PhyslibAlpha.AlgebraicFramework.HilbertSpace.TraceClass.HilbertSchmidt

A bounded linear operator $S: H \to H$ on a complex Hilbert space $H$ is defined to be Hilbert–Schmidt if there exists a Hilbert basis $\{e_i\}_{i \in I}$ of $H$ such that the sum of the squared norms of the images of the basis vectors, $\sum_{i \in I} \|S(e_i)\|^2$, converges.

Let $H$ be a complex Hilbert space and $S: H \to H$ be a continuous linear operator. Let $(e_i)_{i \in w}$ and $(f_j)_{j \in w'}$ be two Hilbert bases of $H$. If the series $\sum_{i \in w} \|S e_i\|^2$ is summable, then the series $\sum_{j \in w'} \|S^* f_j\|^2$ converges to the same value: $$\sum_{j \in w'} \|S^* f_j\|^2 = \sum_{i \in w} \|S e_i\|^2,$$ where $S^*$ denotes the adjoint of $S$.

Let $H$ be a Hilbert space over $\mathbb{C}$ and $S: H \to H$ be a continuous linear operator. If $\{e_i\}_{i \in I}$ is a Hilbert basis for $H$ such that the series $\sum_{i \in I} \|S e_i\|^2$ is summable, then the series $\sum_{i \in I} \|S^* e_i\|^2$ is also summable, where $S^*$ denotes the adjoint of $S$.

Let $H$ be a complex Hilbert space and $S: H \to H$ be a continuous linear operator. Suppose $\{b_i\}_{i \in w}$ and $\{c_j\}_{j \in w'}$ are two Hilbert bases for $H$. If the sum of the squared norms of the images of the first basis under $S$ converges (i.e., $\sum_{i \in w} \|S b_i\|^2 < \infty$), then the sum of the squared norms of the images of the second basis also converges and is equal to the first: $$\sum_{j \in w'} \|S c_j\|^2 = \sum_{i \in w} \|S b_i\|^2$$

Let $H$ be a complex Hilbert space and $S: H \to H$ be a continuous linear operator. If $S$ is a Hilbert–Schmidt operator, then for every Hilbert basis $\{b_i\}_{i \in w}$ of $H$, the series $\sum_{i \in w} \|S b_i\|^2$ is summable.

The zero operator $0: H \to H$ on a complex Hilbert space $H$ is a Hilbert–Schmidt operator.

Let $H$ be a complex Hilbert space and $S: H \to H$ be a bounded linear operator. If $S$ is a Hilbert–Schmidt operator, then for any complex scalar $c \in \mathbb{C}$, the scalar multiple $c S$ is also a Hilbert–Schmidt operator.

Let $H$ be a complex Hilbert space. For any two bounded linear operators $R, S: H \to H$, if $R$ and $S$ are Hilbert–Schmidt operators, then their sum $R + S$ is also a Hilbert–Schmidt operator.

Let $H$ be a complex Hilbert space and let $R, S: H \to H$ be bounded linear operators. If $R$ and $S$ are Hilbert–Schmidt operators (meaning there exists a Hilbert basis $\{e_i\}_{i \in I}$ of $H$ such that the sum of the squared norms $\sum_{i \in I} \|R(e_i)\|^2$ and $\sum_{i \in I} \|S(e_i)\|^2$ converge), then their difference $R - S$ is also a Hilbert–Schmidt operator.

Let $S: H \to H$ be a bounded linear operator on a complex Hilbert space $H$. If $S$ is a Hilbert–Schmidt operator, then its adjoint $S^*$ is also a Hilbert–Schmidt operator.

Let $H$ be a complex Hilbert space. For any bounded linear operators $U, S: H \to H$, if $\|U\| \leq 1$ and $S$ is a Hilbert–Schmidt operator, then the product $U S$ is also a Hilbert–Schmidt operator. (Recall that an operator $S$ is Hilbert–Schmidt if for some Hilbert basis $\{e_i\}_{i \in I}$, the sum $\sum_{i \in I} \|S e_i\|^2$ converges.)

Let $H$ be a complex Hilbert space, and let $S, U: H \to H$ be bounded linear operators. If $S$ is a Hilbert–Schmidt operator and the operator norm of $U$ satisfies $\|U\| \leq 1$, then the composition $S U$ is also a Hilbert–Schmidt operator.

Let $H$ be a complex Hilbert space. For any bounded linear operators $U, S: H \to H$, if $S$ is a Hilbert–Schmidt operator, then the composition $U S$ is also a Hilbert–Schmidt operator.

Let $H$ be a complex Hilbert space and let $S, U: H \to H$ be bounded linear operators. If $S$ is a Hilbert–Schmidt operator, then their composition $S U$ is also a Hilbert–Schmidt operator.

Let $H$ be a complex Hilbert space and $S: H \to H$ be a bounded linear operator. If $S$ is a Hilbert–Schmidt operator, then the operator $S^* S$ is a trace-class operator.

Let $H$ be a complex Hilbert space and $S: H \to H$ be a bounded linear operator. If $S$ is a Hilbert–Schmidt operator, then the operator $SS^*$ (the product of $S$ and its adjoint $S^*$) is a trace-class operator.

Let $w$ be an index set and let $f, g: w \to \mathbb{R}$ be real-valued functions. If $f$ and $g$ are non-negative (i.e., $f(i) \ge 0$ and $g(i) \ge 0$ for all $i \in w$) and their squares are summable ($\sum_{i \in w} f(i)^2 < \infty$ and $\sum_{i \in w} g(i)^2 < \infty$), then the sum of their product satisfies the Cauchy–Schwarz inequality: $$\sum_{i \in w} f(i) g(i) \le \sqrt{\sum_{i \in w} f(i)^2} \sqrt{\sum_{i \in w} g(i)^2}$$

Let $w$ be a set and $f, g: w \to \mathbb{R}$ be functions. If $f(i) \geq 0$ and $g(i) \geq 0$ for all $i \in w$, and the series $\sum_{i \in w} f(i)^2$ and $\sum_{i \in w} g(i)^2$ both converge, then the series $\sum_{i \in w} f(i)g(i)$ also converges.

Let $H$ be a complex Hilbert space and let $\{b_i\}_{i \in w}$ be a Hilbert basis for $H$. For any vectors $x, y \in H$, the sum of the products of the magnitudes of their coefficients with respect to the basis $\{b_i\}$ is bounded by the product of their norms: $$\sum_{i \in w} \|\langle x, b_i \rangle\| \cdot \|\langle b_i, y \rangle\| \leq \|x\| \|y\|,$$ where $\langle \cdot, \cdot \rangle$ denotes the complex inner product and $\|\cdot\|$ denotes the norm.

Let $H$ be a complex Hilbert space and let $R, S: H \to H$ be two Hilbert–Schmidt operators. For any Hilbert basis $\{b_i\}_{i \in w}$ of $H$, the diagonal coefficients of the product operator $RS$, given by $\langle b_i, (RS)b_i \rangle$, are absolutely summable. That is, the sum $\sum_{i \in w} \langle b_i, (RS)b_i \rangle$ converges.

Let $H$ be a complex Hilbert space. Let $R$ and $S$ be Hilbert-Schmidt operators on $H$. Given any two Hilbert bases $\{b_i\}_{i \in I}$ and $\{c_j\}_{j \in J}$ of $H$, the sum of the diagonal entries of the operator $RS$ with respect to the basis $\{b_i\}$ is equal to the sum of the diagonal entries of the operator $SR$ with respect to the basis $\{c_j\}$: $$\sum_{i \in I} \langle b_i, (RS)b_i \rangle = \sum_{j \in J} \langle c_j, (SR)c_j \rangle$$ where $\langle \cdot, \cdot \rangle$ denotes the complex inner product.

Let $H$ be a complex Hilbert space and let $X, W: H \to H$ be bounded linear operators. Suppose that $W$ is a contraction, i.e., $\|W\| \leq 1$, and that $X$ is a Hilbert–Schmidt operator. For any Hilbert basis $\{e_i\}_{i \in I}$ of $H$, the sum of the squared norms of the images of the basis vectors under the product operator $X W$ is bounded by the corresponding sum for $X$: $$\sum_{i \in I} \|(X W) e_i\|^2 \le \sum_{i \in I} \|X e_i\|^2.$$

Let $H$ be a complex Hilbert space. Let $S, A : H \to H$ be bounded linear operators such that $S$ is self-adjoint and Hilbert–Schmidt. For any two Hilbert bases $\{b_j\}_{j \in w}$ and $\{c_i\}_{i \in w'}$ of $H$, the following inequality holds: $$\sum_{i \in w'} \|(S A) c_i\|^2 \leq \|A\|^2 \sum_{j \in w} \|S b_j\|^2$$ where $\|A\|$ denotes the operator norm of $A$.