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Convergence of bounded linear operators

Fermat

Active member
Nov 3, 2013
188
Let \(\displaystyle (T_{n}) \)be a sequence in \(\displaystyle {B(l_2}\) given by
\(\displaystyle T_{n}(x)=(2^{-1}x_{1},....,2^{-n}x_{n},0,0,....). \)Show that \(\displaystyle T_{n}->T\) given by
\(\displaystyle T(x)==(2^{-1}x_{1},2^{-2}x_{2},0,0,....). \)

I get a sequence of geometric series as my answer for the norm, but not sure whether that's correct.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Let \(\displaystyle (T_{n}) \)be a sequence in \(\displaystyle {B(l_2}\) given by
\(\displaystyle T_{n}(x)=(2^{-1}x_{1},....,2^{-n}x_{n},0,0,....). \)Show that \(\displaystyle T_{n}->T\) given by
\(\displaystyle T(x)==(2^{-1}x_{1},2^{-2}x_{2},0,0,....). \)

I get a sequence of geometric series as my answer for the norm, but not sure whether that's correct.
You have written the limit operator wrongly. Its definition should be $T(x)=(2^{-1}x_{1},2^{-2}x_{2},\color{red}{2^{-3}x_{3},2^{-4}x_{4},\ldots}).$ To show that $T_n\to T$, you need to show that \(\displaystyle \sup_{\|x\|\leqslant1}\|T(x) - T_n(x)\| = 0.\)
 

Fermat

Active member
Nov 3, 2013
188
You have written the limit operator wrongly. Its definition should be $T(x)=(2^{-1}x_{1},2^{-2}x_{2},\color{red}{2^{-3}x_{3},2^{-4}x_{4},\ldots}).$ To show that $T_n\to T$, you need to show that \(\displaystyle \sup_{\|x\|=1}\|T(x) - T_n(x)\| = 0.\)
Do you mean \(\displaystyle \sup_{\|x\|=1}\|T(x) - T_n(x)\| -> 0.\) ?

Well, \(\displaystyle \|T(x) - T_n(x)\|\) is a sequence of geometric series, each of whose sum is zero, so the sequence tends to zero. Is this correct?
 

Fermat

Active member
Nov 3, 2013
188
Actually I think that's wrong. Can you help me out?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Do you mean \(\displaystyle \sup_{\|x\|=1}\|T(x) - T_n(x)\| \to 0\) ?
Yes (careless typo on my part).

The coefficients in the definition of $Tx$ form a geometric sequence, but that does not help in the proof that $T_n\to T$. What you have to do is to look at the norm of $\|T(x) - T_n(x)\|$, where $x$ is a unit vector in $l_2$. If $x = (x_1,x_2,x_3,\ldots)$ then $T(x) - T_n(x) = (0,0,0,\ldots,0,2^{-(n+1)}x_{n+1},2^{-(n+2)}x_{n+2},\ldots)$ (where the first $n$ coordinates are all zero), and show that given $\varepsilon$ you can choose $n$ large enough to make the $l_2$-norm of that vector less than $\varepsilon$ for all $x$ in the unit ball of $l_2$.
 

Fermat

Active member
Nov 3, 2013
188
I get that the sup of \(\displaystyle \|T(x) - T_n(x)\|\) over norm vectors is equal to the square root of the series from k=1 to infinity of \(\displaystyle \frac{1}{4^{n+k}}\). So I need to show this sequence of series converges to zero. Agree?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
I get that the sup of \(\displaystyle \|T(x) - T_n(x)\|\) over norm vectors is equal to the square root of the series from k=1 to infinity of \(\displaystyle \frac{1}{4^{n+k}}\). So I need to show this sequence of series converges to zero. Agree?
Yes. (Yes)