Christmas Streak 67/88: Functional Equation #3

Algebra Level 4

Tom and Mot are, again, looking at the functional equation:

$\large \color{#707070} f:\mathbb{R}\longrightarrow\mathbb{R};~\color{#333333} (x+y)f(x-y)=f(x^2-y^2)\color{#707070};~x,~y\in\mathbb{R}$

Now Mot claims that $f(x)=ax$ for any reals $a.$

Then Tom takes an objection pose. Determine if Mot is right or wrong.

This problem is a part of <Christmas Streak 2017> series .

Mot is wrong,

f

is indeed continuous, but it can also be transcendental. Mot is wrong,

f

can also be pathological. Mot is correct. Mot is wrong,

f

is indeed a polynomial, but its degree can get higher than

1.

1 solution

Boi (보이)
Dec 9, 2017

Substitute $x=\dfrac{p+1}{2}$ and $y=\dfrac{p-1}{2}$ to get $f(p)=f(1)p.$

Then plug this into the original equation to see that this holds.

Therefore, $f(x)=ax$ for any reals $a.$

There is no need to do the $x=y=1$ and $-y$ bits of this argument. Just putting $x = \tfrac12(p+1)$ and $y = \tfrac12(p-1)$ into the original equation gives $f(p) = pf(1)$ as required.

Mark Hennings - 3 years, 6 months ago

Thank you!

Boi (보이) - 3 years, 6 months ago

Christmas Streak 67/88: Functional Equation #3

1 solution

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