If $f$ be a decreasing function satisfying $f(x+y)= f(x)+ f(y)- f(x)f(y) ~\forall x, y \in \mathbb R$ and $f'(0)= -1$

Question

If $f$ be a decreasing function satisfying $f(x+y)= f(x)+ f(y)- f(x)f(y) ~\forall x, y \in \mathbb R$ and $f'(0)= -1$ then $\displaystyle\int_0^1 f(x)dx $ is:

A)$1$

B) $1- e$

C) $2- e$

D) None of these.

Attempt:

Put y = 0,

$f(0)(1- f(x)) = 0$

$f(x)\ne 1$ therefore $f(0) = 0$

Differentiating wrt $x$

$f'(x+y) = f'(x) - f(y)f'(x)$

put $x= 0$

$f'(y) = - 1 + f(y) $

$\implies f(y) = e^{y+c} + 1$

put y = 0,

$-1 = e^c$ which ain't possible since $e^x > 0$

How do I solve this problem then? What is the mistake in my solution?

Answer is:

$C$

Answer 1

$\frac{\rm d f}{\rm d y} = -1 + f\\ \rm d f = (-1 + f)\rm d y\\ \rm \frac{df}{-1 + f} = dy \\ \ln\color{red}{|}-1 +f\color{red}{|} = y+C\\f = 1\color{red}{\pm}Ae^y$

Answer 2

$\int_0^1 f(x)dx = \int_0^1 f(x+0)dx = \int_0^1 (f(x) + f(0) - f(x)f(0))dx$. Hence, \begin{align*} 0 &= \int_0^1 (f(0) - f(x)f(0))dx \\ &= f(0)\int_0^1 (1-f(x))dx \end{align*} since $f'(0)=-1 \implies f(x)\neq 1$, so $f(0)=0$. Now, note that $f'(x+y)=f'(x)-f(y)f'(x)$, where we differentiated w.r.t. $x$. Substituting $x=0$, we have that $$f'(y)=-1+f(y) \implies \dfrac{df(y)}{dx} = -1 + f(y) \implies f(y)=c_1e^y + 1,$$ then since $f(0)=0 \implies f(0)=c_1(1)+1 = 0 \implies c_1=-1$. Thus, $$f(x)=-e^x + 1 \implies \int_0^1 (-e^x + 1)dx = -e^x + x \mid_0^1 = -e + 1 - (-1 + 0) = -e+2.$$

Note: The part you were having trouble with was solving the ODE $f'(y)=-1+f(y)$, if you know about integrating factors that is the easiest way to solve it.

Answer 3

The issue is how you integrated the differential equation. $$\frac {df}{f-1}=dy$$ You integrate and you obtain $$\ln(f-1)=y+C$$ The issue is that $f-1$ is negative, at least for $x>0$

Answer 4

Here is a solution which requires very limited analysis.

$$f(x+y)= f(x)+ f(y)- f(x)f(y) \\ 1-f(x+y)= 1-f(x)- f(y)+ f(x)f(y)=(1-f(x))(1-f(y))$$

Define $g(x):=1-f(x)$. Then $$g(x+y)=g(x)g(y)$$

This is the multiplicative version of the Cauchy's functional equation, and we solve it below.

Next, note that for all $x$ you have $$g(x)=g(\frac{x}{2}) g(\frac{x}{2}) \geq 0$$

Moreover, if $g(a)=0$ for some $a$ then $g(x)=g(x-a+a)=g(a)g(x-a)=0 \forall x \in \mathbb R$. This shows that $$g(x) >0 \forall x$$

Finally, let $h(x):= \ln (g(x))$ which is well defined since $g>0$. Then $$h(x+y)=h(x)+h(y)$$

By the Cauchy FE there exists some $a \in \mathbb R$ such that $$h(x)=ax \, \forall x \in \mathbb Q$$

Finally, $f$ monotonic implies $g$ monotonic and hence so is $h$. This implies that $$ h(x)=ax \forall x \in \mathbb R$$

Thus $$ g(x)=e^{ax} \\ f(x)=1- e^{ax}$$

Now, just figure out $a$ from the derivative,