A community for students.
Here's the question you clicked on:
 0 viewing
TuringTest
 4 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 \(x\le0\)
TuringTest
 4 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 \(x\le0\)

This Question is Closed

moneybird
 4 years ago
Best ResponseYou've already chosen the best response.2Setting 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 \[y = \frac{1f(f(0))}{f(0)} \]\[1  f(f(0)) \ge 2f(0)^2\implies f(f(0)) \le 1\]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 \[f(x) = 0\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.