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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.

Mohsen Shahriari
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4 Answers4

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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 $x<0$.

Servaes
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  • ok,but how will you prove for all real values ,induction can't be used ,can it? – Albus Dumbledore Sep 18 '20 at 17:33
  • @Quantum What do you mean? Identity $(1)$ holds for all real $x>0$ and all natural numbers $n$, by induction on $n$. – Servaes Sep 18 '20 at 17:59
  • The natural part is clear, my question is how you used induction to prove for all reals, agreed it is intuitively correct , but i think a more rigorous proof is required for the same – Albus Dumbledore Sep 18 '20 at 18:01
  • @Quantum Again I do not understand. For every real number $x>0$ you have $$f(x)=\sqrt{x}f(\sqrt{x})=x^{2^{-1}}f(x^{2^{-1}}),$$ and so by induction on $n$ it follows that $$f(x)=\left(\prod_{i=1}^nx^{2^{-i}}\right)f(x^{2^{-n}})=x^{\sum_{i=1}^n2^{-i}}f(x^{2^{-n}}),$$ where $x^{\sum_{i=1}^n2^{-i}}=x^{1-2^{-n}}$. – Servaes Sep 18 '20 at 18:10
  • Ok i get it ,somehow i thought induction was restricted to natural numbers ,Anyway nice answer!! – Albus Dumbledore Sep 18 '20 at 18:14
  • @Quantum It is induction on natural numbers. – Servaes Sep 18 '20 at 18:18
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: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

Albus Dumbledore
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    Indeed $f(a) + f(b) = f(a+b)$ for all $a, b>0$ already implies $f(r) = rf(1)$ for all positive rational $r$. Then you have shown $f(x) = xf(1)$ for all $x\ge 0$ by continuity. – Arctic Char Sep 18 '20 at 17:37
  • @ArcticChar please see now,have editted – Albus Dumbledore Sep 18 '20 at 17:41
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    @ArcticChar We aslo have $xf(x)=f(x^2)=-xf(-x)$ so $f(-x)=-f(x)$. Thereby additivity holds throughout. -- But I wonder if we can get rid of continuity as a condition; after all, the original equation is somewhat stronger. We certainly get a pythagorean field (closure under $\sqrt.$) where $f(x)=2x$ instead of just $\Bbb Q$, but perhaps even more, even $\Bbb R$? – Hagen von Eitzen Sep 18 '20 at 20:06
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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

Soumyadwip Chanda
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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 $.

Mohsen Shahriari
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