---
tag: 3F94J43P2C
kind: definition
name: Nuclear operator
msc: [47B10, 46B28]
defines: XPANM7C1D3
uses: [06GZ956657, 03TM135G0N, 4A82S5FVRT, M2NEP1K5AE, Z8FABSYQTS, Z2CP530WDE]
sources:
  - url: https://en.wikipedia.org/wiki/Nuclear_operator
---

## Definition

Let $X$ and $Y$ be [[Banach spaces|06GZ956657]] over $\R$ or $\C$, and let $X'$ be the [[continuous dual|4A82S5FVRT]] of $X$.
A [[bounded linear map|03TM135G0N]] $T \colon X \to Y$ is *nuclear* if there are sequences $(f_n)$ in $X'$ and $(y_n)$ in $Y$ such that
$$\sum_{n=1}^{\infty} \norm{f_n} \norm{y_n} < \infty \quad \text{and} \quad T x = \sum_{n=1}^{\infty} f_n(x) \, y_n \quad \text{for all } x \in X.$$
The *nuclear norm* $\nu(T)$ is the infimum of $\sum_n \norm{f_n} \norm{y_n}$ over all such representations.

## Remarks

Every [[finite-rank operator|Z8FABSYQTS]] is nuclear, with a finite sum.
Example: on $\ell^2$, let $T x = \sum_n d_n x_n e_n$ with $\sum_n \abs{d_n} < \infty$; take $f_n(x) = d_n x_n$ and $y_n = e_n$, so $\norm{f_n} \norm{y_n} = \abs{d_n}$ is summable and $T$ is nuclear.
Non-example: the identity of an infinite-dimensional Banach space is not nuclear, because it is not compact.
Every nuclear operator is a norm limit of the partial sums, which are of finite rank, so it is [[compact|M2NEP1K5AE]].
When $X$ and $Y$ are Hilbert spaces, the nuclear operators are exactly the [[trace class operators|Z2CP530WDE]], and $\nu(T)$ is the trace norm.