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Find all functions $f : \mathbb{R} → \mathbb{R}$ such that $$f \big(x + y f (x)\big) = f (x) + xf (y)$$ for all $x, y \in \mathbb{R}.$

Could someone please provide a solution as well as their reasoning and how they reached the solution? I tried plugging in different values but I don't really seem to be getting anywhere. Maybe that's because I am new to solving functional equations and don't really recognise the best possible subs. Anyway, Thanks a lot.

Mohsen Shahriari
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Aayam Mathur
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    Did you learn anything from setting $x=0?$ Have you found any functions that work? – Ross Millikan May 22 '20 at 04:52
  • Yes, by raw guessing and trying linear functions ( learnt that from chan's handout), i found out f(x)=0 and f(x)=x. Those indeed are the solutions as i checked the answer. – Aayam Mathur May 22 '20 at 05:30
  • Plugging in x,y=0 gives nothing,and i couldn't really find a use for just x=0 and just y=0. – Aayam Mathur May 22 '20 at 05:30
  • Here setting $x=0$ gives $f(yf(0))=f(0)$. Now if $f(0) \neq 0$ we have that $f$ is a constant, call it $c$. Then $c=c+xc$ so $c$ must be $0$. We now know $f(0)=0$. – Ross Millikan May 22 '20 at 13:58
  • Thats very coool. Also could you please recommend some good lectures through which i could learn a bit more about solving such questions? Any books maybe? Basically i just wanna be as well prepared for the imo as possible. Thanks – Aayam Mathur May 22 '20 at 15:03
  • No, what I learned about this I learned long ago. I then noticed that $f(x)=x$ was a solution, but couldn't prove it was the only one. – Ross Millikan May 22 '20 at 15:37
  • Quite a few solutions on AoPS https://approach0.xyz/search/?q=%24f%20%5Cleft(x%20%2B%20y%20f%20(x)%5Cright)%20%3D%20f%20(x)%20%2B%20xf%20(y)%24&p=1 – Sil May 22 '20 at 18:13

1 Answers1

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$$P(x,y) \implies f(x+yf(x))=f(x)+xf(y)$$ Let $a\ne0$ be a real number: $$P(a,0) \implies f(a)=f(a)+af(0) \iff f(0)=0$$ Let $b$ be a real number such that $f(b)=0$ $$P(b,x) \implies f(b+xf(b))=f(b)+bf(x) \implies0=bf(x)$$ $$b \ne 0 \implies f(x)=0 \ \ \ \forall \ x \in \mathbb{R}$$ $$b=0 \implies0 \ \text{is a unique zero of } f \tag{1}$$ $$P(1,y)\implies f(1+yf(1))=f(1)+f(y)$$ $$1+yf(1)=y \iff y= \frac{1}{1-f(1)}, f(1) \ne1$$ $$P\left(1,\frac{1}{1-f(1)}\right) \implies f(1)=0$$ This gives a contradiction with $(1)$ that leads to $f(1)=1$. $$P(1,x) \implies f(x+1)=f(x)+1$$ $$P(x,y+1) \implies f(x+yf(x)+f(x))=f(x)+xf(y)+x=f(x+yf(x))+x \tag{2}$$ $$x+yf(x)=0 \iff y=\frac{-x}{f(x)}, x \ne 0$$ Substituting such $y$ in $(2)$ we get $$f(f(x))=x, x \ne 0 \tag{3}$$ this and the fact that $f(0)=0$ makes the above equation true for all $x$.

Getting back to equation $(2)$, we know that $x+yf(x)$ can attain any real value $b$ by $y=\frac{b-x}{f(x)}$ for any $x \ne 0$, so we can replace $x + yf(x)$ with just $y$: $$f(x + yf(x) + f(x))=f(x + yf(x))+x\iff f(y+f(x))=f(y)+x$$ replacing $x$ with $f(x)$, and using $(3)$ we get $f(x + y) = f(x) + f(y)$,

and so $P(f(x), y) \overset{(3)}{\implies} f(xy)=f(x)f(y)$, so $f$ is additive and multiplicative, which classically give $f(x)=x$ or $f(x)=0$.

Anas A. Ibrahim
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  • 1+yf(1)=y. How did you arrive at this from the previous equation? – Aayam Mathur May 22 '20 at 13:12
  • Also could you pleassee elaborate on what happened in equation (2)? I did not understand how we got to that conclusion of x+yf(x) being 0 – Aayam Mathur May 22 '20 at 13:14
  • $f(x)=0$ works also – Ross Millikan May 22 '20 at 15:38
  • @Ross Millikan It is already added as a solution if you read it carefully – Anas A. Ibrahim May 22 '20 at 17:13
  • @aayam mathur these aren't conclusions, these are mere selections of substitutions to force the cancellation – Anas A. Ibrahim May 22 '20 at 17:14
  • For example $f(1+yf(1))=f(y)+f(1)$, I wanted to select a $y$ such that $f(1+yf(1))$ would cancel $f(y)$, so I set $y=1+yf(1)$ to make such substitution. And I know that $f(1)=1$, I just needed to prove it, so this substitution shows that $f(1)=0$ while $0$ is a unique zero of $f$, a contradiction. So when I made the $y$ substitution, I divided by $f(1)-1$ assuming that $f(1) \ne 1$, that contradiction proves that it does equal – Anas A. Ibrahim May 22 '20 at 17:16
  • In equation $(2)$, I used the fact that $f(x+1)=f(x) + $ from the previous equation and substituted $y$ with $y+1$ and expanded. Now, we know from the original equation that $f(x)+xf(y)=f(x+yf(x))$ so I substituted that and ended with that expression. Now, to prove the surjectivity, I wanted an expression of $f($something$)=x$, so, since the expression $x+yf(x)$ can attain any real value, I chose to eliminate it by some substitution and set it equal to zero, and I did and ended up with $f(f(x))=x, x \ne 0$ – Anas A. Ibrahim May 22 '20 at 17:27
  • Hope this clarifies everything for you. Surjectivity is a main thing here by the way, you have to understand it beforehand. – Anas A. Ibrahim May 22 '20 at 17:29
  • Ahh i see so those were in fact clever substitutions to cancel out stuff. So we basically we confirm ourselves that the function has some value or property and then substitute accordingly to get the desired result ? So to be a able to solve an FE i should have a general outline of how my solution would be so that i can go over each claim step by step and prove them leading to the answer right? – Aayam Mathur May 22 '20 at 17:39
  • Yes, exactly. But not always. Sometimes you can't notice a solution or the algebra becomes very hairy when you try to figure it out. So you keep doing the usual substitutions like $0$ or $1$ or eliminating until it sounds familiar to you what the function should look like. Then you work towards proving some stuff to prove it's the only solution(s) – Anas A. Ibrahim May 22 '20 at 17:49
  • Hmm thanks a lot! You're a nice person! – Aayam Mathur May 22 '20 at 18:27
  • You could get $ f ( - 1 ) = - 1 $ by letting $ x = - 1 $ in the equation $ f ( x + 1 ) = f ( x ) + 1 $. I suggest you edit your answer using this fact, so that it becomes simpler. – Mohsen Shahriari May 22 '20 at 20:54
  • @MohsenShahriari It may be come this way, but I just want to show how useful the technique of forced cancellation might be. You learn nothing new from an answer if you saw it do it that way. – Anas A. Ibrahim May 22 '20 at 21:03
  • Got it. Nice answer anyway. – Mohsen Shahriari May 23 '20 at 06:37
  • @AnasA.Ibrahim I am not obtaining Final Result from $P(-1,-x-1)$. This is what is did $f(-1-(x+1)f(-1))=f(-1)+(-1)f(-x-1)$ $\implies$ $f(x)=-1-f(-x-1)$. After this I am stuck. And Why did you think $f(m)=-1$ for some $m$. What was motivation. It won't come to my mind. – mathophile Feb 06 '22 at 12:39
  • @mathophile I realized the error in my solution. I edited the answer, thanks. – Anas A. Ibrahim Feb 11 '22 at 15:15