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Find all functions over $\mathbb{R}$ such that $$f(t^2+u)=tf(t)+f(u)$$

My progress

Setting $(x,y)$ in the equation where $x,y \geq0$. We have$$f(x+y)=\sqrt{x}f(\sqrt{x})+f(y).\tag{1}$$ Now, setting $(y,x)$ in the equation we have, $$f(y+x)=\sqrt{y}f(\sqrt{y})+f(x). \tag{2} $$Subtracting $(1)$ from $(2)$ we get $$f(x)-\sqrt{x}f(\sqrt{x})=f(y)-\sqrt{y}f(\sqrt{y})$$ This implies that $f(x)-\sqrt{x}f(\sqrt{x})$ is constant for all $x \geq 0$ or $f(x)=\sqrt{x}f(\sqrt{x})+c$ for some constant $c$. Combining this fact with $(1)$ the functional equation can be reduced to $$f(x+y)=f(x)+f(y)+c.\tag{3}$$ for $x,y \geq 0.$ The solution to this Cauchy's functional equation can be found to $f(x)=kx+c(x-1)$ where $k$ is $f(1)$. Now putting $(1,0)$ in the original functional equation we get $$f(1)=f(1)+f(0)$$This implies $f(0)=0$ and the solution to this functional equation is $f(x)=kx$ for all non-negative reals.

Now my question is: how do we extend this to negative reals? (Any hints shall be appreciated.)

Has
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  • Let me know if anything here or here answers your question. The extension is provided by the fact that $f$ can be shown to be odd, which is covered in a few answers. Marking this as a duplicate will make it a useful reference point. – Sarvesh Ravichandran Iyer Mar 13 '22 at 08:27

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If $u <0$ there exist $t>0$ such that $t^{2}+u >0$. So $f(u)=f(t^{2}+u)-tf(t)=k(t^{2}+u)-kt^{2}=ku$.

Kavi Rama Murthy
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