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Difference between revisions of "2011 IMO Problems/Problem 3"

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Let <math>f: \mathbb R \to \mathbb R</math> be a real-valued function defined on the set of real numbers that satisfies <cmath> f(x + y) \le yf(x) + f(f(x)) </cmath> for all real numbers <math>x</math> and <math>y</math>. Prove that <math>f(x) = 0</math> for all <math>x \le 0</math>.
 
Let <math>f: \mathbb R \to \mathbb R</math> be a real-valued function defined on the set of real numbers that satisfies <cmath> f(x + y) \le yf(x) + f(f(x)) </cmath> for all real numbers <math>x</math> and <math>y</math>. Prove that <math>f(x) = 0</math> for all <math>x \le 0</math>.
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==Solution==
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Let <math>P(x,y)</math> be the given assertion.
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Comparing <math>P(x,f(y)-x)</math> and <math>P(y,f(x)-y)</math> yields, <cmath>xf(x)+yf(y)\leq 2f(x)f(y).</cmath>
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<math>y\mapsto 2f(x)\implies xf(x)\leq 0. \qquad (*)</math>
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------------------------
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<math>\textbf{Claim: }f(k)\leq 0~~\forall k.</math>
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<math>Proof.</math> Suppose <math>\exists k:f(k)>0,</math> then <cmath>f(k+y)\leq yf(k)+f(f(k)).</cmath>
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Now <math>y\to -\infty</math> implies that <math>\lim_{x\to -\infty} f(x)=-\infty.</math>
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<math>P(x,z-x)\implies f(z)\leq (z-x)f(x)+f(f(x)).</math>
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Then <math>x\to -\infty,</math> yields a contradiction. <math>\blacksquare</math>
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--------------------------
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From <math>(*)</math> we get <math>f(x)=0,\forall x<0.</math>
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<math>P(0,f(0))\implies f(0)\geq 0,</math> thus we get <math>f(0)=0,</math> as desired. <math>\square</math>
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                            ~ZETA_in_olympiad
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==See Also==
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{{IMO box|year=2011|num-b=2|num-a=4}}
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[[Category:Olympiad Algebra Problems]]
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[[Category:Functional Equation Problems]]

Latest revision as of 00:20, 19 November 2023

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

Solution

Let $P(x,y)$ be the given assertion. Comparing $P(x,f(y)-x)$ and $P(y,f(x)-y)$ yields, \[xf(x)+yf(y)\leq 2f(x)f(y).\] $y\mapsto 2f(x)\implies xf(x)\leq 0. \qquad (*)$

$\textbf{Claim: }f(k)\leq 0~~\forall k.$

$Proof.$ Suppose $\exists k:f(k)>0,$ then \[f(k+y)\leq yf(k)+f(f(k)).\] Now $y\to -\infty$ implies that $\lim_{x\to -\infty} f(x)=-\infty.$ $P(x,z-x)\implies f(z)\leq (z-x)f(x)+f(f(x)).$

Then $x\to -\infty,$ yields a contradiction. $\blacksquare$

From $(*)$ we get $f(x)=0,\forall x<0.$ $P(0,f(0))\implies f(0)\geq 0,$ thus we get $f(0)=0,$ as desired. $\square$ 

                           ~ZETA_in_olympiad

See Also

2011 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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