I have been doing some functional equations and in some of them they just say " WLOG let $f(1)=1$ ", but I don't get why they can do that... Can someone please help me?
I can't find the example of $f(1)=1$ but here is one where they take $f(0)=0$: $f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y))$.
At the beginning they just say WLOG $f(0)=0$.
On
In the example, because the functional equation depends only on the difference between two values of $f$, if the function $g$ satisfies the equation, so will $f=g+c$ where $c$ is an arbitrary constant. Choose $c=-g(0)$ and you will find that $f(0)=0$.
Since you can always choose such a value, you can't lose any generality if you do.
If $f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y))$ and $g(x)=k\; f(x)$
then $g(x^3)-g(y^3)=(x^2+xy+y^2)(g(x)-g(y))$ because $kf(x^3)-kf(y^3)=(x^2+xy+y^2)(kf(x)-kf(y))$ by multiplying both sides of the original expression by $k$.
so you lose very little by assuming that $f(1)=1$ and remembering that you can then multiply any result by a constant.
Added: If you let $h(x)=f(x)-f(0)$ then $h(x^3)-h(y^3)=(x^2+xy+y^2)(h(x)-h(y))$ because $f(x^3)-f(0)-f(y^3)+f(0)=(x^2+xy+y^2)(f(x)-f(0)-f(y)+f(0))$. You know $h(0)=f(0)-f(0)=0$ so again you lose very little by assuming that $f(0)=0$ and remembering that you can then add a constant to any result.