First, the question is:

$f$ is a differentiable function and $f : R \rightarrow R$



My approach for problem is using L'Hospital's rule: $$ \frac{xf(x)-yf(y)}{x-y}=f(x+y) $$ Assuming $y=x$ $$ \lim_{y\to{x}} \frac{xf(x)-yf(y)}{x-y} = f(2x)$$ Taking the limit using l'hospital's rule: $$ (yf'(y)+f(y))|_{y=x} = f(2x) $$ Taking the derivative of both sides: $$ f'(x)+xf''(x)+f'(x)=2f'(2x) $$ But the answer is $f'(2x)=f'(x)$

How is this possible?

  • 1
    $\begingroup$ Could it be that $f$ must have the form $f(x)=ax+b$? (All such functions satisfy the functional equation.) $\endgroup$
    – mickep
    May 18 '15 at 20:10
  • $\begingroup$ @mickep thanks, but what's wrong with my solution? Is there a way not using functional equation techniques ie. finding the function? Maybe a calculus route? $\endgroup$ May 18 '15 at 20:21
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    $\begingroup$ I'm not sure, but if $f$ is as I wrote above, then $f''(x)=0$, and thus $2f'(x)=2f'(2x)$, so the statement follows. I have to sleep, but I'll have a look at this problem tomorrow again... $\endgroup$
    – mickep
    May 18 '15 at 20:25

In equation $$ xf(x)+yf(y)=(x-y)f(x+y), $$ differentiate with respect to $x$ to get $$ xf'(x)+f(x)=(x-y)f'(x+y)+f(x+y). $$ Now, set $y$ to $x$ and to $-x$ to deduce that $$ xf'(x)+f(x)=f(2x)=2xf'(0)+f(0), $$ which proves that $f(x)=xf'(0)+f(0).$

  • $\begingroup$ This is the solution. Thanks. $\endgroup$ May 19 '15 at 5:52

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