Yet Another Functional Equation

by utkarshgupta, Mar 7, 2016, 1:03 PM

Problem (IMO 2011) :
Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies
\[f(x + y) \leq yf(x) + f(f(x))\]for all real numbers $x$ and $y$. Prove that $f(x) = 0$ for all $x \leq 0$.

Proposed by Igor Voronovich, Belarus


Solution :

Let $P(x,y)$ be the assertion $f(x+y) \le yf(x)+f(f(x))$

Lemma 1 : $f(f(x)) \ge f(x)$

Proof :

$P(x,0) \implies$
$$\boxed{f(f(x)) \ge f(x)}$$

Lemma 2 : $f(x) \ge x \iff f(x) \ge 0$

Proof :

$P(x,f(x)-x) \implies$
$$\boxed{(f(x)-x)f(x) \ge 0}$$This gives the lemma directly.

Lemma 3 : $f(f(x)) \ge 0$

Proof :
Follows directly from Lemma 1 and 2.


Lemma 4 : $f(0) \ge 0$

Proof :

$P(0,f(f(0))) \implies$
$$f(f(0))f(0) + f(f(0)) \ge f(f(f(0))) \ge f(f(0))$$$$\implies f(f(0))f(0) \ge 0$$$$\implies \boxed{f(0) \ge 0}$$

Lemma 5 : $f(0)=0$

Proof :

$P(0,f(y)) \implies$
$$f(y)f(0) + f(f(0)) \ge f(f(y)) \ge 0$$
Since $f(0) \ge 0$, using $P(0,y)$
$$(yf(0)+f(f(0)))f(0) +f(f(0)) \ge f(y)f(0)+f(f(0)) \ge 0$$But this is obviously not true for $y \to - \infty$

Hence there must be equality that is
$$\boxed{f(0)=0}$$
Lemma 6 : $f(x) \le 0$
Proof :
$P(0,x) \implies $
$$ \boxed{f(x) \le 0}$$
Lemma 7 : $f(f(x))=0$
Proof :
Obvious from lemmas 3 and 7.

Main Proof :
Using lemma 7, the given inequality $P(x,y)$ can be rewritten as
$$f(x+y) \le yf(x)$$Consider $P(x,-x)$
$$0 \le -xf(x)$$Thus $x \le 0 \implies f(x) \ge 0$
But Lemma 6 implies $f(x) \le 0$

Hence
$$x \le 0 \implies f(x)=0$$
QED

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Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.

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    by math_holmes15, Jan 14, 2025, 8:53 AM

  • First post of 2024

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  • First post of 2023

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  • Nice blog ! Your isogonality lemma is really powerful !

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  • alas,this is ded

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  • Thanks for the nice blog.

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  • I think this might be silly but ... when should we expect to have another post ?? I am very keen to see it :D

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  • Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain :D

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  • hey utkarsh jee is over now ... continue your elementary blog pleaseeeeeee!

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  • INTERSTING BLOG

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  • I have no plans for this blog right now....
    No time here people !
    But lets see....
    I may try some combinatorics :P

    by utkarshgupta, Feb 15, 2017, 12:47 PM

  • Thanks for the nice blog!

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  • Revive it!!!
    Best blog out there, for sure!

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