I need to research on the uniform, weak and strong convergence

$T_n:L_2(\Bbb{R})\rightarrow L_2(\Bbb{R})$

$T_n \;\,f(x) = f(x+n)$

I've concluded that this sequence converges to zero or diverges.

Do you have any ideas?

Thank you for your time.


If you've already ruled out the possibility that $T_n$ converges to something other than 0, then it should be very easy to see that it doesn't converge to 0 strongly. ($\|T_n f\| = \|f\|$ for all $n$.) So it can't converge uniformly (i.e. in operator norm) either.

To show that $T_n \to 0$ weakly, try first showing that for any $f \in L^2(\mathbb{R})$ and for any $\epsilon > 0$ there exists sufficiently large $n$ so that $\int_{[-n,n]^C} f^2 < \epsilon$. (Use dominated convergence.) Given $f,g$ and $\epsilon$, find an $n$ that works for both $f$ and $g$. Then estimate $(T_n f, g)$.

  • $\begingroup$ I've almost solved it. Does operator function as $T_nI$(0,1) =$\begin{cases} -n+1,&\text{if }x\in[0,1]\\ -n,&\text{otherwise}\\\end{cases}$. If yes, then sequence diverges. $\endgroup$ – kushtibargo May 29 '13 at 20:19

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