A function $f:\mathbb{R^{+}}\rightarrow \mathbb{R^{+}}$ which satisfies the functional equation $$f((c+1)x+f(y))=2cx+f(x+2y)$$
Where $c$ is a positive constant, Find all such possible functions.
I tried putting $y=1$ but $f((c+1)x+f(1))$ made things tougher. I am unable to solve this could someone help me?
Best Answer
There is the idea of substituion, which consists in equating subfunctional expressions. And this is the case where this idea helps to see the implicit constraints imposed on the function by the original functional equation: $$(c+1)x+f(y)=x+2y\Rightarrow cx+f(y)=2y\Rightarrow f(y)\ge 2y, \; y\in \mathbb R^+\qquad(1)$$ In deed, if the inequality higher isn't satisfied for some $y_0$, then exists $x_0$ such that equality $(c+1)x_0+f(y_0)=x_0+2y_0$ is reached and it leads to $$f((c+1)x_0+f(y_0))=2cx_0+f(x_0+2y_0)\Rightarrow0=2cx_0$$ what is impossible.
Now one of the natural ideas is to find an upper limiting expression. Let's use $(1)$: $$f((c+1)x+f(y))\ge 2(c+1)x+2f(y)\Rightarrow$$ $$ \Rightarrow 2cx+f(x+2y)\ge 2(c+1)x+2f(y)\Rightarrow $$ $$\Rightarrow f(x+2y)\ge 2x+2f(y) \qquad (2) $$ $$ f((c+1)x+2\frac{f(y)}{2})\ge 2(c+1)x+2f(\frac{f(y)}{2})\Rightarrow $$ $$\Rightarrow 2cx+f(x+2y)\ge 2(c+1)x+2f(\frac{f(y)}{2})\Rightarrow $$ $$\Rightarrow f(x+2y)\ge 2x+2f(\frac{f(y)}{2})\Rightarrow f(y)\le 4y, \; y\in \mathbb R^+ \qquad (3)$$ Let's suppose exists $y_1$ such that $(3)$ isn't satisfied. Then it's obvious exists $x_1$ such that: $$2x_1+4y_1=f(y_1)\Rightarrow x_1+2y_1=\frac{f(y_1)}{2}\Rightarrow$$ $$\Rightarrow f(x_1+2y_1)\ge 2x_1+2f(\frac{f(y_1)}{2})\Rightarrow$$ $$\Rightarrow 0 \ge 2x_1+f(\frac{f(y_1)}{2})$$ what is impossible.
Now let's suppose exists $y_3$ such that $f(y_3)\gt 2y_3 \; (4)$. Then exists $a \in (2;4]$ such that $f(x)\le ax, \; x \in \mathbb R^+$ and for any sufficiently small $\varepsilon \gt 0$ we can choose $x_3$ such that $f(x_3)\ge (a-\varepsilon)x_3$. The idea here is to use the existence of such "critical" points of function to find even stronger restrictions on the function. Let's promote this idea: $$(c+1)x+f(y)=x_3 \Rightarrow$$ $$\Rightarrow f((c+1)x+f(y))=(a-\varepsilon)(c+1)x+(a-\varepsilon)f(y)=2cx+f(x+2y) \le 2cx+ax+2ay \Rightarrow$$ $$\Rightarrow (a-\varepsilon)(c+1)x+(a-\varepsilon)f(y) \le 2cx+ax+2ay \Rightarrow$$ $$\Rightarrow (a-\varepsilon)(c+1)x+2(a-\varepsilon)y \le 2cx+ax+2ay \Rightarrow$$ $$\Rightarrow ((a-2-\varepsilon)c-\varepsilon)x \le 2\varepsilon y \qquad (5)$$ Note that in the $(5)$ for sufficiently small $\varepsilon$ expression $((a-2-\varepsilon)c-\varepsilon)x$ is positive (since $a,c$ are fixed numbers in this context). As $y\to0$ we get $2\varepsilon y\to0$ while $x\to \frac{x_3}{c+1}$ and $((a-2-\varepsilon)c-\varepsilon)x \to ((a-2-\varepsilon)c-\varepsilon)\frac{x_3}{c+1}$, so the (5) isn't satisfied (since $(3)$ we get $f(y)\to 0$ as $y\to0$, it also explains, why substitution $(c+1)x+f(y)=x_3$ is legal).
From above we get assumption $(4)$ is false and assuming $(1)$ we conclude $$f(x) \equiv 2x$$ is the only solution of the original functional equation.