Master of Functional Equations

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Source: Own, based on a problem from Ukrainian Mathematical Olympiad

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#1 h Aug 14, 2018, 9:16 PM • 7 Y

Y by khan.academy, mathlomaniac, Kayak, integrated_JRC, don2001, centslordm, Adventure10

The first person to solve this problem will receive the title of Master of Functional Equations.

Determine all functions $f \colon \mathbb R^+ \to \mathbb R^+$ that satisfy the equation
$f(f(x)+f(y))=xyf(x+y)$ for any $x, y \in \mathbb R^+$ . Note that $\mathbb R^+ \stackrel{\text{def}}{=} \{x \in \mathbb R \mid x > 0\}$ .

Background
Acknowledgements

This post has been edited 1 time. Last edited by Bandera, Aug 15, 2018, 4:53 PM
Reason: Acknowledgements added

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381 posts

Eray

#2 h Aug 14, 2018, 10:15 PM • 12 Y

Y by Happy2020, Supercali, AmSm_9, Smita, Kayak, integrated_JRC, tenplusten, RudraRockstar, centslordm, physics-ka-ashiq, Adventure10, Mango247

Who else wants to bet for pco?

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AmSm_9

#3 h Aug 16, 2018, 1:20 AM • 2 Y

Y by centslordm, Adventure10

I guessed a solution

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Smita

#4 h Aug 16, 2018, 1:47 AM • 3 Y

Y by centslordm, Adventure10, Mango247

Eray wrote:

Who else wants to bet for pco?

I would also bet for pco

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TLP.39

#5 h Aug 17, 2018, 2:35 PM • 21 Y

Y by lminsl, MNJ2357, AlastorMoody, Kayak, toshihiro shimizu, NahTan123xyz, Bandera, don2001, AmSm_9, tenplusten, enthusiast101, Supercali, Aritra12, centslordm, physics-ka-ashiq, L567, Pluto1708, MathLuis, Adventure10, Mango247, Sedro

Bandera wrote:

Determine all functions $f \colon \mathbb R^+ \to \mathbb R^+$ that satisfy the equation
$f(f(x)+f(y))=xyf(x+y)$ for any $x, y \in \mathbb R^+$ . Note that $\mathbb R^+ \stackrel{\text{def}}{=} \{x \in \mathbb R \mid x > 0\}$ .

The only answer is $\boxed{f(x)\equiv \frac{1}{x}}.$ We will prove this through the following claims.

Let $P(x,y)$ be the assertion.

Claim 1 : $x^2f(x)$ is unbounded above.
Proof : Assume the contrary, let $a$ be the upper bound.
For any $k>\sqrt{\frac{2a}{f(2)}},$ we have $f(k)<\frac{f(2)}{2}.$
By $P(1-x,1+x),$ we can see that $(0,f(2)]\in f(\mathbb{R}).$
Thus,there exists $l$ such that $f(l)+f(k)=f(2),$ the assumed condition then imply that $l<\sqrt{\frac{2a}{f(2)}}.$
Finally, $P(k,l)\implies f(f(2))=klf(k+l)\le \frac{a}{\frac{k}{l}+2+\frac{l}{k}},$ thus by letting $k\to\infty,$ we can make the ratio $\frac{k}{l}$ as large as possible, implying that $f(f(2))\le 0,$ a contradiction. $\square$

From Claim 1, we get the following corollary :

Corollary 2 : $f(x)$ is surjective.
Proof : For any positive number $r,$ let $s$ be a positive number satisfying $s^2f(s)\ge 4r.$ Now, let $t=\sqrt{\frac{s^2}{4}-\frac{r}{f(s)}},$ then
$P(\frac{s}{2}-t,\frac{s}{2}+t)\, ;\, f(f(\frac{s}{2}-t)+f(\frac{s}{2}+t))=r$ Hence the desired result.

Claim 3 : If $a,b,c,d$ satisfy $a<b,c<d,f(a)=f(b),$ and $f(c)=f(d),$ then $a=c$ amd $b=d.$
Proof : Assume the contrary, WLOG $a=\min\{a,b,c,d\}.$
If $a\neq c,$ then
$P(c-a,a),P(c-a,b),P(d-a,a),P(d-a,b)\implies f(c+b-a)=f(d+b-a)$
$P(b,d),P(b,c)\implies df(b+d)=cf(b+c)$
$P(a,c+b-a),P(a,d+b-a)\implies (c+b-a)f(b+c)=(d+b-a)f(b+d)\implies c(d+b-a)=d(c+b-a)\implies (c-d)(b-a)=0.$ Clearly a contradiction. Thus $a=c$ and so $b\neq d.$
Finally, by looking at the quadruple $(a',b',c',d')=(a,b,\min\{b,d\},\max\{b,d\})$ instead of $(a,b,c,d),$ the same method applies that $a=\min\{b,d\}$ , still a contradiction. $\square$

Claim 3 implies that there are only one ordered pair of positive real numbers $(a,b)$ such that $a<b$ and $f(a)=f(b).$

Claim 4 : $f(x)$ is injective.
Proof : Assume the contrary, then there exists a pair of positive real numbers $(a,b)$ such that $a<b$ and $f(a)=f(b).$
$P(x,a),P(x,b)\implies af(x+a)=bf(x+b)\implies f(x+(b-a))=\frac{a}{b}f(x)\,\forall\, x>a$
Define $t=b-a>0$ and $c=\frac{a}{b}<1,$ then let $l$ be a positive real number that satisfy the system of inequalities :
$\begin{cases}l>a \\ \frac{l^2}{4}>\frac{c^2(l+2t)^2}{4}>al\\ \frac{(l+t)^2}{4}>\frac{c(l+2t)^2}{4}>a(l+t)\end{cases}$ then there exists $x_1,x_2,x_3,x_4>a$ such that
$\begin{cases}x_1+x_2=l \\ x_3+x_4=l+t\\ x_1x_2=\frac{c^2(l+2t)^2}{4}\\ x_3x_4=\frac{c(l+2t)^2}{4}\end{cases}$ Thus,we have $x_1x_2f(x_1+x_2)=x_3x_4f(x_3+x_4)=h^2f(2h)$ where $h=\frac{l+2t}{2}.$
Hence, by $P(x_1,x_2),P(x_3,x_4),P(h,h)$ and Claim 3, there exists different multisets $\{y_1,y_2\},\{z_1,z_2\}$ such that $y_1,y_2,z_1,z_2>a$ and $f(y_1)+f(y_2)=f(z_1)+f(z_2).$
Finally, we can see that $f(y_1+t)+f(y_2+t)=f(z_1+t)+f(z_2+t)$ and $f(y_1+2t)+f(y_2+2t)=f(z_1+2t)+f(z_2+2t).$ Thus,
$\begin{cases}y_1y_2f(y_1+y_2)=z_1z_2f(z_1+z_2) \\ (y_1+t)(y_2+t)f(y_1+y_2)=(z_1+t)(z_2+t)f(z_1+z_2)\\ (y_1+2t)(y_2+2t)f(y_1+y_2)=(z_1+2t)(z_2+2t)f(z_1+z_2) \end{cases}$ Solving the system, we got that $y_1+y_2=z_1+z_2,y_1y_2=z_1z_2\implies \{y_1,y_2\}=\{z_1,z_2\}$ , a contradiction. $\square$

Claim 5 : $x^2f(x)$ is strictly increasing.
Proof : Assume that there exists $a>b$ such that $a^2f(a)\le b^2f(b),$ then $af(a)<bf(b)$ and $f(a)<f(b).$
Let $p=\frac{bf(b)-af(a)}{f(b)-f(a)},$ then $p<b$ and thus
$P(p,a-p),P(p,b-p)\implies f(f(p)+f(a-p))=f(f(p)+f(b-p))\implies a=b,$ a contradiction. $\square$

Corollary 6 : $f(x)$ is strictly monotone.
Proof : Since $f(x)$ is bijective, it's suffice to prove that $f(x)$ is continuous.
Since $x^2f(x)=4f(2f(\frac{x}{2}))$ is surjective, and also strictly increasing from Claim 5, $x^2f(x)$ must be continuous, and thus $f(x)$ must be continuous as desired.

Claim 7 : $f(x)\equiv \frac{1}{x}$
Proof : We will split into $2$ cases :

Case 1 : $f(x)$ is strictly increasing.
In this case, we have
$\forall y\in (0,1), \, f(f(1))<f(f(1)+f(y))=yf(1+y)<yf(2)$ This clearly implies that $f(f(1))\le 0,$ a contradiction. Hence no answer in this case.

Case 2 : $f(x)$ is strictly decreasing.
In this case, we have
$P(1,1)\implies f(1)=1$
Now, for any pair of positive numbers $a<b$ ,we have
$P(a-k,k),P(b-k,k)\implies (a-k)f(a)<(b-k)f(b)\,\forall\, k\in (0,a)\implies af(a)\le bf(b)$
Hence $f(x)\ge\frac{1}{x}$ for all $x\ge 1$ and $f(x)\le\frac{1}{x}$ for all $x\le 1.$
For any $t>1,$ choose $T>1$ such that $\frac{1}{t}+\frac{1}{T}<1,$ then
$P(t,T)\implies \frac{tT}{t+T}\ge f(\frac{1}{t}+\frac{1}{T})\ge f(f(t)+f(T))=tTf(t+T)\ge \frac{tT}{t+T}$
Thus, $f(\frac{1}{t}+\frac{1}{T})= f(f(t)+f(T))\implies \frac{1}{t}+\frac{1}{T}= f(t)+f(T)\implies f(t)=\frac{1}{t}.$
On the other hand, for any $t<1,$ choose $T<1$ such that $t+T<1,$ then
$P(t,T)\implies \frac{tT}{t+T}\le f(\frac{1}{t}+\frac{1}{T})\le f(f(t)+f(T))=tTf(t+T)\le \frac{tT}{t+T}$
Thus, $f(\frac{1}{t}+\frac{1}{T})= f(f(t)+f(T))\implies \frac{1}{t}+\frac{1}{T}= f(t)+f(T)\implies f(t)=\frac{1}{t}.$
Hence $f(x)\equiv \frac{1}{x}$ as desired. $\square$

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#6 h Aug 18, 2018, 8:05 PM • 3 Y

Y by centslordm, Adventure10, Mango247

TLP.39 wrote:

Define $t=b-a>0$ and $c=\frac{a}{b}<1,$ then let $l$ be a positive real number that satisfy the system of inequalities :
$\begin{cases}l>a \\ \frac{l^2}{4}>\frac{c^2(l+2t)^2}{4}>al\\ \frac{(l+t)^2}{4}>\frac{c(l+2t)^2}{4}>a(l+t)\end{cases}$

Why does the system have any solution?

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778 posts

TLP.39

#7 h Aug 19, 2018, 1:54 AM • 2 Y

Y by centslordm, Adventure10

It suffices to prove that for each inequality of this system:
$\begin{cases}l>a \\ \frac{l^2}{4}>\frac{c^2(l+2t)^2}{4}\\ \frac{c^2(l+2t)^2}{4}>al\\ \frac{(l+t)^2}{4}>\frac{c(l+2t)^2}{4}\\ \frac{c(l+2t)^2}{4}>a(l+t)\end{cases}$ There exists $N>0$ such that for all $l>N,$ $l$ satisfy the equation.
This is clear for the first, third, and fifth as L.H.S. has higher degree than R.H.S., and has positive leading coefficient when consider as polynomials on $l$ .
For the second and fourth, just see that both sides have equal degree but L.H.S. has bigger leading coefficient than R.H.S. when consider as polynomials on $l$ and we are done.

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#8 h Aug 19, 2018, 9:24 AM • 3 Y

Y by centslordm, Adventure10, Mango247

@TLP.39 Congratulations! You have won the title:
$\boxed{\mathfrak f(x+y) = \mathfrak f(x) +\mathfrak f(y) \; \maltese \text{ Master}}$

You may want to post another problem (here or in a new topic) for finding more FE Masters. It should be:

Very hard;
Solvable;
New, i.e. not having published solutions (here or anywhere else on the Internet);
Concerning functional equations.

This post has been edited 1 time. Last edited by Bandera, Aug 19, 2018, 10:51 AM
Reason: Uniqueness requirement added

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#9 h Aug 19, 2018, 10:58 AM • 4 Y

Y by MNJ2357, centslordm, Adventure10, Mango247

Here's my solution.
Answer The only solution is $f(x) \equiv \frac 1 x.$

Plan of proof (all variables below are assumed to be from $\mathbb R^+$ ; moreover, $x$ and $y$ are considered free)
0. $f(f(x)+f(y))=xyf(x+y)$
1. $y \le \tfrac {x^2} 4 f(x) \implies \exists z \colon f(z)=y$
---A0. $x^2f(x)$ is bounded, $\exists C \; \forall x \colon x^2f(x) \le C$
---A1. $\exists a \; \forall x \le a \; \exists z \colon f(z)=x$
---A2. $f$ is bounded, $\exists B \; \forall x \colon f(x) \le B$
---A3. $x \le a \implies \exists z \le 2B \colon f(z)=x$
---A4. $\exists \varepsilon_n \colon \lim_{n \to \infty} \varepsilon_n=0, \lim_{n \to \infty} f(\varepsilon_n)=0$
---A5. $\pmb \bot$ (a contradiction)
2. $x^2f(x)$ is unbounded, $\nexists C \; \forall x \colon x^2f(x) \le C$
3. $f$ is surjective, $\forall x \; \exists z \colon f(z)=x$
4. $x^2f(x)$ is surjective, $\forall x \; \exists z \colon z^2f(z)=x$
---B0. $f$ isn't injective, $\exists \, v<w \colon f(v)=f(w)$
---B1. $(f(x_1)=f(x_2)) \land (y_1\!-x_1\!=y_2\!-x_2\!>0) \implies \tfrac {f(y_1)}{f(y_2)}=\tfrac {x_2}{x_1}$
---B2. $(f(x_1\!+y_1)=f(x_2\!+y_1)) \land (f(x_1\!+y_2)=f(x_2\!+y_2)) \implies (x_1\!=x_2) \lor (y_1\!=y_2)$
---B3. $(\exists T \; \forall t>T \colon x_1f(x_1\!+t)=x_2f(x_2\!+t)) \implies f(x_1)=f(x_2)$
---B4. $\pmb \bot$
5. $f$ is injective, $f(x)=f(y) \implies x=y$
6. $xf(x)$ is non-decreasing, $x<y \implies xf(x) \le yf(y)$
7. $f$ is continuous, $\lim_{n \to \infty} x_n = x \implies \lim_{n \to \infty} f(x_n) = f(x)$
8. $\lim_{x \to +\infty} f(x) = 0$
9. $f(1)=1$
10. $f(f(x))=x$
11. $f(x)=\tfrac 1 x$

Details notation
0.> The original equation.
1.> Let $t = \tfrac y {f(x)} \in (0, \tfrac {x^2} 4]$ . Then $\exists x_1>0, x_2>0 \colon x_1x_2=t, x_1+x_2=x$ (these are solutions of a quadratic equation: $x_{1,2}=\tfrac {x \pm \sqrt {x^2-4t}} 2$ ). 0 $[x \gets x_1, y \gets x_2]$ : $f(f(x_1)+f(x_2))=tf(x)=y$ and we can take $z=f(x_1)+f(x_2)$ .
A0.> Branching condition (assumption).
A1.> 1 $[x \gets 1, y \gets x]$ : $x \le \tfrac {f(1)} 4 \implies \exists z \colon f(z)=x$ , so we can take $a = \tfrac {f(1)} 4$ .
A2.> If $x \le 2a$ , then A1 $[x \gets \tfrac x 2]$ : $\exists z \colon f(z)=\tfrac x 2$ . 0 $[x \gets z, y \gets z]$ : $f(2f(z))=z^2f(2z)$ . A0 $[x \gets 2z]$ : $4z^2f(2z) \le C$ . Thus, $x \le 2a \implies f(x)=f(2f(z))=z^2f(2z) \le \tfrac C 4$ . On the other hand, if $x > 2a$ , then A0 gives $f(x) \le \tfrac C {x^2} < \tfrac C {4a^2}$ . Hence, we can take $B = \max \{\tfrac C 4, \tfrac C {4a^2}\}$ .
A3.> The same as for 1 and A1, just note that due to A2 a value of $z$ at the end of 1's proof does not exceed $2B$ .
A4.> Let $x_n=n, y_n=n^2, \varepsilon_n=f(x_n)+f(y_n)$ . A0 $[x \gets x_n]$ : $f(x_n) \le \tfrac C {n^2}$ , thus $\lim_{n \to \infty} f(x_n)=0$ . Similarly, $\lim_{n \to \infty} f(y_n)=0$ . Hence, $\lim_{n \to \infty} \varepsilon_n=0$ . 0 $[x \gets x_n, y \gets y_n]$ : $f(\varepsilon_n)=x_ny_nf(x_n+y_n)=n^3f(n+n^2)$ . A0 $[x \gets n+n^2]$ : $f(n+n^2) \le \tfrac C {(n+n^2)^2} < \tfrac C {n^4}$ . Thus, $f(\varepsilon_n) < \tfrac {Cn^3}{n^4}$ and $\lim_{n \to \infty} f(\varepsilon_n)=0$ .
A5.> Take any $d \le a$ . Now use $\varepsilon_n$ from A4; let $x_n=d-f(\varepsilon_n)$ , for all $n$ such that $f(\varepsilon_n)<d$ we get: $x_n \in (0,a)$ , A3 $[x \gets x_n]$ : $\exists z_n \le 2B \colon f(z_n)=x_n$ , so $d=f(z_n)+f(\varepsilon_n)$ . 0 $[x \gets z_n, y \gets \varepsilon_n]$ : $f(f(z_n)+f(\varepsilon_n))=z_n \varepsilon_n f(z_n+\varepsilon_n)$ . A2 $[x \gets z_n+\varepsilon_n]$ : $f(z_n+\varepsilon_n) \le B$ , thus $f(d)=z_n \varepsilon_n f(z_n+\varepsilon_n) \le 2B \cdot \varepsilon_n \cdot B$ , a contradiction since $f(d)>0$ but $\varepsilon_n \to 0$ when $n \to \infty$ .
2.> A5, dead branch defined by A0 pruned, this is the rest.
3.> From 2 it follows that $\exists x' \colon 4x \le x'^2f(x')$ . 1 $[x \gets x', y \gets x]$ : $\exists z \colon f(z)=x$ .
4.> 0 $[x \gets \tfrac z 2, y \gets \tfrac z 2]$ : $4f(2f(\tfrac z 2))=z^2f(z)$ . When $z$ runs through all $\mathbb R^+$ , so does $\tfrac z 2$ , from 3 it follows that so does $f(\tfrac z 2)$ , then $2f(\tfrac z 2)$ , $f(2f(\tfrac z 2))$ , and finally, $4f(2f(\tfrac z 2))=z^2f(z)$ .
B0.> Branching condition (assumption).
B1.> 0 $[x \gets x_1, y \gets y_1\!-x_1]$ : $f(f(x_1)+f(y_1\!-x_1))=x_1(y_1\!-x_1)f(y_1)$ . 0 $[x \gets x_2, y \gets y_2\!-x_2]$ : $f(f(x_2)+f(y_2\!-x_2))=x_2(y_2\!-x_2)f(y_2)$ . The left sides are equal, thus $x_1f(y_1)=x_2f(y_2)$ .
B2.> Assume that $y_1 \!\ne y_2$ . WLOG, $y_2 \!> y_1$ . B1 $[x_1 \gets x_1\!+y_1, x_2 \gets x_2\!+y_1, y_1 \gets x_1\!+y_2, y_2 \gets x_2\!+y_2]$ : $\tfrac {f(x_1+y_2)}{f(x_2+y_2)}=\tfrac {x_2+y_1}{x_1+y_1}$ . Since the left side is equal to $1$ , the right side must be $1$ too, thus $x_1\!=x_2$ .
B3.> Let $t>T$ . 0 $[x \gets x_1, y \gets t]$ : $f(f(x_1)+f(t))=x_1tf(x_1+t)$ , 0 $[x \gets x_2, y \gets t]$ : $f(f(x_2)+f(t))=x_2tf(x_2+t)$ . Thus, $\forall t>T \colon f(f(x_1)+f(t))=f(f(x_2)+f(t))$ . Note that $f(t)$ cannot be constant for all $t>T$ , because in such case, for example, $f(T+1)=f(T+2)=f(T+3)$ and B2 $[x_1 \gets T, x_2 \gets T+1, y_1 \gets 1, y_2 \gets 2]$ gives a contradiction, so we can choose $t_1>T,\,t_2>T$ such that $t_1 \ne t_2$ and $f(t_1) \ne f(t_2)$ . B2 $[x_1 \gets f(x_1), x_2 \gets f(x_2), y_1 \gets f(t_1), y_2 \gets f(t_2)]$ : $f(x_1)=f(x_2)$ because $f(t_1) \ne f(t_2)$ .
B4.> Let $d=w-v>0, r=\tfrac w v>1, \; u=\tfrac {2v^2}{v+w}, z=\tfrac {2w^2}{v+w}$ . Clearly, $z-u=2d$ , $\tfrac z u = r^2$ and $0<u<v<w<z$ . Now take any $y_1>v$ and let $y_2=y_1+d$ . Then $y_2-w=y_1-v>0$ . B1 $[x_1 \gets v, x_2 \gets w]$ : $\tfrac {f(y_1)}{f(y_2)}=r$ . Let $y_3=y_1+z-u=y_2+d$ . Then $y_3-w=y_2-v>0$ . B1 $[x_1 \gets v, x_2 \gets w, y_1 \gets y_2, y_2 \gets y_3]$ : $\tfrac {f(y_2)}{f(y_3)}=r$ . Multiplying, we get $\tfrac {f(y_1)}{f(y_3)}=r^2=\tfrac z u$ . Denoting $t=y_1-u$ , we can rewrite this as $uf(u+t)=zf(z+t)$ , that is true for any $t>v-u$ . B3 $[T \gets v-u, x_1 \gets u, x_2 \gets z]$ : $f(u)=f(z)$ . Now let $s=w+d=v+2d=v+z-u$ . Then $s-z=v-u>0$ . B1 $[x_1 \gets u, x_2 \gets z, y_1 \gets v, y_2 \gets s]$ : $\tfrac {f(v)}{f(s)}=\tfrac z u = r^2$ . On the other hand, B1 $[x_1 \gets v, x_2 \gets w, y_1 \gets w, y_2 \gets s]$ : $\tfrac {f(w)}{f(s)}=\tfrac w v = r$ . The left sides are equal due to $f(v)=f(w)$ , but the right sides are not because $r>1$ , a contradiction.
5.> B4, dead branch defined by B0 pruned, this is the rest.
6.> Let $x<y$ . If $f(x) \le f(y)$ , then clearly $xf(x) < yf(y)$ . Assume that $f(x)>f(y)$ but $xf(x)>yf(y)$ . Let $z=\tfrac {xf(x)-yf(y)}{f(x)-f(y)}$ . Obviously, $0<z<x<y$ . Also note that $(x-z)f(x)=(y-z)f(y)$ . 0 $[x \gets z, y \gets x-z]$ : $f(f(z)+f(x-z))=z(x-z)f(x)$ , 0 $[x \gets z, y \gets y-z]$ : $f(f(z)+f(y-z))=z(y-z)f(y)$ . Hence, $f(f(z)+f(x-z))=f(f(z)+f(y-z))$ . From this 5 deduces: $f(z)+f(x-z)=f(z)+f(y-z) \implies f(x-z)=f(y-z) \implies x-z=y-z \implies x=y$ , a contradiction.
7.> From 6 it follows that the function $g(x) \equiv x^2f(x) \equiv x \cdot xf(x)$ is strictly increasing. 4 says that $g(x)$ is surjective. Hence, it is continuous at the whole domain and so is $f(x) \equiv \tfrac {g(x)}{x^2}$ .
8.> From 5 and 7 it follows that $f$ is strictly monotonic. Moreover, from 3 it follows that only two cases are possible: $(\,\lim_{x \to 0^+} f(x) = 0\,) \land (\,\lim_{x \to +\infty} f(x) = +\infty\,)$ when $f$ is increasing and $(\,\lim_{x \to 0^+} f(x) = +\infty\,) \land (\,\lim_{x \to +\infty} f(x) = 0\,)$ when $f$ is decreasing. The first case quickly leads to a contradiction because fixing some $y$ in 0 and taking it to the limit when $x \to 0^+$ we get (employing the continuity): $f(f(y))=0 \cdot yf(y)=0$ .
9.> 0 $[x \gets 1, y \gets 1]$ : $f(2f(1))=f(2)$ . 5 $[x \gets 2f(1), y \gets 2]$ : $2f(1)=2$ .
10.> Dividing the original equation by $x$ and taking it to the limit when $y \to +\infty$ , we get (using 8 and 7): $\tfrac {f(f(x))} x = \lim_{y \to +\infty} yf(x+y)$ . Denote the function at the left side by $h(x)$ . The right side is in fact independent of $x$ . Indeed, whatever $x$ is, we already know that the limit exists (it is equal to $h(x)$ ). Thus, $\lim_{y \to +\infty} yf(x+y)=$ $x \cdot 0 + \!\lim_{y \to +\infty} yf(x+y)=$ $x \!\lim_{y \to +\infty} f(x+y)+\!\lim_{y \to +\infty} yf(x+y)=$ $\lim_{y \to +\infty} xf(x+y)+\lim_{y \to +\infty} yf(x+y)=$ $\lim_{y \to +\infty} (x+y)f(x+y)=$ $\{z=x+y\}=$ $\lim_{z \to +\infty} zf(z),$ that is independent of $x$ . Hence, $h(x)$ is a constant, $h(x) \equiv C$ . From 9 if follows that $C=1$ .
11.> 0 $[y \gets f(x)]$ : $f(f(x)+f(f(x)))=xf(x)f(x+f(x))$ . 10 reduces the left side to $f(f(x)+x)$ . Thus, $xf(x)=1$ .

Direct check shows that $f(x) \equiv \tfrac 1 x$ is indeed a solution.

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#10 h Mar 9, 2019, 4:05 PM • 3 Y

Y by khan.academy, centslordm, Adventure10

I'm reviving this thread since this is(or will be) one of the most difficult FE marathons.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that

$f(xf(x+y)+yf(y-x))=f(x)^2+f(y^2)$ for all $x,y\in\mathbb{R}$
$f$ is continuous on $\mathbb{R}\backslash \{-1,1\}$ . (Continuity at $x=1,-1$ is not provided.)

Prove that if $x\neq 1,-1$ , then $f(x)=-f(-x)$ .

This post has been edited 1 time. Last edited by MNJ2357, Mar 9, 2019, 4:08 PM

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#11 h Mar 23, 2019, 2:47 PM • 3 Y

Y by centslordm, Adventure10, Mango247

Bump!!

:showoff:

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#12 h Mar 23, 2019, 4:28 PM • 3 Y

Y by centslordm, Adventure10, Mango247

Here is a problem I like a lot and at first if one doesn't know the solution it may seem difficult.
(Although pco sir solved it a while earlier -- salut)

Are there any functions $f: \mathbb{R}_{>0} \to \mathbb{R}$ such that $f(xf(y) + yf(x)) = x^2 + y^2$ holds true $\forall (x,y) \in \mathbb{R}_{>0} ^2$ .
(Here $\mathbb{R}_{>0}$ is the set of all positive real numbers only).

Source and remarks

This post has been edited 1 time. Last edited by adityaguharoy, Mar 23, 2019, 4:31 PM

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#13 h Mar 28, 2019, 7:18 PM • 4 Y

Y by MNJ2357, centslordm, Adventure10, Mango247

MNJ2357 wrote:

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that
$f(xf(x+y)+yf(y-x))=f(x)^2+f(y^2)$ for all $x,y\in\mathbb{R}$
$f$ is continuous on $\mathbb{R}\backslash \{-1,1\}$ . (Continuity at $x=1,-1$ is not provided.)

Prove that if $x\neq 1,-1$ , then $f(x)=-f(-x)$ .

I'll demonstrate that we can actually find all such functions.

All functions

Solution

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#14 h Sep 6, 2019, 12:55 PM • 2 Y

Y by centslordm, Adventure10

Here is not so trivial yet good FE, which may take time but try it until you get it.

Find all functions $f:\mathbb{W-{1}}\rightarrow\mathbb{W-{1}}$ such that $xf(x'y) - yx^{y}f(x) = yf(x"xy) + xyf(y'x)$ , where operations $f(x'y)$ is defined as $f(f(f(f(.....y))))...$ $x$ times, i.e, $f(x'2) = f(f(x))$ and $f(x"y) = f(x)^{y}$ .

This post has been edited 2 times. Last edited by RustyFox, Sep 6, 2019, 1:04 PM

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