A weird equation

This is a really hard functional equation by me:

Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$f(x^2+f(y)^2)=xf(x)+yf(y)$ for all $x,y \in \mathbb{R}$

It looks quite simple but it's very difficult. Good luck if you try it!

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Some preliminary observations that might be useful. Will post more things as I keep progressing through.

Switch $x$ and $y$ around to get

$f(x^2+f(y)^2)=x f(x)+y f(y) = f(y^2 + f(x)^2)$

Substitute $y=0$ to get

$f(x^2+f(0)^2)=f(f(x)^2)$

Sharky Kesa - 4 years, 5 months ago

simple observations reveals f(x)=x

hiroto kun - 4 years, 5 months ago

Find all means Find all and prove that you have all of them

Wen Z - 4 years, 3 months ago

Browsing old problems on Sharvil's feed...

Sub $x=y=0$ to get $f(f(0)^2)=0$ , and then sub $y=f(0)^2$ to get $f(x^2)=xf(x)\Longrightarrow f(0)=0$ . Now subbing $x=f(t),y=0$ gives $f(f(t)^2)=f(t)f(f(t))$ . But subbing $x=0,y=t$ gives $f(f(t)^2)=tf(t)$ . So if $f(t)\neq 0$ , then $f(f(t))=t$ .

Now suppose $f(u)\neq 0$ for some $u$ . Since $f(u^2)=uf(u)$ and clearly $u\neq 0$ , we have $f(u^2)\neq 0$ . Hence, $f(f(u^2))=u^2$ . But recall $f(u^2)=uf(u)=f(f(u)^2)$ , so $u^2=f(f(f(u)^2))$ . Clearly $f(f(u)^2)\neq 0$ otherwise $u=0$ , so $f(f(f(u)^2))=f(u)^2$ , and thus $f(u)=\pm u$ . So

$f(x)=\begin{cases}x\\-x\\0\end{cases}$

for each $x\in \mathbb{R}$ .

Now suppose $f(u)=0$ for some $u\neq 0$ . Then subbing $x=u$ gives $f(u^2+f(y)^2)=yf(y)$ for all $y\in \mathbb{R}$ . If $f(y)\neq 0$ for some $y$ , we would have $f(y)=\pm y$ and thus $yf(y)=\pm y^2 \neq 0$ so that $f(u^2+f(y)^2)=\pm \left(u^2+f(y)^2\right)$ , and $u^2+f(y)^2=\pm y^2\Longrightarrow u=0$ or $u^2=-y^2<0$ , a contradiction. So $f(y)=0$ for all $y\in \mathbb{R}$ .

Finally, suppose $f(u)=u$ and $f(v)=-v$ for some $u,v\neq 0$ . Then subbing $x=u,y=v$ gives $f(u^2+v^2)=u^2-v^2\neq 0\Longrightarrow \pm \left(u^2+v^2\right)=u^2-v^2\Longrightarrow u=0$ or $v=0$ , a contradiction.

So there are three possible solutions, namely $f(x)=x$ for all $x\in \mathbb{R}$ , $f(x)=-x$ for all $x\in \mathbb{R}$ , and $f(x)=0$ for all $x\in \mathbb{R}$ . It is easy to check that all three are valid solutions of the given equation.

Miles Koumouris - 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$