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I need to prove that class Sp$(X)$ in Banach space $X$ is equal to $S_1(H)$ , if $X = H$ is Hilbert space.

Definition 1 :

Let $X$ and $Y$ be Banach spaces. Operator $A \in B(X,Y)$ is called nuclear or operator with trace $(A \in \text{Sp}(X,Y)) $ if there exist one-dimentional operators $A_k$ such that $A = \sum_{k=0}^\infty A_k$, where the series converges absolutely in $B(X,Y)$.

Definition 2:

Let $H$ be Hilbert space and operator $A$ on $H$ is compact. We may define $s_n$ numbers of this operator: $$s_n(A) = \sqrt{\lambda_n (A^* A)},n = 1,2,3…, $$ where $\lambda_n(A^*A)$ are eigenvalues of $A^*A$ in descending order. We say that $A \in S_p(H)$ if sequence $(s_n)_{n=1}^\infty \in \ell_p$ . Elements of $S_1(H)$ are called nuclear operators.

I proved that $S_1(H) \subset Sp(H)$:

Let $A \in S_1(H)$, then by Hilbert-Schmidt theorem $ H = Ker(A^*A) \oplus H_1 $, where $H_1$ is closed subspace with orthonormal basis consisting of eigenvectors $\{\varphi_n\}_{n=1}^{\infty}$ of $A^*A$. It is easy to see that $Ker(A^*A)=Ker(A)$.

Then for any $x\in H$ we have $x=x_0+\sum_{n=1}^{\infty}(x,\varphi_n)\varphi_n$ where $x_0\in Ker(A)$. From this $Ax=\sum_{n=1}^{\infty}(x,\varphi_n)A\varphi_n=\sum_{n=1}^{\infty}s_n(x,\varphi_n)\psi_n$, where $\psi_n=A\varphi_n/||A\varphi_n||$ form an orthonormal system. We define $A_n x=s_n(x,\varphi_n)\psi_n$, $||A_n||=s_n$ and we obtain needed result.

How do I prove $Sp(H)\subset S_1(H)$?

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It is not true. Any compact operator $A$ (and not just those in $S_1(H)$) can be written as $\sum_kA_k$ with $A_k$ rank-one. Indeed, if $A$ is compact, then using the Polar Decomposition and the Spectral Theorem (or the Singular Value Decomposition, which amount to the same) you have $A=\sum_k V_k|A_k|$ where the $V_k$ are partial isometries and $|A_k|$ are rank-one positive operators.

And the converse works the same (note that your proof uses "compact" and not "trace-class"). So the equality is $Sp(H)=K(H)$.

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