Continuous functions $f$ that satisfy $f(1)=2$ and $f\left(x^2+y^2\right)=xf(x)+yf(y)$ for all $x,y\in\mathbb{R}$ Original Question:

Find the continuous function $f:\mathbb{R}\to\mathbb{R}$ that satisfies $f(1)=2$ and
$$f\left(x^2+y^2\right)=xf(x)+yf(y)\text,$$
for all $x,y\in\mathbb{R}$.

My Attempt:
I plugged in some numbers and got:

*

*$f(0)=0$

*$f(2)=f(1)+f(1)=4$

*$f(8)=2f(2)+2f(2)=16$
At this point, I conjectured that $f(x)=2x$, substituting it into the condition to verify:
$$2\left(x^2+y^2\right)=x\cdot2x+y\cdot2y$$
which is true.
My problem is that how do I make sure that this is the only possible $f$? And guessing doesn't seem like a proper solution here, so if you're willing to provide a standard solution, it will be much appreciated as well.
This problem may sound a little too simple, but I somehow can't get over it.
 A: First taking $y=0$ shows that for all $x\in\Bbb{R}$ you have
$$f(x^2)=xf(x)\qquad\text{ and }\qquad f(x^2)=f((-x)^2)=-xf(-x),$$
from which it follows that $f(x)=-f(-x)$ for all $x\in\Bbb{R}$. In particular $f(0)=0$, and for every real $x>0$ and every $n\in\Bbb{N}$ we have by induction that
$$f(x)=x^{1-2^{-n}}f(x^{2^{-n}}).\tag{1}$$
As $f$ is continuous it follows that
$$f(x)=\lim_{n\to\infty}x^{1-2^{-n}}f(x^{2^{-n}})=xf(1)=2x.$$
It quickly follows that also $f(x)=2x$ if $x<0$.
A: :setting $y=0$ we have $f(x^2)=xf(x)$
thus $f(x^2)+f(y^2)=f(x^2+y^2)$
$f(a)+f(b)=f(a+b)$ which is cauchy...
$f(x)=kx$ or $k=2$
you can read https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation
A: If $f(x)$ is differentiable:
$$f\left(x^{2}+y^{2}\right)=xf\left(x\right)+yf\left(y\right)$$
Partial differentiation w.r.t. x
$$2xf'\left(x^{2}+y^{2}\right)=xf'\left(x\right)+f\left(x\right)$$
Put $x=1$ and $y=0$ to get $f'(1)=2$
Now, put $x=1$ in that equation. We have:
$$2f'\left(1+y^{2}\right)=f'\left(1\right)+f\left(1\right)=4$$
Which implies $$f'\left(X\right)=2$$
For $X=y^2+1$
Hence, $$f(X)=2X$$
This proves that there the only differentiable function that satisfies the functional rule is $f(x)=2x$
This is probably my first mathematically rigorous answer on Math SE
A: Even without the assumption of continuity, you can show that a function $ f : \mathbb R \to \mathbb R $ satisfies
$$ f \left( x ^ 2 + y ^ 2 \right) = x f ( x ) + y f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ iff there is a constant $ a \in \mathbb R $ such that $ f ( x ) = a x $ for all $ x \in \mathbb R $. It's easy to check that functions of this form are solutions. We try to prove the converse. (The additional assumtion $ f ( 1 ) = 2 $ forces $ a = 2 $ and thus the only solution will be $ f ( x ) = 2 x $.)
Letting $ y = 0 $ in \eqref{0}, we have
$$ f \left( x ^ 2 \right) = x f ( x ) \text , \tag 1 \label 1 $$
which for $ x = 0 $ yields $ f ( 0 ) = 0 $. Using \eqref{1}, we can rewrite \eqref{0} as $ f \left( x ^ 2 + y ^ 2 \right) = f \left( x ^ 2 \right) + f \left( y ^ 2 \right) $, which means that we have $ f ( x + y ) = f ( x ) + f ( y ) $ for all $ x , y \in \mathbb R ^ { 0 + } $. Thus, letting $ a = f ( 1 ) $ and using \eqref{1}, for any $ x \in \mathbb R ^ { 0 + } $ we get
$$ x f ( x ) + 2 f ( x ) + a = f \left( x ^ 2 \right) + \big( f ( x ) + f ( x ) \big) + f ( 1 ) \\
= f \left( x ^ 2 + 2 x + 1 \right) = f \left( ( x + 1 ) ^ 2 \right) = ( x + 1 ) f ( x + 1 ) \\
= ( x + 1 ) f ( x ) + a ( x + 1 ) \text , $$
which shows that $ f ( x ) = a x $ when $ x \ge 0 $. Therefore, by \eqref{1} we have $ x f ( x ) = a x ^ 2 $ for all $ x \in \mathbb R $, and hence $ f ( x ) = a x $ for every $ x $.
