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On the Hilbert space $H=L^2([0,\infty])$,define linear operator $Tf(x)=f(x+1)$ for x>0. How can I prove that the operator T is bounded? Also, how to check whether the operator T is compact and whether it is bijective?

Since I am beginning to learn functional analysis, but I feel a little confused about the question. As for this operator, I think the norm of T should be 1,but I don't know how to give a rigorous proof. Also, how can I know whether this operator is compact and bijective or not? I know the definition of compact operator https://en.wikipedia.org/wiki/Compact_operator. but I don't know how to apply this definition to this problem? Can someone help me solve this problem?

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Note that

$$\sup_{\|f\|_{L^2(0,\infty)}=1} \|Tf\|_{L^2(0,\infty)}= \sup_{\|f\|_{L^2(0,\infty)}=1}\|f\|_{L^2(1,\infty)} \leq 1,$$

so, indeed, the norm is bounded by 1. To show that the norm is exactly $1$, take any $f$ which achieves this, for example an indicator function on $[1,2]$.

It is not compact. To see this we can construct a bounded sequence $\{f_n\}\subset L^2(0,\infty)$ such that $\{Tf_n\}$ does not contain a convergent subsequence. Motivated by the previous example, take $f_n = \chi_{[n,n+1]}.$ Then, $Tf_n = \chi_{[n+1,n+2]}$ and $\|Tf_n-Tf_m\|=1$ for all $n\neq m$.

It is also not bijective. Note that taking $f=0$ and $g=\chi_{[0,1]}$ we have $f\neq g$ but $Tf = Tg = 0$.

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  • $\begingroup$ oh, thank you! That really helps! $\endgroup$
    – user144600
    Nov 23, 2016 at 20:00
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    $\begingroup$ You're welcome. Good luck with your studies! $\endgroup$
    – Matt
    Nov 23, 2016 at 20:00

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