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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 $f(f(f(1)))=(f(1))^2$.
  • $f(-x^2)=f(x)\cdot f(-x), \forall x \in \mathbb R$.
Javiershadow1
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  • Lets see. A differential equation can be made with:$$f’(x)=f\left(\frac 1x\right)-\frac {f’\left(\frac 1x\right)}x $$ Also let $y=\frac1x$:$$ f\left(\frac1y\right)=\frac{f(y)}y$$ – Тyma Gaidash Oct 09 '21 at 17:17
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    The title doesn't really match the question. – Greg Martin Oct 09 '21 at 17:37
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    How far have you come yourself? Any thoughts of how to solve it? – md2perpe Oct 09 '21 at 18:17
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    @md2perpe The bullets are the OP's attempt. They are consequences of setting $ y = 0 $ and $ y = - 1 $ in the original functional equation. – Mohsen Shahriari Oct 09 '21 at 18:26
  • Sorry, I'm a new member and my first attempt was to put the original functional equation in the title but I couldn't do that. As @Mohsen Shahriari says, the bullets are deduced from the original functional equation. – Javiershadow1 Oct 09 '21 at 19:14

1 Answers1

2

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.

Dávid Laczkó
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  • Until now that's the only solution I've found out. – Javiershadow1 Oct 09 '21 at 19:33
  • See my addition for $f$ as polinomial expression. – Dávid Laczkó Oct 09 '21 at 20:39
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    For showing that $ f ( x ) = x + 1 $ is the only polynomial solution, note that $ - f \bigl ( - f ( x ) \bigr ) = x $ implies that $ f $ cannot have a degree more than $ 1 $, i.e. it must be linear. $ f ( - 1 ) = 0 $ and $ f ( 0 ) = 1 $ then imply $ f ( x ) = x + 1 $ for all $ x \in \mathbb R $. – Mohsen Shahriari Oct 10 '21 at 01:11
  • @MohsenShahriari Ah, yes, I wanted to simplify it so much that I oversimplified it. I took a different approach then yours which is more complicated at the end. – Dávid Laczkó Oct 10 '21 at 19:53
  • @Javiershadow1 I think I managed to prove that that is the only solution. – Dávid Laczkó Oct 20 '21 at 19:20
  • Thank you! But why did you consider in (5) that b and c are equal for all x? Aren't they function of x? – Javiershadow1 Oct 21 '21 at 00:22
  • @Javiershadow1 By definition $(1)$ they are actually a function of both $x$ & $a$, funtions of $x + a$, so they are not constants. At $(5)$, $c/a$ must be $1/x$ to get $x + 1$ if that's the only solution, and $(8)$ and the result after $(9)$ gives that, and as $a$ is a constant, $c$ is a function of $x$. And from $(4)$ and $c$ being a function of $x$, $b$ is also a function of $x$. Possibly my notation is not how it should be. – Dávid Laczkó Oct 21 '21 at 06:33
  • @Javiershadow1 Or possibly by definition $(1)$ we should rather say, that for any fixed value of $x$, both $b$ & $c$ are the functions of $a$. This might require to index them with $x$ for clarity, I don't know. – Dávid Laczkó Oct 21 '21 at 06:46
  • Your "new attempt" is very wrong. You've formulated $ f $ being monotonically increasing as "$ f ( x + a ) = f ( x ) + b $ with $ b \ge 0 $ for $ a > 0 $". The precise formulation of this would be "there is a function $ b : \mathbb R \times \mathbb R _ + \to \mathbb R _ { 0 + } $ such that for all $ x \in \mathbb R $ and all $ a \in \mathbb R _ + $, $ f ( x + a ) = f ( x ) + b ( x , a ) $"; i.e. the value of $ b $ would depend on both $ x $ and $ a $ (the situation is similar for $ c $). (to be continued) – Mohsen Shahriari Oct 25 '21 at 08:01
  • (continued) Up to the step when you derive $ ( 3 ) $ everything is OK, as you talk about the same $ x $ and $ a $ all the time. But after that, when you equate $ f ( - 1 ) $ with $ \frac { b - c } a + c $, you treat $ b $ and $ c $ as if they are constant [at least] with respect to $ x $, and everything derived from that is not valid. If it's difficult for you to see what's exactly invalid there, try to use your argument on the monotonically increasing function $ x \mapsto \exp x $ and see where it goes wrong. – Mohsen Shahriari Oct 25 '21 at 08:01
  • @MohsenShahriari It can easily be the case, but why I did that is because plugging in any value to the argument of $f$ makes $x$ a constant, and we get $(8)$ that shows that $a$ is not independent of $x$, and because of $(4)$, $b$ is also not. But I get $c$ as a constant, however due to its original definition it cannot be. – Dávid Laczkó Oct 25 '21 at 13:58
  • The real problem is somewhere that you seem to miss. Let me formulate the equations more precisely to present it clearly. $ ( 3 ) $ precisely means $$ f \left ( \frac 1 x \right ) = \frac { b ( x , a ) + x c ( x , a ) } a + c ( x , a ) \text . $$ What you get by letting $ x = - 1 $ in $ ( 3 ) $ will therefore be $$ 0 = \frac { b ( - 1 , a ) - c ( - 1 , a ) } a + c ( - 1 , a ) \text , $$ while you seem to have taken it to be $$ 0 = \frac { b ( x , a ) - c ( x , a ) } a + c ( x , a ) \text . $$ That's what I meant by treating $ b $ and $ c $ as if they are constants with respect to $ x $. – Mohsen Shahriari Oct 25 '21 at 17:36
  • @MohsenShahriari Yes, but that's why I call it an attempt and not a solution. If $(3)$ is true, we observe that as at the LHS argument $a$ is not present while it is on the RHS, this means that for a particular value of $x$ the RHS must produce the same value $\forall a$, and that is what I wanted to use in the following - I don't know what else could be used than the given paricular $x$ & $f(x)$ values. Maybe this observation is still not enough to provide a solution? – Dávid Laczkó Oct 26 '21 at 06:47