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Solve $f(t^2+u)=tf(t)+f(u)$ on $\mathbb{R}$.

My solution is, if we take $u=t$, then $f(t^2+t)=(t+1)f(t)$ and let $g(x)=\frac{f(x)}{x}$, then $g(t^2+t)=g(t),(t\neq-1,0)$. It means $g(x)=k,\quad k\in\mathbb{R}\tag{*}$ Therefore $\boxed{f(x)=kx\quad\forall k,x\in\mathbb{R}}$. I'm not sure about $(*)$ is true.

Alexander Chew
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Mutse
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  • The leap from $g(t^2+t)=g(t)$ to (*) is unjustified. – Ѕᴀᴀᴅ Apr 21 '20 at 08:04
  • You don't need to distinguish between $g(x) = 0$ and $g(x) = k$ since $k \in \mathbb{R}$, which includes $k = 0$. – Carl Apr 21 '20 at 08:15
  • We can prove that this equation is equivalent to: $$f(x+y)=f(x)+f(y)\qquad\text{and}\qquad f(x^2)=xf(x)\qquad\forall x,y\in\mathbb{R}.$$ Similar to the case of Cauchy functional equation, I suspect that this equation may admit some non-trivial solutions other than those of the form $f(x)=kx$. – Sangchul Lee Apr 21 '20 at 08:27

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I want to succeed the answer given before me by using Cauchy's equation in an actual solution. Let the given assertion be $P(t,u)$ $$P(1,0)\implies f(1)=f(1)+f(0) \Leftrightarrow f(0)=0$$ $$P(t,0) \implies f(t^2)=tf(t)+f(0)=tf(t)\tag{1}$$ Substituting $x=t^2$ and $(1)$ in $P(t,u)$ $$P(t,u) \implies f(t^2+u)=tf(t)+f(u)=f(t^2)+f(u)$$ $$\implies f(x+u)=f(x)+f(u) \tag{2}$$ But this holds for all $x \geq 0$, so $$P(-t,0) \implies f(t^2)=-tf(-t)$$ Now combining that with $(1)$ we get $$f(-t)=-f(t)$$ Or that $f$ is odd. So, substituting $x$ with $-x$ in $(2)$ $$f(u-x)=f(u)+f(-x)=f(u)-f(x) \text{ } \forall x \geq 0$$ or in other words $$f(u+x)=f(u)+f(x) \text{ } \forall x \le 0$$

The fact that $x$ works $(2)$ implies that $-x$ also works in $(2)$ or the fact that $f$ is odd extends $(2)$ to work for all $x,u \in \mathbb{R}$.

Now, we're going to use $(2)$ many times by making partitioning sums inside functions (I mean $f(a_1+a_2+\dots+a_n)=f(a_1)+f(a_2)+\dots+f(a_n)$ for any positive integer $n$ and variables $a_i$) $$P(t+1,0)\implies f(t^2+2t+1)=(t+1)f(t+1)$$ $$\Leftrightarrow f(t^2)+f(2t)+f(1)=(t+1)(f(t)+f(1))$$ and by $(1)$ and distributing the right-hand side, $$\Leftrightarrow tf(t)+f(2t)+f(1)=tf(t)+tf(1)+f(t)+f(1)$$ $$\Leftrightarrow f(2t)=tf(1)+f(t) $$ and since $f(2t)=f(t+t)=f(t)+f(t)=2f(t)$ $$f(2t)=tf(1)+f(t) \Leftrightarrow 2f(t)=tf(1)+f(t)$$ $$\Leftrightarrow f(t)=tf(1)$$

Putting $f(1)=k$ and substituting it in the original equation, we get that it applies for any $k,t,u \in \mathbb{R}$, hence the conclusion. $\Box$

Anas A. Ibrahim
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Given: $$f(t^2+u)=tf(t)+f(u) \tag{1}$$

Substituting $t = 1$ and $u = 0$ into (1), we get $$f(1)=f(1) + f(0) \Rightarrow f(0)=0\tag{2}$$

Substituting $u = 0$ and (2) into (1), we get $$f(t^2)=t f(t)\tag{3}$$

Substituting (3) and $v = t^2$ into (1), we get $$f(u+v)=f(u)+f(v) \tag{4}$$

Note that (4) holds only when $v = t^2 \ge 0$. To incorporate negative values, note that by substituting $t = -x$ in (3),

\begin{align} f((-x)^2) &= (-x) f(-x) \\ \Rightarrow f(x^2) &= -x f(-x) \\ \Rightarrow x f(x) &= -x f(-x) \qquad \textrm{from (3)} \\ \Rightarrow f(-x) &= -f(x) \end{align}

Thus (4) holds for all $u,v\in\mathbb{R}$.

Equation (4) is Cauchy's functional equation and under some light assumptions (for example, continuity at one point), the only solution is $f(x) = cx$ for $c\in\mathbb{R}$.

See this SE post for a lot more detail on the solutions to Cauchy's functional equation.

Carl
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    For this functional equation, you don't need extra assumptions for Cauchy's FE. Let $k=f(1)$. From $(3)$ and $(4)$, $$\begin{align}(t+1)\big(f(t)+k\big)&=(t+1)f(t+1)=f\big((t+1)\big)^2\&=f(t^2+2t+1)=f(t^2)+2f(t)+k\&=tf(t)+2f(t)+k.\end{align}$$ This means $f(t)=kt$. – Batominovski Apr 21 '20 at 18:39
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    Perhaps, for $(4)$ you should say that $$f(u+v)=f(u)+f(v)\ \ \forall u\in\Bbb R\ \ \forall {\color{red}{v\geq 0}}.\tag{4'}$$ (Because you proved this result by setting $v=t^2$.) To allow $v<0$, you can use $(3)$ to show that $f(-x)=-f(x)$. Therefore, if $v<0$, we have by $(4')$ that \begin{align}f(u+v)&=-f\big(-(u+v)\big)=-f\big((-u)+(-v)\big)\&=-\big(f(-u)+f(-v)\big)=-f(-u)-f(-v)=f(u)+f(v). \end{align} Thus, $(4)$ indeed holds for all $u,v\in\Bbb R$. – Batominovski Apr 21 '20 at 18:43
  • Thanks, @WETutorialSchool I updated the answer to include the case where $v < 0$. – Carl Apr 22 '20 at 07:40
  • @WETutorialSchool I disagree that we don't need the Cauchy FE considerations here. You showed in your comment that $f(t) = kt$ is a solution for a particular $k$, namely $k = f(t)$. It does not follow that this is the only solution or even that it holds for general $k$. – Carl Apr 22 '20 at 07:43
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    If you read my first comment completely, you will see that I didn't use any knowledge about Cauchy's FE to derive $f(t)=kt$ $\color{red}{\text{for every }t\in\Bbb R}$, where $k=f(1)$. That comments gives a proof that such functions are the only solutions. That, $f(t)=kt$ is a solution for every choice of $k$ can be proven by plugging this back into the original FE to see which $k$ works (and it turns out: every $k$ does work). – Batominovski Apr 22 '20 at 07:45
  • @WETutorialSchool It looks like you should write up an answer to this question rather than making brief claims in the comments. It also doesn't help that you edit your last comment after I replied to it. Your last comment used to say that $k=f(1)$ is the only solution but now you claim something different. I don't see (from your comments) how you would prove this for every $k$ by "plugging it back into the original FE", in contrast to what you wrote before. I'm happy to delete this answer if you would like to replace it with a better one. – Carl Apr 22 '20 at 08:01
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    The substance of my previous comment did not change after the edit. I merely elaborated on what I said to make it clearer. I am not writing an answer because the same idea has already been written in Anas's answer. You should keep your answer, because Anas's answer is also incomplete (it doesn't prove $f(u+v)=f(u)+f(v)$ for all $u,v\in\mathbb{R}$ like you already handled it). So, together, I think the two answers make one complete answer to the question, and there is no need to add another answer. – Batominovski Apr 22 '20 at 08:39
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Here is a proof without any use of Cauchy's functional equation. First, as other users have observed: $f(0)=0$ and $$f(t^2)=t\,f(t)$$ for all $t\in\mathbb{R}$. Thus, if $x\geq 0$ and $y\in\mathbb{R}$, then $$f(x+y)=f\big((\sqrt{x})^2+y\big)=\sqrt{x}\,f(\sqrt{x})+f(y)=f\big((\sqrt{x})^2\big)+f(y)=f(x)+f(y)\,.$$

Therefore, for any $t\in\mathbb{R}$, we have $$f\big((t+1)^2\big)=f(t^2+2t+1)=f\big((t^2+t+1)+t\big)=f(t^2+t+1)+f(t)\,,$$ as $t^2+t+1>0$ for any $t\in\mathbb{R}$. Next, $$f(t^2+t+1)=f\big((t^2+1)+t\big)=f(t^2+1)+f(t)$$ because $t^2+1>0$ for all $t\in\mathbb{R}$. Finally, $$f(t^2+1)=f(t^2)+f(1)$$ because $t^2\geq 0$ for each $t\in\mathbb{R}$. Consequently, $$\begin{align}f\big((t+1)^2\big)&=f(t^2+t+1)+f(t)=\big(f(t^2+1)+f(t)\big)+f(t)\\&=\Big(\big(f(t^2)+f(1)\big)+f(t)\Big)+f(t)\,.\end{align}$$ Ergo, $$f\big((t+1)^2\big)=f(t^2)+2\,f(t)+f(1)\,.\tag{*}$$

On the other hand, $$f\big((t+1)^2\big)=(t+1)\,f(t+1)\,,$$ but $$f(t+1)=f(1+t)=f(1)+f(t)\,,$$ as $1>0$. Therefore, $$f\big((t+1)^2\big)=(t+1)\,\big(f(1)+f(t)\big)=t\,f(t)+t\,f(1)+f(t)+f(1)\,.$$ Because $f(t^2)=t\,f(t)$, we conclude that $$f\big((t+1)^2\big)=f(t^2)+t\,f(1)+f(t)+f(1)\,.\tag{#}$$ From (*) and (#), we conclude that $$f(t)=t\,f(1)$$ for all $t\in\mathbb{R}$.

Batominovski
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