Easy but strange Olympiad functional equations

Question

Question -

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies 1) $f(f(x))=x$ and 2) $f(\lambda x)=\lambda f(x)$ for all real numbers $x$ and $\lambda .$ Prove that $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$. What if $f(\lambda x)=\lambda^{k} f(x) ?$

My work -

I first put $f(1)=c$ so that $f(c)=1$...

now let $x=c$ in $2)$ so that $f(\lambda c)=\lambda$ now let $ \lambda=c$ so that $f(c^2)=c$ taking $f$ again we get $c=+-1$ ..hence $f(1)=+-1$

if $f(1)=1$ then we easily get $f(x)=x$ and if $-1$ then $f(x)=-x$ for all $x$..

but i am not able to prove what will happen if $f(\lambda x)=\lambda^{k} f(x) ?$....

hint says i have to prove that $k=+1 or -1$ but i am not getting how to prove it?

any hints???

thankyou

Answer 1

It is clear that $f$ is a bijection . If $k$ is even, we get $f(-\lambda x)=(-\lambda)^{k}f(x)=(\lambda)^{k}f(x)=f(\lambda x)$ which implies $-\lambda x =-\lambda x$ for all $\lambda, x$ which is a contradiction. Hence $k$ is odd.

Now $\lambda^{k}=f(\lambda c)$ (because $f(c)=1$). Hence $f(\lambda ^{1/k} c) =\lambda f(c)=\lambda $. Applying $f$ on both sides we get $f(\lambda)=\lambda ^{1/k} c$ for all $\lambda$. Now I leave to you to see that this function satisfies $f(f(x))=x$ only if $k^{2}=1$. It may be noted the hypothesis fails when $k=-1$ so we must have $k=1$.

Answer 2

First part:

1. $ f(0) = f(2 \times 0) = 2f(0) \Rightarrow f(0) = 0$.
2. For $ a, b \neq 0$, we have $ f(ab) = a f(b) = b f(a) \Rightarrow \frac{f(a)}{a} = \frac{f(b)}{b}$, hence this is a constant.
   Thus $f(a) = a f(1)$.
3. If $f(1) = c$, then $ c^2 = cf(1) = f(c) = f(f(1)) = 1$ so $ c = \pm 1$. Verify that $f(x) = x, f(x) = -x$ are solutions to the functional equation.

Second part: (Assuming that $k$ is an positive integer)
1. $ f(0) = f(2 \times 0) = 2f(0) \Rightarrow f(0) = 0$.
2. For $ a, b \neq 0$, we have $ f(ab) = a^k f(b) = b^k f(a) \Rightarrow \frac{f(a)}{a^k} = \frac{f(b)}{b^k}$, hence this is a constant.
Thus $f(a) = a^k f(1)$.
3. If $f(1) = c$, then $ c^{k+1} = c^kf(1) = f(c) = f(f(1)) = 1$.
If $k$ is an even integer, then $ c = 1$. However, verify that $f(x) = x$ is not a solution to the functional equation.
If $k$ is an odd integer, then $ c = \pm 1$. Verify that $f(x) = x, f(x) = -x$ are solutions to the functional equation.

Side note of interest: If we allowed for negative $k$ by restricting $\lambda \neq 0$, then for $k = -1$, $c$ can be any value. Verify that $f(x) = \frac{c}{a}$ are solutions to the functional equation.

Answer 3

I think the easiest way is to put $x=1$ in $$f(\lambda x)=\lambda f(x)$$ Giving (I'll just replace $\lambda$ with $x$ and $f(1)$ with $k$) $$f(x)=kx$$ Now $$f(f(x))=x \implies k^2 x=x$$ $$\Leftrightarrow k^2=1 \Leftrightarrow k= \pm 1$$ $$\implies f(x)=\pm x$$ If $f(\lambda x)=\lambda^k f(x)$, I'll again put $x=1$, replace $\lambda$ with $x$, and put $f(1)=c$, giving Then $$f(x)=cx^k$$ So $$f(f(x))=x \implies c(f(x))^k = x \implies c^{k+1} x^{k^2} = x \tag{1}$$ for all $x$. So $k^2=1$ or $k=\pm 1$.

If $k=1$, we have $$c^2=1 \implies c=\pm 1 \implies f(x)= \pm x$$ If $k=-1$, then equation $(1)$ works for all $c \in \mathbb{R}$ giving $$f(x)=\frac{c}{x} \text{ } \forall x,c \in \mathbb{R}$$ and that's it, we have all the solutions.