Christmas Streak 66/88: Functional Equation #2

Algebra Level 4

Tom and Mot are looking at the functional equation where $f$ is differentiable:

$\large \color{#707070}f:\mathbb{R}\longrightarrow\mathbb{R};~\color{#333333} f(x+y)=f(x)+f(y)\color{#707070};~x,~y\in \mathbb{R}.$

After a few minutes of thinking, Tom concludes that $f(x)=ax$ for any reals $a.$

However, Mot has some doubts. Determine if Tom is right or wrong.

Tom is correct. Tom is wrong,

f

can also be a polynomial whose degree is greater than

1.

Tom is wrong, he didn't include

f(n)=0.

Tom is wrong, it should be

f(n)=an+b

for any reals

a

and

b.

Tom is wrong,

f

can also be a transcendental function.

1 solution

Boi (보이)
Dec 9, 2017

Since $f$ is differentiable, we differentiate both sides with respect to $y:$

$f'(x+y)=f'(y)$

Now we substitute $y=0:$

$f'(x)=f'(0)$

This suggests that $f'(x)$ is a constant function.

Integrating both sides, $f(x)=f'(0)x + C$

However, substituting $x=y=0$ to the original equation yields $f(0)=0,$ which, substituting $x=0$ to the equation above yields $C = 0.$

Therefore, $\boxed{f(x)=f'(0)x}.$

$f(x)=0$ is clearly included, since $f'(0)=0$ yields $f(x)=0.$