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

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.

• On which domain is $f$ defined? Dec 10, 2019 at 8:20
• $1^{+}$ and in real numbers. Dec 10, 2019 at 8:26
• What is $1^+$? Do you mean the interval $[1, \infty)$? Dec 10, 2019 at 8:27
• Martin yes but with open brackets both side. Dec 10, 2019 at 8:27
• @LoveInvariants But you have $f(1)=2$. So it isn't open in both directions. Dec 10, 2019 at 8:28

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.

• but f(0) isn't 2. Dec 10, 2019 at 8:18
• If we let $x=1$, $f(1)=f(0)$. Dec 10, 2019 at 8:20
• oh yes. But the answer was given -2. Shouldn't it be 0. Dec 10, 2019 at 8:22
• Well, unless there's some domain information that you haven't provided, that isn't correct. Dec 10, 2019 at 8:22
• @DonThousand: Unless I am mistaken, there is no such function, see my comment at the question. Or did I make some error? Dec 10, 2019 at 8:59

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: $$6 = f(1+2) = 2f(1) + f(2) = 8$$.

• Maybe d option is correct which is not defined but answer is written as -2. Dec 10, 2019 at 9:06
• can't there be system of equations Dec 10, 2019 at 9:14
• @LoveInvariants: Sorry, I don't get what you mean. You asked about a function having certain properties, not about a system of equations. If your real problem is not what you wrote above then I suggest that you update the question, or post a new one. Dec 10, 2019 at 9:17
• Nope can't there be system of functions with same derivative at x=1. The questions asks limit of derivative at 1 no? Dec 10, 2019 at 9:19
• @LoveInvariants Your edit only changes the question to be about the derivative of a non-existant function rather than the function itself! There is no function $f$ satisfying your condition. Unless you change the functional equation itself, asking any further questions about a non-existant function satisfying it will remain meaningless. Dec 10, 2019 at 9:45

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 $$f(x)$$.

• ok. Let me try then I will accept. Dec 10, 2019 at 8:34
• This doesn't work, since for $h<1$, $f(h)$ is not defined. Dec 10, 2019 at 8:35
• Darn. Missed that. Added another route. Dec 10, 2019 at 8:36
• still this doesn't help me. Dec 10, 2019 at 8:45
• What does $f(1)=2$ mean since it’s given that $1$ isn’t in the domain? Dec 10, 2019 at 9:06

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.