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Let $1\le p \le \infty$ and let $(X,\Omega, \mu)$ be a $\sigma$-finite measure space.If $A\in \mathscr{B_0}(L^P(\mu))$, show that there is a sequence $\{A_n\}$ of finite rank operators such that $||A_n-A||\rightarrow 0$.

I know I need to use (a useful result). But I do not know how to construct those finite rank operators in (a useful result).

Any help would be appreciated!

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David Lee
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  • See here: http://math.stackexchange.com/questions/288992/there-are-compact-operators-that-are-not-norm-limits-of-finite-rank-operators – Math1000 Feb 29 '16 at 16:03
  • what is your definition of a compact operator ? and the best approximation of rank $N$ (in the sense of operator norm of the residual) is constructed by taking the $N$ greatest singular values/vectors – reuns Feb 29 '16 at 20:51
  • @Math1000.Thank you. Is there an answer to this question? I can not find it. could you explain more clear? – David Lee Mar 01 '16 at 00:09
  • @user1952009.Compact operators are bounded linear operators that the closure of its image of the closed unit ball is compact,i.e.closure of ${T(Ball L^p(\mu))}$ is compact. could you expain your seond sentence more clear? – David Lee Mar 01 '16 at 00:12

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