How can one prove P(τ)S(τ) = S(τ)P(0)?

[tutumar]

Assume u:R\rightarrow C^n and define shift operator S(\tau) with

S(\tau)u(t)=u(t-\tau)

and truncation operator P(\tau) with

P(\tau)u(t)=u(t) for t\leq\tau and 0 for t>\tau

Then P(\tau)S(\tau)=S(\tau)P(0) for every \tau>=0.

Can someone please prove last statement..

 

[HallsofIvy]

Looks like pretty direct computation. If u(t) is any such function, then what isSD(\tau)u? What is P(\tau)S(\tau)u? Then turn around and find S(\tau)P(0)u.

 

[tutumar]

Yes, I tried that, and it just doesn't fit..

P(\tau)S(\tau)u(t)=P(\tau)u(t-\tau)=u(t-\tau) if t-\tau<=\tau and 0 for t-\tau>\tau

S(\tau)P(0)u(t)=S(\tau)u(t) for t<=0 and 0 otherwise=u(t-\tau) if t<=0 and 0 otherwise..

Well, something's got to be wrong here, but I can't see what..

 

[Einj]

I think your last equation is wrong. As, if we have:

$$P(0)u(t)=u(t) \mbox{ if } t\leq 0 \mbox{ and } 0 \mbox{ otherwise }$$

than:

$$S(\tau)P(0)u(t)=u(t-\tau) \mbox{ if } t-\tau\leq 0 \mbox{ and } 0 \mbox{ if } t-\tau>0$$

Still, I'm not able to prove the statement as in the first case you have $$t-\tau\leq\tau$$ and in this case there is $$t-\tau\leq 0$$. I'm sorry...