Find $f:\mathbb R\to\mathbb R$ satisfying $f\bigl(yf(x+y)+f(x)\bigr)=4x+2yf(x+y)$.
This is my attempt:
\begin{align}
& \text{Let } f(0)=t \text. \\
&P(0, 0): f(t)=0 \text. \\
&P(0, t): t = 0 \text. \implies f(0)=0 \text. \\
\\
&\text{Assume) } \exists \ t \text{ s.t. } t \ne 0 \text{ and } f(t)=0 \text. \\
&P(t, 0): f(0)=4t \text. \\
&P(0, 0): f(4t)=0 \text. \\
&P(4t, 0): f(0)=16t \text. \\
&\implies 4t=16t \text, \ t=0 \text. \\
&\therefore f(t)=0 \iff t=0 \text. \\
\\
&\text{Let } f(a)=f(b) \text. \\
&P(a, 0):f\bigl(f(a)\bigr)=4a \text. \\
&P(b, 0):f\bigl(f(b)\bigr)=4b \text. \\
&\implies a = b \text. \\
&\therefore f(a)=f(b) \iff a = b \text.
\end{align}
My expectation is the $f(x)=2x$.
Can you help me getting the perfect process of finding the function $f$?
Edit:
I found more info…
\begin{align}
& P(x, x): f\bigl(xf(2x)+f(x)\bigr)=4x+2xf(2x) \text. \\
& P(x, -x): f\bigl(f(x)\bigr)=4x \text. \\
& P(0, x): f\bigl(xf(x)\bigr)=2xf(x) \text. \\
& x=1 \text; \ f\bigl(f(1)\bigr)=2f(1) \text. \\
& \text{From } P(1, -1): f\bigl(f(1)\bigr)=4 \text. \\
& \therefore f(1)=2 \text.
\end{align}
Best Answer
You can prove that the only $ f : \mathbb R \to \mathbb R $ satisfying $$ f \bigl ( y f ( x + y ) + f ( x ) \bigr ) = 4 x + 2 y f ( x + y ) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $ is the doubling function $ f ( x ) = 2 x $, even without further assumptions (like continuity of $ f $). It's straightforward to verify that the doubling function is indeed a solution. To prove the converse, put $ y = 0 $ in \eqref{0} to get $$ f \bigl ( f ( x ) \bigr ) = 4 x \tag 1 \label 1 $$ for all $ x \in \mathbb R $. In particular, \eqref{1} shows that $ f $ is injective. Letting $ a = f \left ( \frac 1 2 \right ) $, \eqref{1} gives $ f ( a ) = 2 $. Setting $ y = a - x $ in \eqref{0} and using \eqref{1} you get $$ f \bigl ( 2 ( a - x ) + f ( x ) \bigr ) = 4 x + 4 ( a - x ) = 4 a = f \bigl ( f ( a ) \bigr ) = f ( 2 ) \tag 2 \label 2 $$ for all $ x \in \mathbb R $. By injectivity of $ f $, \eqref{2} shows $$ 2 ( a - x ) + f ( x ) = 2 \text , $$ or equivalently $$ f ( x ) = 2 x + b $$ for all $ x \in \mathbb R $, where $ b = 2 - 2 a $. Therefore, $$ f \bigl ( f ( x ) \bigr ) = 2 ( 2 x + b ) + b = 4 x + 3 b $$ for all $ x \in \mathbb R $, which together with \eqref{1} yields $ b = 0 $. Hence, $ f ( x ) = 2 x $ for all $ x \in \mathbb R $, as desired.