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 3 of 5: Push y to -infinity to force f(x)≤ 0 everywhere

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In plain words

Applying the global bound to f(f(x)) simplifies the inequality further, and comparing it with itself after swapping x and y squeezes f(x) between two expressions that both vanish as y runs off to minus infinity.

f(x)\le0

Detailed analysis

Since $f(f(x))\le b$ for every $x$ (step 2), the original inequality simplifies to $f(x+y)\le yf(x)+b$ for all $x,y$ . Replacing $(x,y)$ by $(x+y,-y)$ gives $f(x)\le -yf(x+y)+b$ . On the other hand, for $y<0$ multiplying $f(x+y)\le yf(x)+b$ by the negative number $y$ reverses the inequality and rearranges to $-yf(x+y)\le -y^2f(x)-yb$ . Combining the two gives $f(x)\le -y^2f(x)+b(1-y)$ , i.e. $f(x)\le\dfrac{b(1-y)}{1+y^2}$ for every $y<0$ . As $y\to-\infty$ the right side tends to $0$ , so $f(x)$ cannot exceed any positive number, i.e. $f(x)\le0$ for every real $x$ .