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Given a function $f:\mathbb{R}\rightarrow \mathbb{R}$ which satisfies $$f(x\cdot f(y))= \frac{f(x)}{y}$$ $\forall x,y \in \mathbb{R},y≠0$ and is not identically zero. Prove that $$f(x)\cdot f\left( \frac{1}{x}\right)=1$$

Putting $x=0$ gives $f(0)=0$ and in the functional equation putting $y=x$ gives,

$$f(x\cdot f(x))= \frac{f(x)}{x}$$

Assume $g(x)=x\cdot f(x)$

The above functional equation reduce to, $$g(g(x))= \frac{(g(x))^2}{x^2} $$

And we have to prove,$$g(x)\cdot g\left( \frac{1}{x}\right)=1$$

But this substitution doesn't help much it only removes the $x$ from argument. I couldn't solve much in this. I tried $h(x)=\frac{f(x)}{x}$ but that didn't worked too. How could I solve this?

Chesx
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1 Answers1

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Let $f:\mathbb{R}\rightarrow \mathbb{R}$ s.t:

$$f(x\cdot f(y))= \frac{f(x)}{y},\forall x,y \in \mathbb{R},y≠0$$ For $x=0$ we get $f(0)=0$ like you said.
For $x=y=1$ we get: $$f(1f(1))=\frac{f(1)}{1}\Rightarrow f(f(1))=f(1)$$

Now, based on what we have to prove, we can't have $f(1)=0$ since then, if we let $y=1$ we'd get $\forall x\in\mathbb{R}$:

$$f(xf(1))=f(x)\Rightarrow f(0)=f(x)\Rightarrow f(x)=0\Rightarrow f(x)f(1/x)=0\neq1,\forall x\in\mathbb{R}^{*}.$$

So $f(1)\neq 0$ and hence we can set $y=f(1),x=1$ to get:

$$f(f(1))=\frac{f(1)}{f(1)}\Rightarrow f(f(1))=1$$ and since $f(f(1))=f(1)$ we have $f(1)=1$.

Next, let $x=1,y\neq 0$ to get:

\begin{equation} \tag{1} f(f(y))=\frac{f(1)}{y}\Rightarrow yf(f(y))=1 \end{equation} Now let $x=1,y=\frac1y$. Then:

\begin{equation} \tag{2} f\left(f\left(\frac1y\right)\right)= yf(1)=y \end{equation}

Finally, setting to $(1)$ $y = f\left(\frac1y\right)$ we get:

\begin{equation} \tag{3} f\left(\frac1y\right)f\left(f\left(f\left(\frac1y\right)\right)\right)=1 \end{equation}

and since from $(2)$ we have $f\left(f\left(\frac1y\right)\right)=y$, we conclude:

$$ (3)\Rightarrow f\left(\frac1y\right)f\left(y\right)=1$$.

$\textbf{Edit:}$ I should also justify the step $y=f\left(\frac{1}{y}\right)$, since in order for it to be valid, we must have $f\left(\frac{1}{y}\right)\neq 0,\forall y\in\mathbb{R}^{*}$ or equivalently $f(y)\neq 0,\forall y\in\mathbb{R}^{*}$.
If we had some $y_{0}\in\mathbb{R}^{*}$ such that $f(y_{0})=0$, then for all $x\in\mathbb{R}^{*}$ we'd get:

$$f(xf(y_{0}))=\frac{f(x)}{y_0}\Rightarrow f(0)=\frac{f(x)}{y_0}\Rightarrow 0 = f(x)$$ which can't be the case for the same reason we can't have $f(1)=0$

Fotis
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  • Of course when I write $y=\frac{1}{y}$, I mean, $y=\frac{1}{z}$ for $z\in\mathbb{R}^{*}$, but I didn't want to get too messy with more variables. Hope the notation isn't confusing. – Fotis Jun 03 '23 at 16:08
  • Why does your proof hold for all $x \in \mathbb{R}$ and not just those that satisfy $x=1/x$ (step (2))? – mavavilj Jun 03 '23 at 16:34
  • @mavavilj Yeah so like I explained in my comment above, $y=1/y$ is kind of abuse of notation. If I wanted to be more precise, I should write $y=1/z$ for $z\in\mathbb{R}^{}$ or $y\mapsto 1/y$ for $y\in\mathbb{R}^{}$ – Fotis Jun 03 '23 at 16:37
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    Yes but also that you're assuming that $x=1$, when the proof requires $\forall x \in \mathbb{R}$ and then you prove it for $y$ because you mix the symbols. It's usually a good practice to use the same notation as given. – mavavilj Jun 03 '23 at 16:41