Author Topic: What is this?  (Read 700 times)

Jingxuan Zhang

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What is this?
« on: March 08, 2018, 08:45:09 AM »
$$F^*$$The one that suffices $FF^*=F^*F=I$?

Victor Ivrii

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Re: What is this?
« Reply #1 on: March 08, 2018, 09:10:26 AM »
Adjoint operator. We have inner product $(u,v)=\int u\bar{v}\,dx$ (over $\mathbb{R}$) and $F^*$ is defined as $(Fu,v)=(u, F^*v)$ for all $u,v\in L^2$.

Jingxuan Zhang

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Re: What is this?
« Reply #2 on: March 08, 2018, 09:37:24 AM »
But I assume its just another way to write $F^{-1}$?

Victor Ivrii

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Re: What is this?
« Reply #3 on: March 08, 2018, 12:08:14 PM »
We stay here that $F$ is an unitary operator, i.e. $F^*=F^{-1}$