3
$\begingroup$

Let $x,y \in \mathbb{R}$ where $y=x-t$. Translation-invariant (or shift-invariant) kernel $\kappa(\cdot,\cdot)$ is defined as $\kappa(x,y) = \kappa(x,x-t) = \kappa(t)$.

Can I say that the function $\kappa$ is symmetric ?

I think "yes", if I can define the translation-invariance as "DIFFERENCE between $x$ and $y$" so that $y=x+t$. And $\kappa(x,y)=\kappa(x,x+t)=\kappa(-t)$.

But, wanted make sure the relation between translation-invariance and symmetry (i.e., $\kappa(x,y)=\kappa(y,x)$).

Can I simply say that translation invariant kernel is $\kappa(x,y) = \kappa(x-y) = \kappa(y-x)$ ? translation-invariant $\to$ symmetric. Is this always correct??? Or, is the symmetry required for the second equality?

  • 2
    I don't understand what you're trying to do. You introduce two different functions $f$, one with two arguments, one with one argument. Then you refer to "the function $f$" with a definite article; but it's not clear (to me) which of these two $f$s you're referring to.2012-11-27
  • 0
    I edited my question. I was trying to understand the relation between translation invariance and symmetry...2012-11-28
  • 0
    You can always split into symmetric and antisymmetric part. invariance under a group action is sometimes also considered as a sort of symmetry, but symmetric operators is a sort of Z/2 invariance not R invariance2013-10-01

2 Answers 2