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Let $T_n$ such that $T_n(e_k)=\begin{cases}\lambda_ke_k&\mbox{ if }k\leq n\\\ 0&\mbox{ if }k>n \end{cases}$. Then $T_n$ is finite ranked hence compact and for $v\in\ell^2$, $v=(v_0,v_1,\ldots)$ $$\lVert (T-T_n)v\rVert^2=\sum_{k=0}^{+\infty}|\langle((T-T_n)v)_k\rangle|^2=\sum_{k\geq n+1}|(T-T_n)(v_k)|^2=\sum_{k\geq ... 9 A very nice and short proof of Pitt's theorem (a bit more than a page and covering the case you're interested in as well) was recently given by Sylvain Delpech MR review here, online article here. Added: Let me adress your questions in the comments in a pedestrian way since I find this more illuminating than appealing to heavy artillery. Lemma. Let ... 7 In order to distinguish the new definition of  {\mu_{n}}(T)  from the old one, let us call it  {\mu^{\text{New}}_{n}}(T) . We shall assume throughout this discussion that  \displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T) < \infty . By the Divergence Test from calculus (it’s hard to believe that something so simple can crop up here!), we have  ... 6 a) Let$$f_n(x)=\left\{\begin{array}{cl} 2^nx, & x\in[0,2^{-n}]\\1\ ,& x\in[1-2^{-n},1]\end{array}\right..$$Then f_n\in C([0,1]) and \|f_n\|_{C([0,1])}=1. Denote g_n=Hf_n. Then$$g_n(x)=\left\{\begin{array}{cl} 2^{n-1}x, & x\in[0,2^{-n}]\\1-2^{-n-1}x^{-1},& x\in[1-2^{-n},1]\end{array}\right..$$Note that if m<n, then ... 6 Suppose, there is some \epsilon > 0 such that M_\epsilon = \{ x \;\vert\; g(x) > \epsilon\} has positive measure \mu(M_\epsilon) > 0. Now pick a sequence of sets M_n \subset M_\epsilon with M_{n+1} \subset M_n and \mu(M_{n+1}) < \frac{1}{2}\mu(M_n) for all n \in \mathbb N. Note that 2. implies \mu(M_n) > 0 for all n\in ... 6 Suppose that g is not zero a.e. and let \epsilon>0 be such that E=\{x:|g(x)|>\epsilon\} has nonzero measure. Consider the orthogonal projection p=T_{\chi_{E}}. Assume that T_{g} is a compact operator. The operator T_{g} naturally induces a continuous linear operator of L^2(\mathbb{R}) into the Hilbert subspace pL^{2}(\mathbb{R})=\{\xi ... 5 Define$$T_j(x):=\left(x_1,\frac{x_2}2,\dots,\frac{x_j}j,0,\dots,0\right).$$It's a compact operator (because it's finite ranked) and$$T(x)-T_j(x)=\left(0,\dots,0,\frac{x_{j+1}}{j+1},\dots\right),$$hence$$\lVert T(x)-T_j(x)\rVert_{\ell^1}=\sum_{k=j+1}^{+\infty}\frac{|x_k|}{k}\leq \frac 1{j+1}\sum_{k=j+1}^{+\infty}|x_k|\leq \frac 1{j+1}\lVert ...

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Let $P_\epsilon$ be the orthogonal projection onto $V_\epsilon$ and $$T_n := T(\mathbb I - P_{1/n})$$ $T_n$ is a finite-rank operator and $$\lVert (T - T_n)v \rVert = \lVert T P_{1/n} v \rVert \leq \frac 1 n \lVert v \rVert$$ and so $$\lVert T - T_n \rVert \leq \frac 1 n \to 0$$ We can conclude $T$ is compact because it is limit in the operator norm ...

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