Operator norm

Given two normed vector spaces ${\displaystyle V}$ and ${\displaystyle W}$ (over the same base field, either the real numbers ${\displaystyle \mathbb {R} }$ or the complex numbers ${\displaystyle \mathbb {C} }$), a linear map ${\displaystyle A:V\to W}$ is continuous if and only if there exists a real number ${\displaystyle c}$ such that^[2] $${\displaystyle \|Av\|\leq c\|v\|\quad {\text{ for all }}v\in V.}$$

The norm on the left is the one in ${\displaystyle W}$ and the norm on the right is the one in ${\displaystyle V}$. Intuitively, the continuous operator ${\displaystyle A}$ never increases the length of any vector by more than a factor of ${\displaystyle c.}$ 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 ${\displaystyle A,}$ one can take the infimum of the numbers ${\displaystyle c}$ such that the above inequality holds for all ${\displaystyle v\in V.}$ This number represents the maximum scalar factor by which ${\displaystyle A}$ "lengthens" vectors. In other words, the "size" of ${\displaystyle A}$ is measured by how much it "lengthens" vectors in the "biggest" case. So we define the operator norm of ${\displaystyle A}$ as $${\displaystyle \|A\|_{\text{op}}=\inf\{c\geq 0:\|Av\|\leq c\|v\|{\text{ for all }}v\in V\}.}$$

The infimum is attained as the set of all such ${\displaystyle c}$ is closed, nonempty, and bounded from below.^[3]

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

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

If we specifically choose the Euclidean norm on both ${\displaystyle \mathbb {R} ^{n}}$ and ${\displaystyle \mathbb {R} ^{m},}$ then the matrix norm given to a matrix ${\displaystyle A}$ is the square root of the largest eigenvalue of the matrix ${\displaystyle A^{*}A}$ (where ${\displaystyle A^{*}}$ denotes the conjugate transpose of ${\displaystyle A}$).^[4] This is equivalent to assigning the largest singular value of ${\displaystyle A.}$

Passing to a typical infinite-dimensional example, consider the sequence space ${\displaystyle \ell ^{2},}$ which is an L^p space, defined by $${\displaystyle \ell ^{2}=\left\{(a_{n})_{n\geq 1}:\;a_{n}\in \mathbb {C} ,\;\sum _{n}|a_{n}|^{2}<\infty \right\}.}$$

This can be viewed as an infinite-dimensional analogue of the Euclidean space ${\displaystyle \mathbb {C} ^{n}.}$ Now consider a bounded sequence ${\displaystyle s_{\bullet }=\left(s_{n}\right)_{n=1}^{\infty }.}$ The sequence ${\displaystyle s_{\bullet }}$ is an element of the space ${\displaystyle \ell ^{\infty },}$ with a norm given by $${\displaystyle \left\|s_{\bullet }\right\|_{\infty }=\sup _{n}\left|s_{n}\right|.}$$

Define an operator ${\displaystyle T_{s}}$ by pointwise multiplication: $${\displaystyle \left(a_{n}\right)_{n=1}^{\infty }\;{\stackrel {T_{s}}{\mapsto }}\;\ \left(s_{n}\cdot a_{n}\right)_{n=1}^{\infty }.}$$

The operator ${\displaystyle T_{s}}$ is bounded with operator norm $${\displaystyle \left\|T_{s}\right\|_{\text{op}}=\left\|s_{\bullet }\right\|_{\infty }.}$$

This discussion extends directly to the case where ${\displaystyle \ell ^{2}}$ is replaced by a general ${\displaystyle L^{p}}$ space with ${\displaystyle p>1}$ and ${\displaystyle \ell ^{\infty }}$ replaced by ${\displaystyle L^{\infty }.}$

Let ${\displaystyle A:V\to W}$ be a linear operator between normed spaces. The first four definitions are always equivalent, and if in addition ${\displaystyle V\neq \{0\}}$ then they are all equivalent:

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

If ${\displaystyle V=\{0\}}$ then the sets in the last two rows will be empty, and consequently their supremums over the set ${\displaystyle [-\infty ,\infty ]}$ will equal ${\displaystyle -\infty }$ instead of the correct value of ${\displaystyle 0.}$ If the supremum is taken over the set ${\displaystyle [0,\infty ]}$ instead, then the supremum of the empty set is ${\displaystyle 0}$ and the formulas hold for any ${\displaystyle V.}$

Importantly, a linear operator ${\displaystyle A:V\to W}$ is not, in general, guaranteed to achieve its norm ${\displaystyle \|A\|_{\text{op}}=\sup\{\|Av\|:\|v\|\leq 1,v\in V\}}$ on the closed unit ball ${\displaystyle \{v\in V:\|v\|\leq 1\},}$ meaning that there might not exist any vector ${\displaystyle u\in V}$ of norm ${\displaystyle \|u\|\leq 1}$ such that ${\displaystyle \|A\|_{\text{op}}=\|Au\|}$ (if such a vector does exist and if ${\displaystyle A\neq 0,}$ then ${\displaystyle u}$ would necessarily have unit norm ${\displaystyle \|u\|=1}$). R.C. James proved James's theorem in 1964, which states that a Banach space ${\displaystyle V}$ is reflexive if and only if every bounded linear functional ${\displaystyle f\in V^{*}}$ achieves its norm on the closed unit ball.^[5] It follows, in particular, that every non-reflexive Banach space has some bounded linear functional (a type of bounded linear operator) that does not achieve its norm on the closed unit ball.

If ${\displaystyle A:V\to W}$ is bounded then^[6] $${\displaystyle \|A\|_{\text{op}}=\sup \left\{\left|w^{*}(Av)\right|:\|v\|\leq 1,\left\|w^{*}\right\|\leq 1{\text{ where }}v\in V,w^{*}\in W^{*}\right\}}$$ and^[6] $${\displaystyle \|A\|_{\text{op}}=\left\|{}^{t}A\right\|_{\text{op}}}$$ where ${\displaystyle {}^{t}A:W^{*}\to V^{*}}$ is the transpose of ${\displaystyle A:V\to W,}$ which is the linear operator defined by ${\displaystyle w^{*}\,\mapsto \,w^{*}\circ A.}$

The operator norm is indeed a norm on the space of all bounded operators between ${\displaystyle V}$ and ${\displaystyle W}$. This means $${\displaystyle \|A\|_{\text{op}}\geq 0{\mbox{ and }}\|A\|_{\text{op}}=0{\mbox{ if and only if }}A=0,}$$ $${\displaystyle \|aA\|_{\text{op}}=|a|\|A\|_{\text{op}}{\mbox{ for every scalar }}a,}$$ $${\displaystyle \|A+B\|_{\text{op}}\leq \|A\|_{\text{op}}+\|B\|_{\text{op}}.}$$

The following inequality is an immediate consequence of the definition: $${\displaystyle \|Av\|\leq \|A\|_{\text{op}}\|v\|\ {\mbox{ for every }}\ v\in V.}$$

The operator norm is also compatible with the composition, or multiplication, of operators: if ${\displaystyle V}$, ${\displaystyle W}$ and ${\displaystyle X}$ 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, that is: $${\displaystyle \|BA\|_{\text{op}}\leq \|B\|_{\text{op}}\|A\|_{\text{op}}.}$$

For bounded operators on ${\displaystyle V}$, this implies that operator multiplication is jointly continuous.

It follows from the definition that if a sequence of operators converges in operator norm, it converges uniformly on bounded sets.

By choosing different norms for the codomain, used in computing ${\displaystyle \|Av\|}$, and the domain, used in computing ${\displaystyle \|v\|}$, we obtain different values for the operator norm. 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 ${\displaystyle N^{2}}$ operations (for an ${\displaystyle N\times N}$ matrix), with the exception of the ${\displaystyle \ell _{2}-\ell _{2}}$ norm (which requires ${\displaystyle N^{3}}$ operations for the exact answer, or fewer if you approximate it with the power method or Lanczos iterations).

Computability of Operator Norms^[7]

		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 ${\displaystyle \ell _{2}}$ norm 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 ${\displaystyle p,q,}$ that is, ${\displaystyle 1/p+1/p'=1}$ and ${\displaystyle 1/q+1/q'=1.}$

Suppose ${\displaystyle H}$ is a real or complex Hilbert space. If ${\displaystyle A:H\to H}$ is a bounded linear operator, then we have $${\displaystyle \|A\|_{\text{op}}=\left\|A^{*}\right\|_{\text{op}}}$$ and $${\displaystyle \left\|A^{*}A\right\|_{\text{op}}=\|A\|_{\text{op}}^{2},}$$ where ${\displaystyle A^{*}}$ denotes the adjoint operator of ${\displaystyle A}$ (which in Euclidean spaces with the standard inner product corresponds to the conjugate transpose of the matrix ${\displaystyle A}$).

In general, the spectral radius of ${\displaystyle A}$ is bounded above by the operator norm of ${\displaystyle A}$: $${\displaystyle \rho (A)\leq \|A\|_{\text{op}}.}$$

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 ${\displaystyle A}$ has spectrum ${\displaystyle \{0\}.}$ So ${\displaystyle \rho (A)=0}$ while ${\displaystyle \|A\|_{\text{op}}>0.}$

However, when a matrix ${\displaystyle N}$ 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 $${\displaystyle \rho (N)=\|N\|_{\text{op}}.}$$

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

The space of bounded operators on ${\displaystyle H,}$ with the topology induced by operator norm, is not separable. For example, consider the Lp space ${\displaystyle L^{2}[0,1],}$ which is a Hilbert space. For ${\displaystyle 0<t\leq 1,}$ let ${\displaystyle \Omega _{t}}$ be the characteristic function of ${\displaystyle [0,t],}$ and ${\displaystyle P_{t}}$ be the multiplication operator given by ${\displaystyle \Omega _{t},}$ that is, $${\displaystyle P_{t}(f)=f\cdot \Omega _{t}.}$$

Then each ${\displaystyle P_{t}}$ is a bounded operator with operator norm 1 and $${\displaystyle \left\|P_{t}-P_{s}\right\|_{\text{op}}=1\quad {\mbox{ for all }}\quad t\neq s.}$$

But ${\displaystyle \{P_{t}:0<t\leq 1\}}$ is an uncountable set. This implies the space of bounded operators on ${\displaystyle L^{2}([0,1])}$ is not separable, in operator norm. One can compare this with the fact that the sequence space ${\displaystyle \ell ^{\infty }}$ is not separable.

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

• Banach–Mazur compactum – Concept in functional analysis
• Continuous linear operator – Function between topological vector spaces
• Contraction (operator theory) – Bounded operators with sub-unit norm
• Discontinuous linear map
• Dual norm – Measurement on a normed vector space
• Matrix norm – Norm on a vector space of matrices
• Norm (mathematics) – Length in a vector space
• Normed space – Vector space on which a distance is definedPages displaying short descriptions of redirect targets
• Operator algebra – Branch of functional analysis
• Operator theory – Mathematical field of study
• Topologies on the set of operators on a Hilbert space
• Unbounded operator – Linear operator defined on a dense linear subspace