How can I solve $f(t^2+u)=tf(t)+f(u)$

Question

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.

Answer 1

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. $\Box$

Answer 2

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.

Answer 3

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 $t\in\mathbb{R}$.