Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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. –  joriki Nov 27 '12 at 23:11
    
I edited my question. I was trying to understand the relation between translation invariance and symmetry... –  Jon Lee Nov 28 '12 at 2:03
    
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 invariance –  plusepsilon.de Oct 1 '13 at 9:10

1 Answer 1

Unfortunately I still don't understand the parts of your question where you try to motivate the idea that translation invariance might imply symmetry, so I can only state that this implication doesn't hold. If a kernel $\kappa(x,y)$ is translation-invariant and is thus given by $\kappa(x,y)=g(x-y)$ for some function $g$, then $\kappa$ is symmetric in the sense $\kappa(x,y)=\kappa(y,x)$ if and only if $g$ is symmetric in the sense $g(t)=g(-t)$, since $\kappa(x,y)=g(x-y)$ and $\kappa(y,x)=g(y-x)=g(-(x-y))$. For instance, for symmetric $g(t)=t^2$, the kernel is symmetric with $\kappa(x,y)=\kappa(y,x)=(x-y)^2$, while for antisymmetric $g(t)=t$, the kernel is antisymmetric with $\kappa(x,y)=-\kappa(y,x)=x-y$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.