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I came across the following problem concerning functional equations.

Let $ f : \mathbb R \to \mathbb R $ be an additive function such that $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $. Prove that $ f ( x ) = c x $, $ c \in \mathbb R $.

The following solution was given.

Since $ f $ is an additive function we have $ f ( x - y ) = f ( x ) - f ( y ) $ for all $ x , y \in \mathbb R $. Let $ x \ne 0 , 1 $. Then $$ \frac 1 { x - 1 } - \frac 1 x = \frac 1 { x ( x - 1 ) } \text , $$ and we get $$ f \left( \frac 1 { x - 1 } \right) - f \left( \frac 1 x \right) = f \left( \frac 1 { x ( x - 1 ) } \right) \text . $$ Hence $$ \frac { f ( x - 1 ) } { ( x - 1 ) ^ 2 } - \frac { f ( x ) } { x ^ 2 } = \frac { f \big( x ( x - 1 ) \big) } { x ^ 2 ( x - 1 ) ^ 2 } \text , $$ which is equivalent to the identity $$ x ^ 2 f ( x - 1 ) - ( x - 1 ) ^ 2 f ( x ) = f \left( x ^ 2 - x \right) \text . $$ Thus $$ x ^ 2 \big( f ( x ) - f ( 1 ) \big) - ( x - 1 ) ^ 2 f ( x ) = f \left( x ^ 2 \right) - f ( x ) \text , $$ and therefore $$ f \left( x ^ 2 \right) = 2 x f ( x ) - x ^ 2 f ( 1 ) \text . \tag 1 \label 1 $$ Now, replace $ x $ by $ x + x ^ { - 1 } $ in this identity to obtain $$ f \left( x ^ 2 + x ^ { - 2 } + 2 \right) = 2 \left( x + x ^ { - 1 } \right) \left( x + x ^ { - 1 } \right) - \left( x ^ 2 + x ^ { - 2 } + 2 \right) f ( 1 ) \text . $$ Thus $$ f \left( x ^ 2 \right) + x ^ { - 4 } f \left( x ^ 2 \right) + f ( 2 ) = \\ 2 x f ( x ) + 4 x ^ { - 1 } f ( x ) + 2 x ^ { - 3 } f ( x ) - \left( x ^ 2 + x ^ { - 2 } + 2 \right) f ( 1 ) \text , $$ and using \eqref{1} again we get $$ f ( x ) = \left( \frac { f ( 2 ) + 2 f ( 1 ) } 4 \right) x $$ for $ x \ne 0 , 1 $. But $ f ( 2 ) = 2 f ( 1 ) $ and therefore $ f ( x ) = f ( 1 ) x $, which holds also for $ x = 0 , 1 $.

I am actually not able to understand the following:

  • Since $ f $ is additive, why did the author choose to consider $ f ( x - y ) = f ( x ) - f ( y ) $ rather than $ f ( x + y ) = f ( x ) + f ( y ) $?
  • How did the author conclude from $ \frac 1 { x - 1 } - \frac 1 x = \frac 1 { x ( x - 1 ) } $ that $ f \left( \frac 1 { x - 1 } \right) - f \left( \frac 1 x \right) = f \left( \frac 1 { x ( x - 1 ) } \right) $?

I spent a considerable amount of time to understand the aforementioned, but failed. A detailed and easiest possible explanation would really be helpful.

Mohsen Shahriari
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user832197
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    The second bullet point is the reason for the first and vice versa. The author used the identity $f(u-v) = f(u)-f(v)$ with $u=1/(x-1)$ and $v=1/x$. I suppose they wanted to do something with the relation $1/(x-1) - 1/x = 1/(x(x-1))$, so it made sense to look at the subtractive formula $f(u-v) = f(u)-f(v)$ so they could use this as the starting point. $$ $$ That said, this solution seems more complicated than is really necessary. I would be surprised if there were not a cleaner argument. – Glare Dec 12 '20 at 07:05
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    You can take $-y$ as $y$ and have it your way! – balaji Dec 12 '20 at 07:07
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    Oh, yeah. Good point @balaji. I guess if the question is "why is the subtractive formula true", check these facts user832197: $f(x+y) = f(x)+f(y)$ implies $f(0) = 0$ (use that $0+0=0$), and then use that to show $f(-y) = -f(y)$. – Glare Dec 12 '20 at 07:12
  • Additive directly implies $f$ is odd – user832197 Dec 13 '20 at 07:41

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Second bullet first. In $$ f(u - v) = f(u) - f(v) $$ make the substitution \begin{align*} u &\mapsto \frac{1}{x-1} \\ v &\mapsto \frac{1}{x} \text{.} \end{align*} Then $$ f\left(\frac{1}{x-1} - \frac{1}{x}\right) = f\left(\frac{1}{x-1}\right) - f\left(\frac{1}{x}\right) \text{.} $$ Of course, the argument to $f$ on the left-hand side is $\displaystyle \frac{1}{x(x-1)}$.

First bullet: (1) depends on a great deal of cancellation among the terms of the previous equation. This cancellation requires subtraction; addition does not exhibit the necessary cancellation.

Finding the useful expression usually takes some experimentation. Pick a few simple expressions and see that happens when you replace $x$ with those expressions. You have a mysterious operator in front of you start studying how it distorts the inputs.

Notice that the functional equation is equivalent to $$ x f(x) = \frac{1}{x} f(1/x) \text{.} $$

If additivity is going to be of any use on the left, you need something like $f(a-x)$ to cancel against the $x f(x)$. \begin{align*} (a-x) f(a-x) &= \frac{1}{a-x} f(1/(a-x)) \\ (a-x)^2 (f(a) - f(x)) &= f(1/(a-x)) \end{align*} This doesn't work because we do not get $-x f(x)$, the minus sign from the two instances of $-x$, always cancel. So try more things: raise the degree, try rational functions. Look at logs and exponentials -- try things.

Eric Towers
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  • I got it....but I want to prove that $f $ is odd only with the help of information given in the problem I.e. $f: \mathbb {R} \to \mathbb {R} $ is additive and $f (\frac {1}{x})=\frac {f (x)}{x^2}$. – user832197 Dec 13 '20 at 06:30
  • If you used the only fact that $f $ is additive, then $f (0)=f (x-x)$ should be equal to $f (x)+f (-x) $, but you wrote $f(0)=f(x)-f (x) $..............this is only possible if and only if we assume the fact that $f (-x)=-f (x) $ or $f $ is odd. .....but how to prove it ?? – user832197 Dec 13 '20 at 06:45
  • @user832197 : Ah, yes. Deleted. – Eric Towers Dec 13 '20 at 06:47