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This is one of my practice problems, which may be solved with a substitution.

Find all functions $ f : \mathbb R \to \mathbb R $ which verify the relation $$ ( x - 2 ) f ( y ) + f \big( y + 2 f ( x ) \big) = f \big( x + y f ( x ) \big) $$ for all $ x , y \in \mathbb R $.

I have tried to find whether such function is surjective, or injective, but unsuccessfully.

Mohsen Shahriari
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Emomali Murodzoda
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  • Hint: the constant zero function satisfies the equation, and is neither surjective nor injective. So you won't be able to do that. – Mohsen Shahriari Dec 17 '20 at 19:38
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    As you're new to the site, I think you'd find these helpful. Isolated questions without any context and showing your efforts are not welcome on this site. Without editing your question and showing what you've done and where exactly you're stuck, other may not find your question suitable for being answered, an the question may get closed by the reviewers. Please put some more time and effort in solving the problem, edit the question and add your attempt, and maybe then you''ll get some good answers. – Mohsen Shahriari Dec 17 '20 at 19:44
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    Checking out these links might be useful: How to ask a good question and How to ask a homework question. – Mohsen Shahriari Dec 17 '20 at 19:46
  • Hint: $ f ( x ) = x - 1 $ satisfies the equation. – Mohsen Shahriari Dec 17 '20 at 21:00
  • I have edited your post. Your comment and a part from the title are now included in the post itself, so that the title and the post make more sense, and the attempt you made is included for the users to see. I also improve the mathJax part. You can check out MathJax basic tutorial and quick reference for learning to use it better in your future posts. I hope my edits give you an idea about how a good question on MSE should look like. Hope you find MSE a good place, and see you more in the future. – Mohsen Shahriari Dec 22 '20 at 11:19

1 Answers1

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You can show that the only functions $ f : \mathbb R \to \mathbb R $ satisfying $$ ( x - 2 ) f ( y ) + f \big( y + 2 f ( x ) \big) = f \big( x + y f ( x ) \big) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $ are those of the form $ f ( x ) = 0 $ and $ f ( x ) = x - 1 $. It's straightforward to verify that these are indeed solutions. To see the converse, let $ a = f ( 0 ) $ and $ b = f ( 2 ) $. Setting $ y = 0 $ in \eqref{0} gives $$ a ( x - 2 ) = f ( x ) - f \big( 2 f ( x ) \big) \text , \tag 1 \label 1 $$ while putting $ x = 2 $ in \eqref{0} shows that $$ f ( y + 2 b ) = f ( b y + 2 ) \text . \tag 2 \label 2 $$ By respectively setting $ x = y + 2 b $ and $ x = b y + 2 $ in \eqref{1} and using \eqref{2} you get $ a ( y + 2 b - 2 ) = a b y $, which for $ y = 3 $ can be simplified to $ a ( b - 1 ) = 0 $.

Now, assume that for some $ x _ 0 \ne 1 $ we have $ f ( x _ 0 ) = 0 $. Setting $ x = x _ 0 $ in \eqref{0} you can see that $ f $ will be the constant zero function, which is one of the mentioned solutions. This, in particular, happens if we take $ x _ 0 = 0 $, which settles the case $ a = 0 $. So, from now on, assume that we have $ b = 1 $. This, in particular, shows that you can't have $ a = 0 $, as then $ f $ must be constantly zero, which contradicts $ b = 1 $.

Consider any $ x _ 1 $ with $ f ( x _ 1 ) = 1 $. By the facts that $ b = 1 $ and $ a $ cannot be equal to zero, you can put $ x = x _ 1 $ in \eqref{1} to see that you must have $ x _ 1 = 2 $. This means that for any $ x \ne 2 $ you have $ f ( x ) \ne 1 $, and you can let $ y = \frac { 2 f ( x ) - x } { f ( x ) - 1 } $ in \eqref{0}. In that case, you'll have $ y + 2 f ( x ) = x + y f ( x ) = \frac { 2 f ( x ) ^ 2 - x } { f ( x ) - 1 } $, and since $ x \ne 2 $, \eqref{0} gives you $ f \left( \frac { 2 f ( x ) - x } { f ( x ) - 1 } \right) = 0 $. As $ f $ is not constantly zero, by the above argument we must have $ \frac { 2 f ( x ) - x } { f ( x ) - 1 } = 1 $, which can be simplified to $ f ( x ) = x - 1 $. We also have $ b = 2 - 1 $, and hence we can conclude that $ f ( x ) = x - 1 $ for all $ x \in \mathbb R $, which gives the other mentioned solution, and we're done.

Mohsen Shahriari
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