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 is$SD(\tau)u$? 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...