# Equivalence of two definitions of nuclear operators on Hilbert spaces

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 $$Sp(H)\subset S_1(H)$$?

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 $$Sp(H)=K(H)$$.