Problem 4

Let

c>0

be a given positive real and

\mathbb{R}_{>0}

be the set of all positive reals. Find all functions

f:\mathbb{R}_{>0}\to\mathbb{R}_{>0}

such that

f((c+1)x+f(y))=f(x+2y)+2cx\quad\text{for all }x,y\in\mathbb{R}_{>0}.

Step 1 of 3: Prove the lower bound f(y)≥2y

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f(y)<2y\ \Longrightarrow\ x=\frac{2y-f(y)}{c}>0\ \Longrightarrow\ (c+1)x+f(y)=x+2y\ \Longrightarrow\ 2cx=0,\text{ contradiction}\ \Longrightarrow\ f(y)\ge2y

Detailed analysis

Suppose $f(y)<2y$ for some $y>0$ . Then $x=(2y-f(y))/c>0$ makes the two arguments $(c+1)x+f(y)$ and $x+2y$ equal, so the functional equation forces $2cx=0$ , contradicting $c>0$ and $x>0$ . Thus $f(y)\ge2y$ for all $y>0$ .