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Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R} $ satisfying $f(x-f(y)) = f(f(y)) + xf(y) + f(x) - 1$

I have solved this by doing the substitution $x \rightarrow 2f(y)$ and then replacing $x$ back. I get $f(x) = 1 - \frac{x^2}{2}$

Now, I want to find all the other functions satisfying the equation or prove that this is the only solution. This is just like the question I have already asked, with the difference being that this functional equation has two variables alongwith some nested functions.

From the approach in the answer to that question, I tried assuming another function $g$ satisfying the equation and then try to prove that $h = f - g$ is a constant function with value $0$. But, I am unable to do that in this case.

Also, there aren't many constraints on the function except it being from $\mathbb{R} \rightarrow \mathbb{R} $. How can I even approach the problem in this case? Are there any methods involving calculus?

Raahil Rai
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1 Answers1

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I realized that my previous answer was incorrect, since $f$ may not be onto (in fact, it actually isn't). I have an alternate way to find $f$ which doesn't require you to show uniqueness.

Substituting $x=f(y)$ in original equation: $$f(x)=\frac{1+f(0)-x^2}{2} \qquad (1).$$

Substituting $x=y=0$ in original equation: $$f(-f(0))=f(f(0))+f(0)-1,$$ therefore $f(0)\neq 0.$

Substituting $y=0$ in original equation: $$f(x-f(0))=f(f(0))+xf(0)+f(x)-1,$$ that is $$f(x-f(0))-f(x)=f(f(0))+xf(0)-1.$$ This shows $$\{f(x-f(0))-f(x): x \in \mathbb R\}=\mathbb R.$$

Finally, let $x\in \mathbb R,$ then there exists $x_1 \in f(\mathbb R),x_2 \in \mathbb R$ such that $x=x_1-f(x_2).$ So

$$f(x)=f(x_1-f(x_2))=f(f(x_2))+x_1f(x_2)+f(x_1)-1.$$ This allows us to now apply equation $(1).$ After simplyifying you obtain: $$f(x)=f(0)-\frac{1}{2}\left(x_1-f(x_2)\right)^2=f(0)-\frac{x^2}{2}.$$

It is also easy to see that $f(0)=1$ and hence $f(x)=1-\frac{x^2}{2}.$

Sahiba Arora
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  • I have understood your answer. Can you just elaborate on how your approach of solving also proves uniqueness. Also, why does'nt substituting $x \rightarrow 2f(y)$ and solving justify that $f(x) = 1 - \frac{x^{2}}{2}$ is a unique solution. I am new to functional functional equations. Thank you. – Raahil Rai Apr 27 '20 at 14:44
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    @RaahilRai My proof is for an arbitrary $x \in \mathbb{R}.$ If you take $x=2f(y).$ Then it is only for those $x$ which are of this form. Since $f$ need not be onto, it is not sufficient. That's why my previous answer was incorrect. I had only shown $h'(g(y))=0$ for all $y.$ – Sahiba Arora Apr 27 '20 at 15:12