Let $ f : [0, 1] \to\ R $ be a function such that :- $1.)$ $f(x) \ge 0$ for all $x$ in $[0,1]$ . $2.)$ $f(1) = 1$ . $3.)$ If $x_1 , x_2$ are in $[0,1]$ such that $x_1 + x_2 \le 1$ , then $f(x_1) + f(x_2) \le f(x_1 + x_2)$ . Show that $f(x) \le 2x $ for all $x$ in $ [0,1] $.
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Tags: functional equation, Functional inequality, inequalities
AdBondEvent
11.10.2016 14:59
Anyone ?
SidVicious
11.10.2016 15:41
$P(x,x) \implies$ $2f(x) \le f(2x) \implies 2^kf(x) \le f(2^kx) \le 1...(*)$ $\forall x \in [0,1], \exists k \in \mathbb N, 2^kx \le 1 \le 2^{k+1}x \stackrel{*}{\implies} 2^kf(x) \le f(2^kx) \le 1 \le 2^{k+1}x \implies \boxed{f(x) \le 2x}$
AdBondEvent
11.10.2016 18:04
SidVicious wrote: $P(x,x) \implies$ $2f(x) \le f(2x) \implies 2^kf(x) \le f(2^kx) \le 1...(*)$ $\forall x \in [0,1], \exists k \in \mathbb N, 2^kx \le 1 \le 2^{k+1}x \stackrel{*}{\implies} 2^kf(x) \le f(2^kx) \le 1 \le 2^{k+1}x \implies \boxed{f(x) \le 2x}$ @above do note that the solution has been asked for all $x$ in $[0,1]$ . The first inequality by you is only possible when $x$ is in $[0,0. 5]$ .
SidVicious
11.10.2016 18:14
Well yeah, then for $\frac{1}{2} \le x \le 1$ we have $f(x) \le 1 \le 2x$ (I considered this trivially).
Korgsberg
18.06.2017 12:32
SidVicious wrote: Well yeah, then for $\frac{1}{2} \le x \le 1$ we have $f(x) \le 1 \le 2x$ (I considered this trivially). how you considered that $f (x)\leq 1$ ?