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Math Help - Operator Convergence

  1. #1
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    Operator Convergence

    I need to show that the operator T_n:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C}) defined by (T_nx)(j)=\frac{1}{2^j}x(j) for j\leq n and (T_nx)(j)=0 for j>n converges to T in the operator norm \|T_n-T\|_{op}\rightarrow 0 where (Tx)(j)=\frac{1}{2^j}x(j) for all j

    Any help would be great
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  2. #2
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    Quote Originally Posted by ejgmath View Post
    I need to show that the operator T_n:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C}) defined by (T_nx)(j)=\frac{1}{2^j}x(j) for j\leq n and (T_nx)(j)=0 for j>n converges to T in the operator norm \|T_n-T\|_{op}\rightarrow 0 where (Tx)(j)=\frac{1}{2^j}x(j) for all j
    Notice that if y = (T-T_n)x then y(j) = 0 if j\leqslant n and y(j) = 2^{-j}x(j) for j>n. So \|y\|^2 = \sum_{j=n+1}^\infty 2^{-2j}|x(j)|^2 < 2^{-2n}\sum_{j=n+1}^\infty |x(j)|^2 < 2^{-2n}\|x\|^2. Hence \|T-T_n\| \leqslant 2^{-n}.
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