# Product of sgn functions

I can't prove the identity in Grafako's book Classical Fourier analysis page 256 directly: for $\xi$ fixed

$\operatorname{sgn}(\xi-y)\cdot\operatorname{sgn}(y)=1-\operatorname{sgn}(\xi)\cdot[\operatorname{sgn}(y)+\operatorname{sgn}(\xi-y)]$, where $\operatorname{sgn}(x)=\begin{cases}1&\mbox{ if }x>0\\\ 0&\mbox{ if }x=0\\\ -1&\mbox{ if }x<0.\end{cases}$

I would like to direct proof, without going through several cases.

Thank's

-

As you've written it, this identity appears false.

For example, if $y$ is not zero and $\xi = 0$, then $\text{LHS} = \operatorname{sgn}(-y) . \operatorname{sgn}(y) = -1$ but $\text{RHS} = 1 - 0 = 1$.

-
adding a $-$ at the left should cure this. – Raymond Manzoni Jan 14 '12 at 19:01
Oops, I forgot the minus sign in the first term... Sorry! :( – user22942 Jan 14 '12 at 19:03

Let's rewrite this as (setting $x=\xi-y$ and ... correcting the sign mistake) : $$\operatorname{sgn}(x)\cdot\operatorname{sgn}(y)=\operatorname{sgn}(x+y)\cdot[\operatorname{sgn}(y)+\operatorname{sgn}(x)]-1,$$

If the sign of x and y are different then $\operatorname{sgn}(y)+\operatorname{sgn}(x)=0$ and the equation is right.
If the sign of x and y are equal then the right becomes $2-1=1$ as it should.

-
Brian you beat me on this correction! :-) – Raymond Manzoni Jan 14 '12 at 19:08