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Functional equations challenge 1


This is a challenging problem !!

Find all continuous  functions  f: R \longrightarrow R    such that for all x\inR:

f(1 - x) = 1 - f(f(x))

Solution : ( first try it , if not then you can read )

1) replacing the equation by a simpler one :   g(g(x))=-g(-x)
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Let   g(x)=\frac 12-f(\frac 12-x)  and so  f(x)=\frac 12-g(\frac 12-x) and the equation becomes -g(x-\frac 12)=g(g(\frac 12-x)) and so :

g(x) is a continuous function such that g(g(x))=-g(-x)
Q.E.D.

2) g(x)=x \forall x\in g(\mathbb R)
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Let A=g(\mathbb R)
From 1) above, we get that a\in A\implies -a\in A and so, since g(x) is continuous :
Either A=\mathbb R, either A=[-u,u] for some u\ge 0.

g(g(x))=-g(-x) \implies g(g(g(g(x))))=-g(-g(g(x)))=-g(g(-x))=g(x) and so g(g(g(x)))=x \forall x\in A

Let then g_r(x) : A\to A the restriction of g(x) to A :
g(g(g(x)))=x \forall x\in A \implies g_r(g_r(g_r(x)))=x \forall x\in A and so g_r(x) is a bijection, so is monotonous
g(g(x))=-g(-x) \implies g_r(g_r(x))=-g_r(-x) implies then -g_r(-x) is increasing and so g_r(x) is increasing.

So, g_r(a)> a \implies g_r(g_r(a))>g_r(a)>a and so g_r(g_r(g_r(a)))>g_r(a)>a which is impossible and so g_r(a)\le a
Same, g_r(a)< a \implies g_r(g_r(a))<g_r(a)<a and so g_r(g_r(g_r(a)))<g_r(a)<a which is impossible and so g_r(a)\ge a

So g_r(x)=x
Q.E.D.

3) general solution of g(g(x))=-g(-x) with g(x) continuous
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Then the original equation g(g(x))=-g(-x) becomes g(x)=-g(-x) (since g(g(x))=g_r(g(x))=g(x)) and so :

The general continuous solutions of g(g(x))=-g(-x) are any odd continuous function g(x) such that g(x)=x \forall x\in g(\mathbb R)
A simple way to build them is :
3.1) If g(\mathbb R)=\mathbb R : g(x)=x \forall x

3.2) If g(\mathbb R)=[-u,u] for some u\ge 0 :
Let any continuous fonction h(x) from [u,+\infty)\to[-u,+u] such that h(u)=u and define g(x) as :
\forall x\in(-\infty,-u) : g(x)=-h(-x)
\forall x\in[-u,+u] : g(x)=x
\forall x\in(u,+\infty) : g(x)=h(x)

Notice that u=0 gives the solution g(x)=0

4) general solution of f(1-x)=1-f(f(x)) with f(x) continuous
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The general solution of original equation is f(x)=\frac 12-g(\frac 12-x)) where g(x) is any function described in 3) above

5) examples of solutions
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g(x)=0 gives the solution f(x)=\frac 12

g(x)=x gives the solution f(x)=x

u=1 and h(x)=\sin(\frac{\pi}2x) gives the solution :
If |x-\frac 12|<1 : f(x)=x
If |x-\frac 12|\ge 1 : f(x)=\frac 12+\sin(\frac{\pi}2x-\frac{\pi}4)

u=2 and h(x)=4|2\{\frac{x-2}8\}-1|-2 gives the solution : f(x)=\frac 52-4|2\{\frac{-2x-3}{16}\}-1| :)

Notice that the only differentiable solutions are f(x)=\frac 12 and f(x)=x

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IMO 2007 Short list problem


Find all functions  \large f:\mathbb{R+} \rightarrow \mathbb{R+} such that

f\left(x + f\left(y\right)\right) = f\left(x + y\right) + f\left(y\right)

for all x,y \large \in  \large \mathbb{R+}

Solution :

For any positive real numbers z, we have that

f(x+f(y))+z=f(x+y)+f(y)+z
f(f(x+f(y))+z)=f(f(x+y)+f(y)+z)
f(x+f(y)+z)+f(x+f(y))=f(x+y+f(y)+z)+f(x+y)
f(x+y+z)+f(y)+f(x+y)+f(y)=f(x+2y+z)+f(y)+f(x+y)
f(x+y+z)+f(y)=f(x+2y+z)
f(a)+f(b)=f(a+b)

and by Cauchy in positive reals, then f(x)=\alpha x for all x\in (0,\infty)
Now it’s easy to see that \alpha=2 , then f(x)=2x    for all positive real numbers x

Problem of the day 3


Nice problem !

Functional Equations !

Q)
Determine all functions ƒ satisfying  the functional relations

ƒ(x) + ƒ(1/(1-x)) =( 2(1-2x)/(x(1-x))

where x is a real number x ≠ 0 , x ≠ 1

Post your answer in comments with whole solution 🙂
person who give best answer of problem of the day more than 4 times will be awarded a book