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 4 of 5: Pin down that both special values vanish

Previous step Next step

In plain words

Feeding cleverly chosen numbers back into the earlier inequalities traps f(0) between 0 from above and 0 from below.

a=b=0

Detailed analysis

Put $x=2a-1,y=1-a$ in step 2's inequality $f(x+y)\le (y+a)f(x)+b$ : the coefficient becomes $y+a=1$ and $x+y=a$ , giving $f(a)\le f(2a-1)+b$ , i.e. $f(2a-1)\ge f(a)-b=0$ (using $f(a)=b$ ). Combined with the universal bound $f(2a-1)\le0$ from step 3, this forces $f(2a-1)=0$ . Setting $y=0$ in the original inequality gives $f(x)\le f(f(x))$ for every $x$ ; taking $x=2a-1$ and using $f(2a-1)=0$ yields $0\le f(f(2a-1))=f(0)=a$ . But step 3 also gives $a=f(0)\le0$ , so $a=0$ , and then $b=f(a)=f(0)=a=0$ ; that is, $a=b=0$ .