MHB Convergence of bounded linear operators

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The discussion focuses on the convergence of a sequence of bounded linear operators \( T_n \) to a limit operator \( T \) defined on \( l_2 \). Participants clarify the correct definition of \( T \) and emphasize the need to show that the supremum of the norm difference \( \|T(x) - T_n(x)\| \) approaches zero as \( n \) increases. The convergence is analyzed through the behavior of a sequence of geometric series, with the goal of demonstrating that the norm difference can be made arbitrarily small for unit vectors in \( l_2 \). The conversation highlights the importance of correctly applying the definitions and limits in the context of operator convergence. Overall, the key takeaway is the necessity of proving that the supremum of the norm difference converges to zero.
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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