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Find all functions $ f : \mathbb R \to \mathbb R $ that satisfy $$ f \left( x ^ 2 + 2 y f ( x ) \right) + f \left( y ^ 2 \right) = f ( x + y ) ^ 2 $$ for all $ x , y \in \mathbb R $.

I tried a couple of standard approaches (for example, making $ f ( x ) = 0 $, or $ f ( x ) = 1 $), but non of them yielded a result. Could someone help me?

Mohsen Shahriari
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aksjdnkajsnd
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2 Answers2

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You can show that the only functions $ f : \mathbb R \to \mathbb R $ satisfying $$ f \left( x ^ 2 + 2 y f ( x ) \right) + f \left( y ^ 2 \right) = f ( x + y ) ^ 2 \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $, are the identity function $ f ( x ) = x $, and the constant functions $ f ( x ) = 0 $ and $ f ( x ) = 2 $. It's straightforward to verify that these are indeed solution. We try to prove the converse.

Setting $ y = 0 $ in \eqref{0} gives $$ f \left( x ^ 2 \right) + f ( 0 ) = f ( x ) ^ 2 \text , \tag 1 \label 1 $$ while plugging $ x = 0 $ in it shows $$ f \big( 2 y f ( 0 ) \big) + f \left( y ^ 2 \right) = f ( y ) ^ 2 \text . \tag 2 \label 2 $$ We can see that $ f ( 0 ) \in \{ 0 , 2 \} $ by putting $ x = 0 $ in \eqref{1}. If $ f ( 0 ) = 2 $, comparing \eqref{1} and \eqref{2}, we get that $ f $ is the constant function with the value $ 2 $, which is one of the mentioned solutions. So, from now on, we consider the case $ f ( 0 ) = 0 $, which in turn makes both \eqref{1} and \eqref{2} equivalent to $$ f \left( x ^ 2 \right) = f ( x ) ^ 2 \tag 3 \label 3 $$ for all $ x \in \mathbb R $. Assume that there is some $ a \in \mathbb R \setminus \{ 0 \} $ with $ f ( a ) = 0 $. Without loss of generality, we can assume that $ a > 0 $; since by \eqref{3} we have $ f ( - a ) ^ 2 = f \left( ( - a ) ^ 2 \right) = f \left( a ^ 2 \right) = f ( a ) ^ 2 = 0 $, and we could take $ - a $ instead of $ a $. Letting $ x = a $ in \eqref{0} and using \eqref{3} we have $ f ( y + a ) ^ 2 = f ( y ) ^ 2 $ for all $ y \in \mathbb R $. Using this and induction, we get $ | f ( x + n a ) | = | f ( x ) | $ for all $ x \in \mathbb R $ and all $ n \in \mathbb Z $, and in particular $ f ( n a ) = 0 $ for all $ n \in \mathbb Z $. Given this, we can substitute $ x + n a $ for $ x $ and $ - x $ for $ y $ in \eqref{0} and use \eqref{3} to get $ f \left( ( x + n a ) ^ 2 - 2 x f ( x + n a ) \right) + f ( x ) ^ 2 = 0 $. Note that \eqref{3} shows that $ f $ takes nonnegative values at nonnegative points. Thus, for any $ x \in \mathbb R $, if we choose $ n $ large enough, namely $ n \ge \frac { \left| \sqrt { 2 | x f ( x ) | } - x \right| } a $, then both the terms on the left-hand side of the last equation will be nonnegative, and thus both of them must be equal to $ 0 $. This shows that in this case, $ f $ must be the constant function with the value $ 0 $, which is another one of the mentioned solutions. Lastly, consider the other case in which $ 0 $ is the only point at which $ f $ takes the value $ 0 $. Letting $ y = - \frac x 2 $ in \eqref{0} we have $ f \left( x ^ 2 - x f ( x ) \right) = 0 $, and hence $ x ^ 2 - x f ( x ) = 0 $, which gives $ f ( x ) = x $ for all $ x \in \mathbb R \setminus \{ 0 \} $. As we also had $ f ( 0 ) = 0 $, $ f $ must be the identity function, which is the remaining mentioned solution, and we're done.

Mohsen Shahriari
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Setting $y = 0$ yields $$f(x^2) + f(0) = f(x)^2$$

Setting $x = 0$ yields $$f(2f(0)\cdot y) + f(y^2) = f(y)^2$$

Set $x = y = t$ then subtract the two equations to get that

$$f(2f(0)\cdot t) = f(0)$$ for all $t$. This means that $f$ must be a constant function or $f(0) = 0$. Assuming it's a constant function, and plugging in $f(x) = c$ into the original functional equation yields $$2c = c^2$$ for which the only solutions are $c = 0$ and $c = 2$. So the solutions from these are $f(x) = 0$ and $f(x) = 2$

If $f(0) = 0$, then the condition from setting $y = 0$ is $$f(x^2) = f(x)^2$$ for which the only solutions are functions of the form $f(x) = x^c$. Set $y = -x$ and plug in $f(x) = x^c$ to get

$$(x^2-2x^{c+1})^c+x^{2c}=0$$

The only value of $c$ that this holds true for all $x$ is $c = 1$. So the only solution with $f(x)$ not constant is $f(x) = x$.

Thus the set of solutions is $$f(x) = \{ 0, 2, x \}$$

Maffred
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Varun Vejalla
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    Why does $f(x^2) = f(x)^2$ imply $f(x) = x^c$? – Maffred Apr 07 '21 at 12:52
  • @Maffred There might be a simpler way to do it, but if you let $f(x) = \exp\left( 2^{g(\log_2(\ln(x)))} \right)$, then the recurrence is $g(1+\log_2(\ln(x))) = g(\log_2(\ln(x)))+1 \to g(t+1) = g(t)+1 \to g(t) = t+c$, which means $f(x) = x^{\exp\left(2^c\right)} = x^C$ – Varun Vejalla Apr 07 '21 at 14:30
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    @VarunVejalla The last step of your argument in the last comment is flawed, and in fact your claim about $ f \left( x ^ 2 \right ) = f ( x ) ^ 2 $ implying $ f ( x ) = x ^ c $ is wrong (unless we have more restrictions on $ f $, for example its continuity, which is not an assumption here). Note that your argument only guaranties $ g ( t + n ) = g ( t ) + n $ for integer values of $ n $. For a counterexample, consider any additive function $ g $ with $ g ( 1 ) = 1 $. (to be continued) – Mohsen Shahriari Apr 10 '21 at 20:47
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    (continued) There are uncountably many wild functions satisfying this. For reference, see here. These aside, in the case of the current problem, we have additional information which lets us prove $ f ( x ) = x $ where you've used the wrong claim. I've posted an answer accordingly. – Mohsen Shahriari Apr 10 '21 at 20:47
  • @MohsenShahriari That's true. I suppose I was assuming that $f(x)$ was sufficiently nice. – Varun Vejalla Apr 12 '21 at 21:07