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 2 of 5: Bound f(f(x)) and extract a global constant bound

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In plain words

Applying step 1's bound to the input f(x) tames the mysterious term f(f(x)); choosing y to cancel the f(x) term then produces a bound with no f(x) left at all.

f(x+y)\le (y+a)f(x)+b

Detailed analysis

Applying step 1's bound with $y$ replaced by $f(x)$ gives $f(f(x))\le af(x)+b$ . Substituting this into the original inequality $f(x+y)\le yf(x)+f(f(x))$ yields $f(x+y)\le (y+a)f(x)+b$ for all $x,y$ . Now set $y=-a$ : the coefficient $y+a$ vanishes, so $f(x-a)\le b$ ; since $x$ ranges over all reals so does $x-a$ , hence $f(z)\le b$ for every real $z$ .