Functional equation

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Source: Bulgarian NMO 2015 P4

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#1 h Apr 29, 2018, 9:01 AM • 4 Y

Y by huyaops, itslumi, Adventure10, Mango247

Find all functions $f:\mathbb{R^+}\to\mathbb {R^+}$ such that for all $x,y\in R^+$ the followings hold:
$i)$ $f (x+y)\ge f (x)+y$
$ii)$ $f (f (x))\le x$

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#2 h Apr 29, 2018, 9:12 AM • 6 Y

Y by Polynom_Efendi, Assassino9931, itslumi, Timmy456, Adventure10, Mango247

tenplusten wrote:

Find all functions $f:\mathbb{R^+}\to\mathbb {R^+}$ such that for all $x,y\in R^+$ the followings hold:
$i)$ $f (x+y)\ge f (x)+y$
$ii)$ $f (f (x))\le x$

i) implies $f(x)$ is strictly increasing
ii) implies then $f(x)\le x$
i) implies $f(x)\ge f(x-y)+y>y$ $\forall y\in(0,x)$
Setting there $y\to x^-$ , we get $f(x)\ge x$ $\forall x$

And so $\boxed{f(x)=x\quad\forall x>0}$ which indeed is a solution.

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#3 h Apr 29, 2018, 9:21 AM • 2 Y

Y by Adventure10, Mango247

I got the first 3 steps ,can you please eloborate what you do in your final step?

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#4 h Apr 29, 2018, 9:24 AM • 2 Y

Y by Adventure10, Mango247

tenplusten wrote:

I got the first 3 steps ,can you please eloborate what you do in your final step?

We have $f(x)>y$ $\forall y\in(0,x)$
Set there $y$ as near as you want of $x$ (but below) and this gives $f(x)\ge x$

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#6 h May 25, 2020, 6:33 AM

Y by

another easy solution
x≥f(f(x))≥f(f(x)-k)+k
if k≥x we are done so x≥f(x)
if there is at least one x wich x-s≥f(x) where s≠0 so we write x-s≥f(x)≥f(x-y)+y now we put x-s=y so that means f(x)=x

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#7 h Jan 4, 2022, 12:58 AM

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Clearly $f(x+y) \geq f(x) + y > f(x)$ , i.e. $f$ is strictly increasing. If we suppose that $f(x_0) > x_0$ for some $x_0$ , then the monotonicity and the second condition yield $x_0 \geq f(f(x_0)) > f(x_0) > x_0$ , contradiction. Hence $f(x) \leq x$ for all $x$ .

Suppose $f(x_0) < x_0$ for some $x_0$ . Then in the first condition replace $x$ with $x_0-f(x_0)$ and $y$ with $f(x_0)$ to obtain $f(x_0) \geq f(x_0 - f(x_0)) + f(x_0) > f(x_0)$ , impossible. Hence $f(x) \geq x$ for all $x$ .

In conclusion, $f(x) = x$ for all $x$ . Direct verification shows that this function is indeed a solution.

This post has been edited 1 time. Last edited by Assassino9931, Feb 26, 2024, 10:58 PM

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ZETA_in_olympiad

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#9 h Nov 7, 2022, 9:29 AM

Y by

Clearly $f$ is strictly increasing and so $f(x)\leqslant x$ . If $f(x)<x$ for some $x$ , then $(x,y)=(x-f(x),f(x))$ gives contradiction. So $f\equiv \text{Id}$ , which works.

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#10 h Jun 17, 2024, 3:22 AM

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Let $P(x,y)$ the assertion of the F.E. (ineq actually). Clearly as $f(x+y) \ge f(x)+y>f(x)$ we have $f$ strictly increasing.
Suppose there exists $x$ with $f(x)>x$ , then by $P(x,f(x)-x)$ we get $x \ge f(f(x)) \ge 2f(x)-x$ which implies $x \ge f(x)$ contradiction!.
Therefore $x \ge f(x)$ for all positive reals $x$ , which also means that as $x$ becomes smaller so does $f(x)$ , so now lets consider $P(x-y,y)$ for $x>y$ this gives $f(x) \ge f(x-y)+y$ , if we assume FTSOC $x>f(x)$ for some $x$ then this gives $0 \ge f(x-f(x))$ , contradiction!.
Therefore $x \ge f(x) \ge x$ so $f(x)=x$ is the only solution, thus we are done :cool:

:cool:

.

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cursed_tangent1434

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#11 h Jun 17, 2024, 1:49 PM • 1 Y

Y by GeoKing

Pretty straightforward. The answer is $f(x) = x$ for all $x \in \mathbb{R}^+$ . It’s easy to see that these functions satisfy the given conditions. We now show these are the only solutions.

Let $P(x,y)$ denote the assertion that $f (x+y)\ge f (x)+y$ for positive reals $x$ and $y$ . We first show the following property of $f$ .

Claim : The function $f$ is (strictly) increasing.
Proof : For any positive $\epsilon$ , $P(x,\epsilon)$ yield,
$f(x+\epsilon)/ge f(x)+\epsilon > f(x)$ which implies the claim.

Say there exists some positive real $x_0$ such that $f(x_0)>x_0$ . Then, $P((f(x_0)-x_0),x_0)$ implies,
$f(f(x_0)) = f((f(x_0)-x_0)+x_0) \ge f(f(x_0)-x_0)+x_0 > x_0$ which contradicts the second condition. Thus, for all $x \in \mathbb{R}^+$ , $f(x) \le x$ .

Say there exists some positive real $x_0$ such that $f(x_0) < x_0$ . Then, $P((x_0-f(x_0)),x_0)$ implies,
$f(x_0)=f((x_0-f(x_0))+f(x_0)) \ge f(x_0-f(x_0))+f(f(x_0))> f(x_0-f(x_0))$ which contradicts the strictly increasing nature of $f$ . Thus, for all $x \in \mathbb{R}^+$ , $f(x) \ge x$ .

Combining these two results, we have that the only possible solution is the desired one.

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#12 h Apr 15, 2025, 10:37 PM

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The only solution is $f(x) = x$ . With i), we get that $f((x + y - \epsilon) + \epsilon) \geq f(\epsilon) + x + y - \epsilon$ , $\epsilon > 0$ .
With ii), we have $x + y \geq f(f(x+y)) \geq f(x + y) - \epsilon + f(\epsilon) \geq x + y - 2\epsilon + 2f(\epsilon)$ . Now, let $\epsilon$ tend to zero, so $f(x + y) - \epsilon$ tends to $f(x + y)$ and $x + y - 2\epsilon + 2f(\epsilon)$ tends to $x + y$ , because $2f(\epsilon) \leq 2\epsilon$ , so $x + y \geq f(x + y) \geq x + y$ , giving us $f(x) = x \ \ \forall \ \ x$ , wich is indeed a solution.

This post has been edited 2 times. Last edited by Jakjjdm, Apr 15, 2025, 10:39 PM

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