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  • 5 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 \(x\le0\)

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  1. anonymous
    • 5 years ago
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    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. anonymous
    • 5 years ago
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