# Nuclear norm

trace norm

A norm on the space $N ( X, Y)$ of nuclear operators (cf. Nuclear operator) mapping a Banach space $X$ into a Banach space $Y$.

Let $X$ and $Y$ be Banach spaces over the field of real or complex numbers, let $L ( X, Y)$ be the space of all continuous linear operators mapping $X$ into $Y$, and let $F ( X, Y)$ be the linear subspace consisting of operators of finite rank (that is, with finite-dimensional range). The Banach dual of $X$ is denoted by $X ^ \prime$, and the value of a functional $x ^ \prime \in X ^ \prime$ at a vector $x \in X$ by $\langle x, x ^ \prime \rangle$.

Every nuclear operator $A \in N ( X, Y)$ can be represented in the form

$$\tag{1 } x \mapsto Ax = \ \sum _ {i = 1 } ^ \infty \langle x, x _ {i} ^ \prime \rangle y _ {i} ,$$

where $\{ x _ {i} ^ \prime \}$ and $\{ y _ {i} \}$ are sequences in $X ^ \prime$ and $Y$, respectively, such that

$$\sum _ {i = 1 } ^ \infty \| x _ {i} ^ \prime \| \| y _ {i} \| < \infty ;$$

such representations are called nuclear. The quantity

$$\tag{2 } \| A \| _ {1} = \ \inf \ \sum _ {i = 1 } ^ \infty \| x _ {i} ^ \prime \| \| y _ {i} \| ,$$

where the infimum is taken over all possible nuclear representations of the form (1), is called the nuclear norm of $A$. The space $N ( X, Y)$ with this norm is a Banach space that contains $F ( X, Y)$ as a dense linear subspace. If $A \in N ( X, Y)$, then the adjoint operator $A ^ \prime$ belongs to $N ( Y ^ \prime , X ^ \prime )$, and $\| A ^ \prime \| _ {1} \leq \| A \| _ {1}$. Let $\| \cdot \|$ denote the usual operator norm in $L ( X, Y)$. Then $\| A \| \leq \| A \| _ {1}$ for all $A \in N ( X, Y)$. If $A \in L ( Y, Z)$ and $B \in N ( X, Y)$, then $AB \in N ( X, Z)$, and $\| AB \| _ {1} \leq \| A \| \| B \| _ {1}$; if $A \in N ( Y, Z)$ and $B \in L ( X, Y)$, then $AB \in N ( X, Z)$, and $\| AB \| _ {1} \leq \| A \| _ {1} \| B \|$. Any operator $F \in F ( X, Y)$ can be represented in the form

$$\tag{3 } x \mapsto Fx = \ \sum _ {i = 1 } ^ { n } \langle x, x _ {i} ^ \prime \rangle y _ {i} .$$

The quantity

$$\tag{4 } \| F \| _ {1} ^ {0} = \ \inf \ \sum _ {i = 1 } ^ { n } \| x _ {i} ^ \prime \| \| y _ {i} \| ,$$

where the infimum is taken over all possible finite representations of the form (3), is called the finite nuclear norm of $F$. The space $F ( X, Y)$ can be identified with the tensor product $X ^ \prime \otimes Y$. Here, to an operator $F$ of the form (3) there corresponds the element

$$\tag{5 } u = \ \sum _ {i = 1 } ^ { n } x _ {i} ^ \prime \otimes y _ {i} \in \ X ^ \prime \otimes Y,$$

and the finite nuclear norm (4) goes into the norm

$$\tag{6 } \| u \| = \ \inf \ \sum _ {i = 1 } ^ { n } \| x _ {i} ^ \prime \| \| y _ {i} \| ,$$

where the infimum is taken over all finite representations of $u$ in the form (5). This norm is called the tensor (or cross) product of the norms in $Y$ and in $X ^ \prime$. The completion of $X ^ \prime \otimes Y$ with respect to the norm (6) is denoted by $X ^ \prime \widehat \otimes Y$. The mapping $X ^ \prime \otimes Y \rightarrow L ( X, Y)$, under which the element (5) is mapped to the operator (3), can be extended to a continuous linear operator $\Gamma : X ^ \prime \widehat \otimes Y \rightarrow L ( X, Y)$. The range of $\Gamma$ is $N ( X, Y)$. If $\Gamma$ establishes a one-to-one correspondence between $X ^ \prime \widehat \otimes Y$ and $N ( X, Y)$, then $N ( X, Y)$ coincides with the closure of $F ( X, Y)$ with respect to the norm (4); in this case the restriction of the nuclear norm to $F ( X, Y)$ is the same as the finite nuclear norm. But, in general, $\Gamma$ may have a non-trivial kernel, so that the nuclear norm is a quotient of the norm in $X ^ \prime \widehat \otimes Y$( see Nuclear operator).

Let $X = Y = H$, where $H$ is a separable Hilbert space, let $L ( H) = L ( H, H)$ be the algebra of bounded operators on $H$, and let $L _ {1} ( H) = N ( H, H)$ be the ideal of nuclear operators in $L ( H)$. In this case $\Gamma$ is one-to-one, for operators of finite rank the nuclear norm coincides with the finite nuclear norm, and each $A \in L _ {1} ( H)$ has a trace $\mathop{\rm tr} A$( see Nuclear operator). The nuclear norm of an operator $A \in L _ {1} ( H)$ coincides with $\mathop{\rm tr} [( A ^ {*} A) ^ {1/2} ]$, where $A ^ {*}$ is the adjoint of $A$ in $H$. The nuclear norm is connected with the Hilbert–Schmidt norm $\| \cdot \| _ {2}$ by $\| A \| _ {2} \leq \| A \| _ {1}$. The general form of a continuous linear functional on the Banach space $L _ {1} ( H)$ is given by

$$\tag{7 } A \rightarrow \mathop{\rm tr} AB,$$

where $B$ is an arbitrary operator from $L ( H)$, and the norm of the functional (7) coincides with $\| B \|$. Consequently, $L ( H)$ is isometric to the dual of $L _ {1} ( H)$. Formula (7) also gives the general form of a linear functional on the closed subspace $L _ \infty ( H)$ of $L ( H)$ that consists of all completely-continuous (compact) operators; here $A \in L _ \infty ( H)$ and $B$ ranges over $L _ {1} ( H)$. In this case the norm of the functional (7) coincides with $\| B \| _ {1}$, that is, the space $L _ {1} ( H)$ of nuclear operators with the nuclear norm is isometric to the dual of $L _ \infty ( H)$ in the usual operator norm. These results have non-trivial generalizations to the case of operators on Banach spaces.

Example. Let $X = Y = l _ {1}$ be the space of summable sequences. An operator $A \in L ( l _ {1} , l _ {1} )$ is contained in $N ( l _ {1} , l _ {1} )$ if and only if there is an infinite matrix $( \sigma _ {ik} )$ such that $A$ sends $\{ \xi _ {k} \} \in l _ {1}$ to $\{ \eta _ {i} \} = \{ \sum _ {k = 1 } ^ \infty \sigma _ {ik} \xi _ {k} \} \in l _ {1}$, and $\sum _ {i = 1 } ^ \infty \sup _ {k} | \sigma _ {ik} | < \infty$. In this case, $\| A \| _ {1} = \sum _ {i = 1 } ^ \infty \sup _ {k} | \sigma _ {ik} |$.

How to Cite This Entry:
Nuclear norm. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Nuclear_norm&oldid=48025
This article was adapted from an original article by G.L. Litvinov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article