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It is given that $f(x)$ is a function defined on $\mathbb{R}$, satisfying $f(1)=1$ and for any $x\in \mathbb{R}$, $$f(x+5)\geq f(x)+5,$$ $$f(x+1)\leq f(x)+1.$$ If $g(x)=f(x)+1-x$ then find $g(2002)$.

Here, $$f(x+5)\leq f(x+4) +1,$$ I didn't get any idea..

pi-π
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  • $f(x)=x$ trivially satisfies the requirement and this example gives $g(x)$ constantly equal to $1$. Are there other examples of solution? – Piquito Feb 04 '18 at 17:19

2 Answers2

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Rewriting the given inequalities in terms of $\,f(x)=g(x)+x-1\,$:

$$\require{cancel} \begin{align} f(x+5)\geq f(x) + 5 \quad&\iff\quad g(x+5)+\bcancel{x}+\cancel{5}- \xcancel{1} \ge g(x) + \bcancel{x} -\xcancel{1} + \cancel{5} \\ &\iff\quad g(x+5) \ge g(x) \\ f(x+1)\leq f(x)+1 \quad&\iff\quad g(x+1)+\bcancel{x}+\cancel{1}- \xcancel{1} \le g(x) + \bcancel{x} -\xcancel{1} + \cancel{1} \\ &\iff\quad g(x+1) \le g(x) \\ \end{align} $$

It follows that $\,\color{blue}{g(x)} \ge g(x+1) \ge g(x+2) \ge g(x+3) \ge g(x+4) \ge g(x+5) \ge \color{blue}{g(x)}\,$, so equalities must hold throughout, then $\,g(x+1)=g(x)\,$, and $\,g(2002)=g(2001)=\ldots =g(1)\,$.

dxiv
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You are in the right direction, just carry on:

$$f(x+5)\leq f(x+4) +1\leq f(x+3) +2\leq f(x+2) +3\leq f(x+1) +4\leq f(x) +5$$

Observe something fishy between the LHS and the RHS here? What can you say about $f(x)$ now?

Gaurang Tandon
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  • From the condition given in the ques, $$f(x+5)\geq f(x)+5$$ and from your answer $$f(x+5)\leq f(x)+5$$? – pi-π Feb 04 '18 at 14:56
  • @blue_eyed_... What does that imply? Equality holds instead of inequality – user471651 Feb 04 '18 at 15:29
  • @user471651, I didn't understand. Could you please elabotate? – pi-π Feb 04 '18 at 15:33
  • @MathLover, How do we get $f(x)$ from this? – pi-π Feb 04 '18 at 15:38
  • @blue_eyed_... Note that whatever methods I used to derive at my inequality were valid, so the end inequality is also valid. And, obviously, the other inequality given in the question is valid as well. So, what is the only way to validate both the inequalities simultaneously? – Gaurang Tandon Feb 04 '18 at 16:15