Operator norm

Given two normed vector spaces V and W (over the same base field, either the real numbers R or the complex numbers C), a linear map A : V → W is continuous if and only if there exists a real number c such that[1]

${\displaystyle \|Av\|\leq c\|v\|\quad {\mbox{ for all }}v\in V.}$

The norm on the left is the one in W and the norm on the right is the one in V. Intuitively, the continuous operator A never increases the length of any vector by more than a factor of c. Thus the image of a bounded set under a continuous operator is also bounded. Because of this property, the continuous linear operators are also known as bounded operators. In order to "measure the size" of A, it then seems natural to take the infimum of the numbers c such that the above inequality holds for all v in V. In other words, we measure the "size" of A by how much it "lengthens" vectors in the "biggest" case. So we define the operator norm of A as

${\displaystyle \|A\|_{op}=\inf\{c\geq 0:\|Av\|\leq c\|v\|{\mbox{ for all }}v\in V\}.}$

The infimum is attained as the set of all such c is closed, nonempty, and bounded from below.[2]

It is important to bear in mind that this operator norm depends on the choice of norms for the normed vector spaces V and W.

Every real m-by-n matrix corresponds to a linear map from Rⁿ to R^m. Each pair of the plethora of (vector) norms applicable to real vector spaces induces an operator norm for all m-by-n matrices of real numbers; these induced norms form a subset of matrix norms.

If we specifically choose the Euclidean norm on both Rⁿ and R^m, then the matrix norm given to a matrix A is the square root of the largest eigenvalue of the matrix A^*A (where A^* denotes the conjugate transpose of A).[3] This is equivalent to assigning the largest singular value of A.

Passing to a typical infinite-dimensional example, consider the sequence space l² defined by

${\displaystyle l^{2}=\{(a_{n})_{n\geq 1}:\;a_{n}\in \mathbb {C} ,\;\sum _{n}|a_{n}|^{2}<\infty \}.}$

This can be viewed as an infinite-dimensional analogue of the Euclidean space Cⁿ. Now take a bounded sequence s = (s_n). The sequence s is an element of the space l ^∞, with a norm given by

${\displaystyle \|s\|_{\infty }=\sup _{n}|s_{n}|.}$

Define an operator T_s by simply multiplication:

${\displaystyle (a_{n})\ {\stackrel {T_{s}}{\mapsto }}\ (s_{n}\cdot a_{n}).}$

The operator T_s is bounded with operator norm

${\displaystyle \|T_{s}\|_{op}=\|s\|_{\infty }.}$

One can extend this discussion directly to the case where l² is replaced by a general L^p space with p > 1 and l ^∞ replaced by L^∞.

The first four definitions are always equivalent, and if in addition ${\displaystyle V\neq \{0\}}$ then they are all equivalent:

${\displaystyle {\begin{alignedat}{4}\|A\|_{op}&=\inf &&\{c\geq 0~&&:~\|Av\|\leq c\|v\|~&&~{\mbox{ for all }}~&&v\in V\}\\&=\sup &&\{\|Av\|~&&:~\|v\|\leq 1~&&~{\mbox{ and }}~&&v\in V\}\\&=\sup &&\{\|Av\|~&&:~\|v\|<1~&&~{\mbox{ and }}~&&v\in V\}\\&=\sup &&\{\|Av\|~&&:~\|v\|\in \{0,1\}~&&~{\mbox{ and }}~&&v\in V\}\\&=\sup &&\{\|Av\|~&&:~\|v\|=1~&&~{\mbox{ and }}~&&v\in V\}\;\;\;{\text{ this equality holds if and only if }}V\neq \{0\}\\&=\sup &&{\bigg \{}{\frac {\|Av\|}{\|v\|}}~&&:~v\neq 0~&&~{\mbox{ and }}~&&v\in V{\bigg \}}\;\;\;{\text{ this equality holds if and only if }}V\neq \{0\}.\\\end{alignedat}}}$

If ${\displaystyle V=\{0\}}$ then the sets in the last two rows will be empty, and consequently their supremums will equal ∞ instead of the correct value of 0.

The operator norm is indeed a norm on the space of all bounded operators between V and W. This means

${\displaystyle \|A\|_{op}\geq 0{\mbox{ and }}\|A\|_{op}=0{\mbox{ if and only if }}A=0,}$

${\displaystyle \|aA\|_{op}=|a|\|A\|_{op}{\mbox{ for every scalar }}a,}$

${\displaystyle \|A+B\|_{op}\leq \|A\|_{op}+\|B\|_{op}.}$

The following inequality is an immediate consequence of the definition:

${\displaystyle \|Av\|\leq \|A\|_{op}\|v\|\ {\mbox{ for every }}\ v\in V.}$

The operator norm is also compatible with the composition, or multiplication, of operators: if V, W and X are three normed spaces over the same base field, and ${\displaystyle A:V\to W}$ and ${\displaystyle B:W\to X}$ are two bounded operators, then it is a sub-multiplicative norm, i.e.:

${\displaystyle \|BA\|_{op}\leq \|B\|_{op}\|A\|_{op}.}$

For bounded operators on V, this implies that operator multiplication is jointly continuous.

It follows from the definition that a sequence of operators converges in operator norm means they converge uniformly on bounded sets.

Some common operator norms are easy to calculate, and others are NP-hard. Except for the NP-hard norms, all these norms can be calculated in N² operations (for an N × N matrix), with the exception of the ${\displaystyle \ell _{2}-\ell _{2}}$ norm (which requires N³ operations for the exact answer, or fewer if you approximate it with the power method or Lanczos iterations).

Computability of Operator Norms[4]

		Co-domain
		${\displaystyle \ell _{1}}$	${\displaystyle \ell _{2}}$	${\displaystyle \ell _{\infty }}$
Domain	${\displaystyle \ell _{1}}$	Maximum ${\displaystyle \ell _{1}}$ norm of a column	Maximum ${\displaystyle \ell _{2}}$ norm of a column	Maximum ${\displaystyle \ell _{\infty }}$ norm of a column
	${\displaystyle \ell _{2}}$	NP-hard	Maximum singular value	Maximum ${\displaystyle \ell _{2}}$ of a row
	${\displaystyle \ell _{\infty }}$	NP-hard	NP-hard	Maximum ${\displaystyle \ell _{1}}$ norm of a row

The norm of the adjoint or transpose can be computed as follows. We have that for any ${\displaystyle p,q}$, then ${\displaystyle \|A\|_{p\rightarrow q}=\|A^{*}\|_{q'\rightarrow p'}}$ where ${\displaystyle p',q'}$ are Hölder conjugate to ${\displaystyle p,q}$, i.e., ${\displaystyle 1/p+1/p'=1}$ and ${\displaystyle 1/q+1/q'=1}$.

Suppose H is a real or complex Hilbert space. If A : H → H is a bounded linear operator, then we have

${\displaystyle \|A\|_{op}=\|A^{*}\|_{op}}$

and

${\displaystyle \|A^{*}A\|_{op}=\|A\|_{op}^{2}}$,

where A^* denotes the adjoint operator of A (which in Euclidean Hilbert spaces with the standard inner product corresponds to the conjugate transpose of the matrix A).

In general, the spectral radius of A is bounded above by the operator norm of A:

${\displaystyle \rho (A)\leq \|A\|_{op}.}$

To see why equality may not always hold, consider the Jordan canonical form of a matrix in the finite-dimensional case. Because there are non-zero entries on the superdiagonal, equality may be violated. The quasinilpotent operators is one class of such examples. A nonzero quasinilpotent operator A has spectrum {0}. So ρ(A) = 0 while ${\displaystyle \|A\|_{op}>0}$.

However, when a matrix N is normal, its Jordan canonical form is diagonal (up to unitary equivalence); this is the spectral theorem. In that case it is easy to see that

${\displaystyle \rho (N)=\|N\|_{op}.}$

This formula can sometimes be used to compute the operator norm of a given bounded operator A: define the Hermitian operator B = A^*A, determine its spectral radius, and take the square root to obtain the operator norm of A.

The space of bounded operators on H, with the topology induced by operator norm, is not separable. For example, consider the Hilbert space L²[0, 1]. For 0 < t ≤ 1, let Ω_t be the characteristic function of [0, t ], and P_t be the multiplication operator given by Ω_t, i.e.

${\displaystyle P_{t}(f)=f\cdot \Omega _{t}.}$

Then each P_t is a bounded operator with operator norm 1 and

${\displaystyle \|P_{t}-P_{s}\|_{op}=1\quad {\mbox{for all}}\quad t\neq s.}$

But {P_t : 0 < t ≤ 1} is an uncountable set. This implies the space of bounded operators on L²[0, 1] is not separable, in operator norm. One can compare this with the fact that the sequence space l ^∞ is not separable.

The set of all bounded operators on a Hilbert space, together with the operator norm and the adjoint operation, yields a C*-algebra.

• Continuous linear operator
• Discontinuous linear map
• Dual norm
• Matrix norm – Norm on a vector space of matrices
• Norm (mathematics) – Length in a vector space
• Normed space
• Operator algebra – Branch of functional analysis
• Operator theory
• Topological vector space – Vector space with a notion of nearness
• Topologies on the set of operators on a Hilbert space
• Unbounded operator

• Aliprantis, Charalambos D.; Border, Kim C. (2007), Infinite Dimensional Analysis: A Hitchhiker's Guide, Springer, p. 229, ISBN 9783540326960.
• Conway, John B. (1990), "III.2 Linear Operators on Normed Spaces", A Course in Functional Analysis, New York: Springer-Verlag, pp. 67–69, ISBN 0-387-97245-5