Convergence of bounded linear operators

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Discussion Overview

The discussion revolves around the convergence of a sequence of bounded linear operators \( T_n \) in the space \( B(l_2) \) to a limit operator \( T \). Participants explore the definitions and properties of these operators, particularly focusing on the norms and conditions for convergence. The conversation includes mathematical reasoning and technical clarifications related to operator limits.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that \( T_n(x) = (2^{-1}x_{1},...,2^{-n}x_{n},0,0,...) \) converges to \( T(x) = (2^{-1}x_{1},2^{-2}x_{2},0,0,...) \).
  • Others argue that the definition of the limit operator \( T \) was initially written incorrectly, suggesting it should include terms \( 2^{-3}x_{3}, 2^{-4}x_{4}, \ldots \).
  • A later reply questions the correctness of the convergence proof, specifically the need to show that \( \sup_{\|x\|=1}\|T(x) - T_n(x)\| \to 0 \).
  • Some participants mention that the coefficients in the definition of \( T(x) \) form a geometric series, but express uncertainty about how this relates to proving convergence.
  • One participant calculates that the supremum of \( \|T(x) - T_n(x)\| \) over unit vectors equals the square root of a series and seeks agreement on whether this series converges to zero.

Areas of Agreement / Disagreement

Participants express differing views on the correct definition of the limit operator and the approach to proving convergence. There is no consensus on the correctness of the convergence proof or the implications of the geometric series.

Contextual Notes

Some participants highlight the need to consider the norm of the difference \( \|T(x) - T_n(x)\| \) and the implications of the geometric series, but the discussion remains unresolved regarding the specifics of the proof and the conditions for convergence.

Fermat1
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Let $$(T_{n}) $$be a sequence in $${B(l_2}$$ given by
$$T_{n}(x)=(2^{-1}x_{1},...,2^{-n}x_{n},0,0,...). $$Show that $$T_{n}->T$$ given by
$$T(x)==(2^{-1}x_{1},2^{-2}x_{2},0,0,...). $$

I get a sequence of geometric series as my answer for the norm, but not sure whether that's correct.
 
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Fermat said:
Let $$(T_{n}) $$be a sequence in $${B(l_2}$$ given by
$$T_{n}(x)=(2^{-1}x_{1},...,2^{-n}x_{n},0,0,...). $$Show that $$T_{n}->T$$ given by
$$T(x)==(2^{-1}x_{1},2^{-2}x_{2},0,0,...). $$

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 $$\sup_{\|x\|\leqslant1}\|T(x) - T_n(x)\| = 0.$$
 
Opalg said:
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 $$\sup_{\|x\|=1}\|T(x) - T_n(x)\| = 0.$$

Do you mean $$\sup_{\|x\|=1}\|T(x) - T_n(x)\| -> 0.$$ ?

Well, $$\|T(x) - T_n(x)\|$$ is a sequence of geometric series, each of whose sum is zero, so the sequence tends to zero. Is this correct?
 
Actually I think that's wrong. Can you help me out?
 
Fermat said:
Do you mean $$\sup_{\|x\|=1}\|T(x) - T_n(x)\| \to 0$$ ?
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 $l_2$.
 
I get that the sup of $$\|T(x) - T_n(x)\|$$ over norm vectors is equal to the square root of the series from k=1 to infinity of $$\frac{1}{4^{n+k}}$$. So I need to show this sequence of series converges to zero. Agree?
 
Fermat said:
I get that the sup of $$\|T(x) - T_n(x)\|$$ over norm vectors is equal to the square root of the series from k=1 to infinity of $$\frac{1}{4^{n+k}}$$. So I need to show this sequence of series converges to zero. Agree?
Yes. (Yes)
 

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