## TuringTest 3 years ago Let $$f: \mathbb{R} \to \mathbb{R}$$ be a real valued function define on the set of real numbers that satisfies$\begin{matrix}&&&&&&&&&(f(x+y)\le yf(x) +f((f(x))\end{matrix}$for all real numbers $$x$$ and $$y$$. Prove the $$f(x) = 0$$ for all $$x\le0$$

1. moneybird

Setting y = 0 $f(x) \le f(f(x))$Setting x = 0$f(y) \le yf(0) + f(f(0))$Setting y = f(x) - x, $f(f(x)) \le f(x)^2 - xf(x) +f(f(x)) \implies f(x)^2 \ge xf(x)$Setting y = f(0) -x$f(f(0)) \le f(x) f(0) - xf(x) + f(f(x))$Substitute the below value for f(f(x)), $f(f(x)) \le f(x)^2 - xf(x) +f(f(x)) \implies f(x)^2 \ge xf(x)$So you get $f(f(0)) \le f(x)f(0) -xf(0) - xf(x) + f(f(x)) \le f(x)f(0) -xf(x) + f(x)f(0) + f(f(0) \implies$$2f(0)f(x) \ge xf(x)$If $i) f(x) \ge 0 \implies 2f(0) \ge x$$ii) f(x) \le 0 \implies 2f(0) \le x$Assume $f(0) \neq 0$, Case 1: $f(0) >0$Setting $y = \frac{-1-f(f(0))}{f(0)}$$-1 - f(f(0)) \ge 2f(0)^2\implies f(f(0)) \le -1$This contradicts with the first equation, therefore $f(0) \le 0$Case 2 f(0) < 0 Assume $f(f(0)) \ge 0$, but it contradicts with the first equation. Therefore,$f(f(0))<0$The second equation can now becomes $f(y) \le yf(0)$from the above equation, set y = f(x)$f(f(x))\le f(0)f(x)$Thus, the first equation can be rewritten as $f(x) \le f(0)f(x)$If $f(0) < 0 \implies f(x) \le 0$But this contracts to the statement$f(x) \le 0 \implies 2f(0) \le x$Hence,$f(0) = 0$So$f(x) \le f(x) f(0) \implies f(x) \le 0$and $2f(0)f(x) \ge xf(x) \implies xf(x) \le 0$If x is a negative number, then the above equation,$xf(x) \le 0 \implies f(x) \ge 0$Therefore, if $x \le 0$$f(x) \le 0$ and$f(x) \ge 0$, which means $f(x) = 0$

2. malevolence19

Damnnnnn

Find more explanations on OpenStudy