# Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$

Let $\mathbb R$ denote the set of all real numbers. Find all function $f: R\to \ R$ such that $$f\left(x^2+f(y)\right)=y+(f(x))^2$$ It is the problem. I tried to it by putting many at the place of $x$ and $y$ but I can't proceed. Please somebody help me.

• $f(x)=x$ is a solution – hamam_Abdallah Oct 13 '16 at 8:13

Let me summarize the proposed ideas and give a self-contained answer: Let $f : \Bbb{R} \to \Bbb{R}$ satisfy the functional equation

$$f(x^2 + f(y)) = y + f(x)^2 \tag{*}$$

for all $x, y \in \Bbb{R}$.

Step 1. $f(0) = 0$.

(This solution is due to @Leo163.) Let $a \in \Bbb{R}$ be such that $f(a) = 0$. For instance, plugging $y = -f(x)^2$ to $\text{(*)}$ confirms the existence of such $a$. Then plugging $(x, y) = (a, a)$ to $\text{(*)}$ gives

$$f(a^2) = f(a^2 + f(a)) = a + f(a)^2 = a.$$

Finally, plugging $(x, y) = (0, a^2)$ to $\text{(*)}$ gives

$$0 = f(a) = f(f(a^2)) = a^2 + f(0)^2.$$

This shows that $f(0) = 0$.

Step 2. $f$ satisfies various identities:

1. $f(x^2) = f(x)^2$.
2. $f(f(x)) = x$. In particular, $f$ is bijective.
3. $f(-x) = -f(x)$.
4. $f(x+y) = f(x)+f(y)$.

The identity 1 (resp. 2) follow by plugging $y = 0$ (resp. $x = 0$) to $\text{(*)}$. For 3, we may assume that $x \geq 0$. Then replacing $(x, y)$ in $\text{(*)}$ by $(x^{1/2}, f(-x))$, we have

$$0 = f(x - x) = f(-x) + f(x^{1/2})^2 = f(-x) + f(x)$$

and 3 follows. Finally, replacing $(x, y)$ in $\text{(*)}$ by $(x^{1/2}, f(y))$ gives

$$f(x + y) = f(x) + f(y), \qquad \forall x \geq 0, \ y \in \Bbb{R}.$$

The restriction that $x \geq 0$ can be removed by combining this with 3: if $x < 0$, then

$$f(x+y) = -f(-x-y) = -(f(-x) + f(-y)) = f(x) + f(y).$$

Step 3. $f(x) = x$.

If $x > 0$ then by the first identity of the previous step,

$$f(x) = f((x^{1/2})^2) = f(x^{1/2})^2 > 0.$$

(We can exclude the equality because $f$ is bijective and $f(0) = 0$.) Combining this with the fact that $f$ is an odd function, we have:

$$f(x) \geq 0 \quad \Longleftrightarrow \quad x \geq 0.$$

Now for any $x \in \Bbb{R}$,

$$f(x) -x = f(x) - f(f(x)) = f(x - f(x)).$$

Therefore $f(x) - x \geq 0$ if and only if $x - f(x) \geq 0$, which implies that $f(x) = x$.

Remark. The Cauchy functional equation has pathological solutions which are also involution, and we need to use the extra structure given by the equation $\text{(*)}$ to exclude such possibilities. The first identity of Step 2 was essential in our solution.

Suppose $a$ is such that $f(a)=0$ (we know there is such $a$, it is enough to consider $y=-f(x)^2$ for some $x$), then substituting $x=a$ and $y=a$, we have $$f(a^2)=a.$$ Then, set $x=0$ and $y=a^2$, to obtain $$0=f(f(a^2))=a^2+f(0)^2.$$ Since we are only dealing with real numbers, it means that $f(0)=0$, and no other number has $0$ as image.

As consequences, we have $f(f(y))=y$ and $f(y^2)=f(y)^2$ for every $y\in\mathbb{R}$.

Here is a short approach without Cauchy equation. Starting from $f(0)=0$ by @Leo163, we obtain $f(f(x))=x$ by plugging $y=0$. This shows $f$ is a bijection. Now plug $y=f^{-1}(\zeta)$, we see $$f(x^2+\zeta)=f^{-1}(\zeta)+(f(x))^2\geq f^{-1}(\zeta)=f(\zeta)$$ Hence $f$ is monotone increasing. Now suppose for some $x_0$ we have $f(x_0)> x_0$, then $x_0=f(f(x_0))\geq f(x_0)> x_0$, a contradiction. Thus we have $f(x)\leq x$. Similarly we have $f(x)\geq x$, so $f(x)=x$.

• My solution is just a verbose version of this solution. Nice! (+1) – Sangchul Lee Oct 18 '16 at 13:14

Observe

1. $f(x^2+f(0)) = [f(x)]^2$
2. $f(f(x)) = x+[f(0)]^2$.

Plugging $0$ into expression 1 yields \begin{align} f_2(0):=f(f(0)) = [f(0)]^2 \end{align} and plugging $f(0)$ into expression 1 yields \begin{align} f(f(0)^2+f(0)) =[f_2(0)]^2 = [f(0)]^4. \end{align} Moreover, by expression 2, we have \begin{align} f(f(0)^2+f(0))= f([f(0)]^2)=f(f_2(0)) = f_3(0). \end{align} Hence we have \begin{align} f_3(0) = [f(0)]^4. \end{align}

Next, using both 1 and 2, we have \begin{align} f(f(x^2+f(0)))=x^2+f(0)+[f(0)]^2. \ \ (\ast) \end{align} Plugging $0$ into $(\ast)$ yields \begin{align} f_3(0) = f(0)+[f(0)]^2. \end{align} Combining everything yields \begin{align} [f(0)]^4= f_3(0) = f(0)+[f(0)]^2 \ \ \Rightarrow \ \ f(0)[f(0)^3-f(0)-1] =0. \end{align} Solving for $f(0)$ yields that either $f(0) = 0$ or $f(0)$ is a root of the polynomial $p(x) = x^3-x-1$, which only has one real root.

Moreover, observe \begin{align} f_4(0) = [f(0)]^4 \end{align} and \begin{align} f_4(0) = f(f(f_2(0))) = f_2(0) + [f(0)]^2 = 2[f(0)]^2 \end{align} which means \begin{align} [f(0)]^4-2[f(0)]^2 = 0. \end{align} Thus, $f(0)$ is also a root of $x^4-2x^2 = x^2(x^2-2)$.

In conclusion, $f(0)$ has to be $0$. Thus, $f(f(x)) = x$ and $f(x^2) = [f(x)]^2$.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$$\mrm{f}\pars{x^{2} + \mrm{f}\pars{y}} =y + \bracks{\mrm{f}\pars{x}}^{2} \label{1}\tag{1}$$

Deriving both members of \eqref{1} respect of $\ds{x}$ and $\ds{y}$ yield, respectively: $$\left\{\begin{array}{rcl} \ds{\mrm{f}'\pars{x^{2} + \mrm{f}\pars{y}}\pars{2x}} & \ds{=} & \ds{2\mrm{f}\pars{x}\mrm{f}'\pars{x}} \\[2mm] \ds{\mrm{f}'\pars{x^{2} + \mrm{f}\pars{y}}\mrm{f}'\pars{y}} & \ds{=} & \ds{1} \end{array}\right. \label{2}\tag{2}$$

By comparing both expressions in \eqref{2}: $${\mrm{f}\pars{x}\mrm{f}'\pars{x} \over x} = {1 \over \mrm{f}'\pars{y}} = \alpha\,, \qquad \pars{~\alpha\ \mbox{is independent of}\ x\ \mbox{and}\ y~} \label{2.a}\tag{2.a}$$
The second one is quite trivial: $${1 \over \mrm{f}'\pars{y}} = \alpha \implies \mrm{f}\pars{y} = {1 \over \alpha}\, y + \beta\,,\qquad \pars{~\beta\ \mbox{is a constant}~}\label{3}\tag{3}$$
\eqref{2.a} and \eqref{3} yield: \begin{align} &{\mrm{f}\pars{x}\mrm{f}'\pars{x} \over x} = \alpha \,\,\,\stackrel{\mrm{see}\ \eqref{3}}{\implies}\,\,\, {\bracks{x/\alpha + \beta}/\alpha \over x} = \alpha \implies {1 \over \alpha^{2}}\, x + {\beta \over \alpha} = \alpha x \implies \left\{\begin{array}{rcl} \ds{1 \over \alpha^{2}} & \ds{=} & \ds{\alpha} \\[2mm] \ds{\beta \over \alpha} & \ds{=} & \ds{0} \end{array}\right. \\[5mm] &\ \stackrel{\mrm{see}\ \eqref{3}}{\implies}\,\,\, \bbx{\ds{\mrm{f}\pars{x} = {x \over \alpha}\,,\qquad \alpha^{3} = 1}} \label{4}\tag{4} \end{align}
By inserting \eqref{4} into \eqref{1} we conclude that $\ds{\alpha = 1}$: $$\bbx{\ds{\mrm{f}\pars{x} = x}}$$

• I see no assumption that $f$ is continuous, much less differentiable. – Sangchul Lee Oct 18 '16 at 13:12
• @SangchulLee Thanks for your remark. I knew that from the very beginning. However, it's a valid proof whenever we assume the conditions $\left(\mathbf{C}\ \mbox{and}\ \mathbf{D}\right)$ you mentioned above. At least, the final result is consistent with the assumed conditions. I am aware that, in general, we can find ( maybe ) some 'weird' function without $\left(\mathbf{C}\ \mbox{and}\ \mathbf{D}\right)$. Indeed, I was expecting the down-vote because it happens to me before with this kind of questions. Also, I put this answer because I see nobody follow this road. – Felix Marin Oct 18 '16 at 21:50
• I agree that solution by differentiation adds a new perspective to the problem. It seems to me, however, that regularity issue is often the hardest part for this kind of problems. It would have been much better if you somehow suggested a way of establishing the regularity of $f$. Also, I think you need not worry about downvote since every answer was downvoted for some unknown reason. – Sangchul Lee Oct 19 '16 at 0:23