Functional equations [Sample paper of Indian Mathematical Olympiad]

Question

Edit- There was information missing (lack of clear printing in my book) in the book through which I referred the question. Confirming with my friend's book I have made a small change. I am really sorry.

Edit 2- Guys, this question is meant for an olympiad where calculus isn't allowed. Try solving it without calculus.

If $f: \mathbb{R} \to \mathbb{R}^+$ satisfies:

1. $f(xf(y)) = yf(x)$ for all $x,y \in\Bbb{R}^+$
2. $\lim_{x \to \infty} f(x) = 0$

Find $f(x)$.

I tried putting $x=y$ and got $f(y(f(y))=yf(y)$ which indicates $f(x) = x$ which doesn't follow the second statement. Help.

Answer 1

Sadly I don't have enough rep to comment:

With the question as written do we not get $f(x) = 0$ for all $x$? It seems inputting any negative value of $y$ and using that $\text{Im} f \subset [0,\infty)$ leads to this conclusion.

Answer 2

If $f$ is a solution, with $x=1$ we have : $$f\circ f=f(1)id$$ so $$\frac{f}{f(1)}\circ f=id.$$ Since $$f(x)=\frac{f(xf(y))}{y}(*)$$ with $x=\frac{1}{f(1)},y=1$ : $$f(\frac{1}{f(1)})=f(1),$$ so $$\frac{f}{f(1)}\circ \frac{f}{f(1)}(x)=\frac{1}{f(1)}f(\frac{1}{f(1)}f(x))=\frac{1}{f(1)}xf(\frac{1}{f(1)})=x,$$ so $\frac{f}{f(1)}$ is an involutory function and by unicity of the inverse : $f=\frac{f}{f(1)}$ so $f(1)=1$ (because $f\neq 0)$ and $f$ is bijective with $f^{-1}=\frac{f}{f(1)}=f$. Moreover, if $x\in\mathbb{R}$, with $y=x$, $xf(x)$ is a fixed point of $f$, and $f(x_0^2)=f(x_0f(x_0))=x_0f(x_0)=x_0^2$, and by induction : $$f(x_0^n)=x_0^n\rightarrow +\infty$$ : impossible. If $x_0<1$, then by $(*)$ : $$f(\frac{1}{x_0})=\frac{f(\frac{1}{x_0}f(x_0))}{x_0}=\frac{1}{x_0}$$ so $\frac{1}{x_0}>1$ is a fixed point of $f$ : impossible.

So $1$ is the only fixed point of $f$ et $f:x\mapsto \frac{1}{x}$, which is a solution to the problem.

Answer 3

Partial Answer

Take the partial derivatives (assuming differentiability): \begin{align*} \partial_x:\qquad f'(xf(y))f(y)&=yf'(x)\\ \partial_y:\qquad f'(xf(y))x f'(y)&=f(x)\\ \frac{yf'(x)}{f(y)}&=\frac{f(x)}{x f'(y)}\\ \frac{x f'(x)}{f(x)}&=\frac{f(y)}{y f'(y)}=C \\ x f'(x)&=Cf(x)\\ f(x)&=B x^C. \end{align*} Assuming $0\not\in\mathbb{R}^+,$ then using this functional form of $f,$ we plug back into the equation \begin{align*} \frac{f(xf(y))}{y f(x)}&=1\\ \frac{B^C x^C y^{C^2}}{x^C y}&=1\\ B^C y^{C^2-1}&=1\\ C^2&=1\\ C&=\pm 1 \\ B&=1. \end{align*} But $C=1$ cannot satisfy the limit criterion, so we have that $C=-1, B=1$ satisfies the criteria. This shows that of differentiable functions, $f(x)=\dfrac1x$ is the only function satisfying the functional equation.

Answer 4

Assume the domain $f$ is $\mathbb{R}^+ = \{x \in \mathbb{R}: x > 0\}$, not $\mathbb{R}$. This approach is similar to Mishikumo2019's.

In the functional equation $f(x f(y)) = y f(x)$, setting $x = 1$ and rewriting $y$ as $x$ gives $$f(f(x)) = x f(1).$$ The RHS increases without bound but $f(x) \to 0$ as $x \to \infty$, so for every value $Y \in \mathbb{R}^+$ there exists some $\delta > 0$ such that if $x < \delta$ and $x \in f(\mathbb{R}^+)$, then $f(x) > Y$: that is, $\lim_{x \to 0^+} f(x) = +\infty$ if the domain of $f$ is restricted to the range of $f$. (With the additional hypothesis that $f$ is continuous, this is equivalent to $\lim_{x \to 0^+} f(x) = +\infty$.)

We know from setting $x = y$ in the functional equation that any number of the form $x f(x)$ is a fixed point of $f$; as a corollary, $f$ has at least one fixed point. If $x_0$ is a fixed point, then so is $x_0 f(x_0) = x_0^2$ and, more generally, $x_0^{2^n}$ for any $n \in \mathbb{N}$. If $x_0 \neq 1$, then we have a sequence of fixed points approaching zero (contradicting the fact that $\lim_{x \to 0^+} f(x) = +\infty$ when the domain of $f$ is restricted to the range of $f$) or extending out to infinity (contradicting $\lim_{x \to \infty} f(x) = 0$). Thus, the only fixed point of $f$ is $1$, so $x f(x) = 1$ for all $x \in \mathbb{R}^+$, and thus the only possible $f$ is $f(x) = 1/x$.