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
Best Answer
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 $k=1$.