Problem 3

Let

f:\mathbb{R}\to\mathbb{R}

be a real-valued function defined on the set of real numbers that satisfies

f(x+y)\le yf(x)+f(f(x))

for all real numbers

x

and

y

. Prove that

f(x)=0

for all

x\le0

Step 5 of 5: Finish with y = -x

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In plain words

With both constants gone, comparing a point to its own negative shift shows negative inputs cannot go strictly below zero either, pinning them at exactly zero.

f(x)=0\ \text{for all}\ x\le0

Detailed analysis

With $a=b=0$ , step 3's inequality $f(x+y)\le yf(x)+b$ becomes $f(x+y)\le yf(x)$ . Setting $y=-x$ gives $f(0)\le -xf(x)$ , and since $f(0)=a=0$ this says $xf(x)\le0$ for every real $x$ . For $x<0$ , dividing by the negative number $x$ reverses the inequality to give $f(x)\ge0$ ; together with the universal bound $f(x)\le0$ from step 3, this forces $f(x)=0$ for every $x<0$ , and $f(0)=0$ was already known. Hence $f(x)=0\ \text{for all}\ x\le0$ .