $f(1)=1$, the above function evaluates to $\frac 1{e-1}$ at $0$ and $1$, providing a contradiction to the strictly decreasing nature of this function in $(0,1)$.

However, leaving the Darboux theorem aside, one may also observe that the MVT may be applied to this function on $[0,1]$, since retracing steps leads to the equality we want. Therefore, we see where the required function originates from in this answer.