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 1 of 5: Name the two special values and bound f everywhere

In plain words

Plugging in x=0 turns the inequality into a plain linear upper bound for f, expressed through f(0) and f(f(0)).

f(y)\le ay+b

Detailed analysis

Write $a=f(0)$ and $b=f(f(0))=f(a)$ . Setting $x=0$ in the given inequality gives $f(y)\le ay+b$ for every real $y$ .