Solving functional equation; $f(x+y)=yf(x)+f(y)$ also given that $f(1)=2$

Question

Question: If the function $f(x)$ satisfies the relation $(f(x+y)=yf(x)+f(y))$ with $f(1)=2$, then $\lim_{x\to 1^+}f'(x)$ is:
Domain of function is $(1,\infty)$

I need to find $f'(1)$ to solve this question. But can't seem to find the solution of the equation.
Now this functional equation is giving me problems. Solving functions without calculus is often too easy for me but idk why this is hard. Maybe this requires calculus.

Till now I've tried putting $1$ at the places of $x$ and $y$ and then trying to solve which results in $x=1$.
Calculus doesn't help in my case. I tried differentiating which resulted in long differential equation which simplified to $x=1$ again. XD

Thank you.

Answer 1

If we set $y=0$, we get $f(x)=f(0)$. This means that the function is a constant. So, $f(x)=2$. However, this clearly doesn't satisfy the condition set by $x=y=1$, which makes $f(2)=4$, so there is no solution.

Edit: Upon OP's comments, where it was specified that the domain of the function is $[1,\infty)$. I don't have a full solution, but here are some thoughts.

Note that by plugging in $x=1$, we have $f(y+1)=2y+f(y)$. So, we have $f(n)=n^2-n+2$ for $n\in\mathbb N$.

When we let $x=y$, our functional equation becomes $f(2x)=(x+1)f(x)$.

So, we will compute $f\left(\frac{2^n+1}{2^n}\right)$ for natural $n$.

Note that $f(2^n+1)=2^{2n}+2^n+2$ for $n\in\mathbb N$

So, $$f\left(\frac{2^n+1}{2^n}\right)=(2^{2n}+2^n+2)\cdot\prod\limits_{i=1}^n\frac{2^i}{2^n+2^i+1}$$So consider the limit as $n\to\infty$.

Edit 2: As @MartinR notes, since setting $y=1$ forces the function to be linear, which clearly contradicts the work above, there is no such function.

Answer 2

Hint: try the definition of the derivative. $$f’(x)=\displaystyle\lim_{h\to 0}{f(x+h)-f(x)\over h}$$

$$=\displaystyle\lim_{h\to 0}{hf(x)+f(h)-f(x)\over h}$$

Another route you could try is since the expression $f(x+y)$ is symmetric in $x,y$ but the RHS is not, $$yf(x)+f(y)-xf(y)-f(x)=0$$

$$f(x)(y-1)+f(y)(1-x)=0$$

This would indicate $f(x)=k(x-1)$ since $\frac{f(x)}{f(y)}= \frac{x-1}{y-1}$ and $f(x)$ doesn’t depend on $y$. But, in this case, even when taking limits, $f(1) \not = 2$ and $f(1)\not = -2$. Hence a contradiction. (It has been shown in another way by Martin R at the time of making this edit). It is anyway unclear what the question means by $f(1)=2$ when $1$ is not in the domain of $f(x)$.

Answer 3

There is no function $f$ with $f(1) = 2$ and $f(x+y) = yf(x) + f(y)$ for $x, y \ge 1$.

Setting $y=1$ gives $f(x+1) = f(x) + 2$, so that $f(2) = 4$ and $f(3) = 6$.

Then set $x=1, y=2$ to get a contradiction: $6 = f(1+2) = 2f(1) + f(2) = 8$.

Answer 4

You can show that the only function $ f $ with $ f : ( 1 , + \infty ) \to \mathbb C $ and $$ f ( x + y ) = y f ( x ) + f ( y ) \tag 0 \label 0 $$ is the constant zero function. To see this, substitute $ x $ for $ y $ and $ y $ for $ x $ in \eqref{0} to get $ f ( y + x ) = x f ( y ) + f ( x ) $. Combining this with \eqref{0} and rearranging the terms, we have $ ( y - 1 ) f ( x ) = ( x - 1 ) f ( y ) $. Letting $ y = 2 $ in this eqation, we have $$ f ( x ) = f ( 2 ) ( x - 1 ) \text . \tag 1 \label 1 $$ Now, if we let $ x = 3 $ and $ y = 2 $ in \eqref{0} we'll have $ f ( 5 ) = 2 f ( 3 ) + f ( 2 ) $, which combining with \eqref{1} shows that $ 4 f ( 2 ) = 5 f ( 2 ) $. Thus $ f ( 2 ) = 0 $, which using \eqref{1} gives what was claimed. Therefore $ f $ is differentiable everywhere with $ f ' ( x ) = 0 $ for every $ x $ in $ ( 1 , + \infty ) $. In particular, you get $ \lim _ { x \to 1 ^ + } f ' ( x ) = 0 $.

As the domain of $ f $ is $ ( 1 , + \infty ) $, $ f ( 1 ) $ has no meaning. But if you rephrase the question so that the domain of $ f $ is $ [ 1 , + \infty ) $ with $ f ( 1 ) $ equal to anything, for example $ f ( 1 ) = 2 $, and ask for the functional equation \eqref{0} to hold only for $ x , y \in ( 1 , + \infty ) $ and not for every $ x $ and $ y $ in the domain, it will make sense, and the answer is as above. If you want \eqref{0} to hold for every $ x $ and $ y $ in the domain $ [ 1 , + \infty ) $, the same reasoning forces you to have $ f ( 1 ) = 0 $, as \eqref{1} says so.