I'm trying to find out the function $f$ that satisfies the functional equation $f\biggl(x\cdot f(\frac yx)\biggr)=x\cdot f\biggl(\frac{f(y)}{x}\biggr),\forall x \in \mathbb R\setminus\{0\},\ \forall y \in \mathbb R$, where $f:\mathbb R \to \mathbb R$ is a continuous monotonically increasing function, with $f(-1)=0$, $f(0)=1$, and $f(x)>x,\forall x \in \mathbb R$.
Besides, $f$ satisfies this two functional equations:
$-f(-f(x))=x, \forall x \in \mathbb R$ (It may be useful to consider the function $g(x)=-f(x)$, so $g(g(x))=x$).
$f(x)=x\cdot f(1/x),\forall x \in \mathbb R\setminus\{0\}$.
Thanks
Note: I've found out that using the three equations there are 2 new relationships:
- $f(-x)=f(1)\cdot f\biggl(-\frac{f(x)}{f(1)}\biggr),\forall x \in \mathbb R$, with $f(1)$ unkown, but satisfying $f(f(f(1)))=(f(1))^2$.
- $f(-x^2)=f(x)\cdot f(-x), \forall x \in \mathbb R$.
New attempt:
I don't know if this problem should be solved by calculus as per @TymaGaidash's comment suggests, but I attempted solve it with more elementary logic, i.e. to show that $x + 1$ is the only expression suitable:
We know that the function is monotonically increasing. This means that
$f(x + a) = f(x) + b \tag{1}$
and $f( \frac{1}{x + a}) = f(1/x) - c$, where $a, b, c \in \mathbb R^+ \cup \{0 \}$, except that $a \ne 0$ as with that we want to change the value of $x$ and see the function values.
Using $f(x)=x\cdot f(1/x) \tag{2}$
we get
$f(x + a) = (x + a) \cdot f( \frac{1}{x + a})$
$f(x) + b = (x + a) \biggl(f(1/x) - c \biggr)$
$f(x) + b = x \cdot f(1/x) + a \cdot f(1/x) - xc - ac$
$b = a \cdot f(1/x) - xc - ac$
$f(1/x) = \frac {b + xc}{a} + c \tag{3}$
Using $(3)$ and $f(-1)=0$ we get
$f(-1) = \frac {b - c}{a} + c = 0$
$b - c + ac = 0$
$b = c - ac \tag{4}$
Using $(3)$ and $(4)$ we get
$f(1/x) = \frac {c - ac + xc}{a} + c = \frac {c + xc}{a} = \frac {1 + x}{a} c \tag{5}$
Using $(2)$ and $(5)$ we get
$f(x) = x \cdot f(1/x) = x \frac {1 + x}{a} c \tag{6}$
We have $f(x + a) = f(x) + b$ from $(1)$, and using $(6)$ for $f(x + a)$ and $f(x)$, and $(4)$ for $b$ we get
$(x + a) \frac{1 + x + a}{a} c = x \frac{1 + x}{a} c + c - ac$
$(x + a)(1 + x + a)c = x(1 + x)c + ac - a^2c$
$x(1 + x)c + a(1 + x)c + (x + a)ac = x(1 + x)c + ac - a^2c$
$ac + axc + axc + a^2c = ac - a^2c$
$2axc = -2a^2c \tag{7}$
before division we need to consider, that by definition $a$ can not be zero, however $c$ might be, so let $c = 0$ for $(6)$ and using $f(0)=1$ we get
$f(0) = 0 \frac {1 + 0}{a} 0 = 1$ and this is impossible so $c \ne 0$ so we divide $(7)$ by $2ac$ to get
$x = -a \tag{8}$
Substituting $(8)$ into $(6)$ we get
$f(x) = x \frac {1 + x}{-x} c = -(1 + x)c \tag{9}$
and using $f(0)=1$ with $(9)$ we get
$f(0) = -(1 + 0)c = 1$
$c = -1$
and substituting it into $(9)$ we get
$f(x) = -(1 + x)(-1) = 1 + x$
Original attempt:
From $f(−1)=0, f(0)=1$ it is suspicious to check $f(x) = x + 1$:
For $f(x)>x,\forall x \in \mathbb R$: $x + 1 > x$ true.
For $-f(-f(x))=x, \forall x \in \mathbb R$: $-f(-f(x)) = -f(-(x + 1)) = -f(-x - 1) = - (-x) = x$ true.
For $f(x)=x\cdot f(1/x)$:
$x + 1 = x\cdot \frac{x + 1}{x}$
$x + 1 = x + 1$ true.
For $f\biggl(x\cdot f(\frac yx)\biggr)=x\cdot f\biggl(\frac{f(y)}{x}\biggr),\forall x \in \mathbb R\setminus\{0\}, \forall y \in \mathbb R$:
$f\biggl(x\cdot \frac {y + x}{x}\biggr)=x\cdot f\biggl(\frac{y + 1}{x}\biggr)$
$f(y + x) = x\cdot \biggl(\frac{y + 1 + x}{x}\biggr)$
$y + x + 1 = y + 1 + x$ true.
To show that $f$ can not be any other polinomial expression, one can reason like @MohsenShahriari did in the below comment. My version is to consider $f(x) = P(x)$ and the condition $f(−1)=0$: the latter means that $-1$ is a root of $P(x) \Rightarrow P(x) = (x + 1) \cdot Q(x)$. Using $f(x)=x\cdot f(1/x)$ we get
$(x + 1) \cdot Q(x) = x \cdot \biggl( \frac{1}{x} + 1 \biggr) \cdot Q \biggl( \frac{1}{x} \biggr)$
$(x + 1) \cdot Q(x) = (1 + x) \cdot Q \biggl( \frac{1}{x} \biggr)$
At this point we can exclude the values $\pm 1$ of $x$, because we want to confirm $\forall x$, so we can divide by $(x + 1)$:
$Q(x) = Q \biggl( \frac{1}{x} \biggr)$
If $Q(x)$ is of degree $> 0$ then
$(x - x_0)(x - x_1)...(x - x_n) = \biggl( \frac{1}{x} - x_0 \biggr) \biggl( \frac{1}{x} - x_1 \biggr) ... \biggl( \frac{1}{x} - x_n \biggr)$
Again, as we want to confirm $\forall x$, we can use
$\lim_{x \rightarrow \infty}(x - C) = \infty$ and $\lim_{x \rightarrow \infty}(\frac{1}{x} - C) = -C$ where C is a real constant and with the Extended Sum Rule we get
$\lim_{x \rightarrow \infty} \biggl((x - x_0)(x - x_1)...(x - x_n) \biggr) = \infty \ne \lim_{x \rightarrow \infty} \biggl( \biggl( \frac{1}{x} - x_0 \biggr) \biggl( \frac{1}{x} - x_1 \biggr) ... \biggl( \frac{1}{x} - x_n \biggr) \biggr) = (-x_0)(-x_1)...(-x_n)$
therefore $Q(x)$ is of degree $0$.
Using the condition $f(0)=1$ we have $(0 + 1) \cdot Q(0) = 1 \Rightarrow Q(0) = 1, \forall x$, therefore the only polinomial expression suitable is $x + 1$.
I didn't take the time to show if there is any solution as a non-polinomial expression.