Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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 \(x\le0\)

Meta-math
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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 \[f(x) = 0\]
Damnnnnn

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question