Properties of the Operator Norm

How Operator Norms Behave

The operator norm measures the largest amplification a bounded linear map can produce. Once that size is defined, a natural next question is how it behaves when maps are added, rescaled, or composed. These properties are useful both for calculating norms and for controlling errors: if two maps are close in operator norm, their individual norms cannot be far apart.

Throughout, $V$, $W$, and $Z$ are normed vector spaces over $\mathbb{R}$. The operator norm of a bounded linear map $T:V\to W$ is the supremum of $\|T(x)\|_W$ over inputs satisfying $\|x\|_V\leq1$. We will use the Operator Norm Is the Least Uniform Bound theorem from earlier in the course: for every $x\in V$, $$ \|T(x)\|_W\leq\|T\|\|x\|_V. $$ When several maps have the same domain and codomain, their sum and scalar multiples are defined pointwise.

Sums and Scalar Multiples

Adding two maps can combine their effects, but the triangle inequality in the codomain limits the size of the combined output. Scaling a map by a scalar scales every output by the absolute value of that scalar. These facts give corresponding estimates, and in the scalar case an exact formula.

Theorem (Basic Operator-Norm Estimates): Let $S,T:V\to W$ be bounded linear maps and let $a\in\mathbb{R}$. Then $S+T$ and $aT$ are bounded linear maps, and $$ \|S+T\|\leq\|S\|+\|T\|, \qquad \|aT\|=|a|\|T\|. $$

Proof. For every $x\in V$, the triangle inequality and the least-uniform-bound property give $$ \|(S+T)(x)\|_W \leq\|S(x)\|_W+\|T(x)\|_W \leq(\|S\|+\|T\|)\|x\|_V. $$ Thus $S+T$ is bounded, and the least-uniform-bound property implies $\|S+T\|\leq\|S\|+\|T\|$. For the scalar multiple, for every $x\in V$, $$ \|(aT)(x)\|_W=|a|\|T(x)\|_W \leq |a|\|T\|\|x\|_V. $$ This proves boundedness and gives $\|aT\|\leq |a|\|T\|$. If $a=0$, then $aT$ is the zero map and both sides of the claimed equality are zero. If $a\neq0$, applying the already-proved inequality to $T=(1/a)(aT)$ gives $$ \|T\|\leq\frac{1}{|a|}\|aT\|. $$ Therefore $|a|\|T\|\leq\|aT\|$, which proves equality. $\square$

The sum estimate is an upper bound, not an assertion that the norm of a sum must equal the sum of the norms. The maps may partly cancel. By contrast, scalar multiplication has an exact norm formula, including when the scalar is zero.

Worked Example: Scaling a Coordinate Projection

Give $\mathbb{R}^3$ the norm $\|(x,y,z)\|_1=|x|+|y|+|z|$, and define $P(x,y,z)=(x,0,0)$. Since $$ \|P(x,y,z)\|_1=|x|\leq |x|+|y|+|z|=\|(x,y,z)\|_1, $$ we have $\|P\|\leq1$. The input $(1,0,0)$ has norm $1$ and its image also has norm $1$, so $\|P\|=1$. For the map $-3P$, the scalar-multiple formula gives $\|-3P\|=3$. Directly, its output is $(-3x,0,0)$, whose norm is $3|x|$, so this value is reached at $(1,0,0)$. For the scalar $0$, $0P$ sends every input to zero and has norm $0=|0|\|P\|$.

Composition and Operator Norms

For composable maps, the output of the first map becomes the input to the second. Applying the least-uniform-bound estimate twice gives a bound for the whole composition.

Theorem (Operator Norm of a Composition): Suppose $T:V\to W$ and $S:W\to Z$ are bounded linear maps. Then $S\circ T:V\to Z$ is bounded and $$ \|S\circ T\|\leq\|S\|\|T\|. $$

Proof. The composition is linear. For each $x\in V$, first apply the least-uniform-bound property to $S$, using $T(x)\in W$, and then to $T$: $$ \|(S\circ T)(x)\|_Z =\|S(T(x))\|_Z \leq\|S\|\|T(x)\|_W \leq\|S\|\|T\|\|x\|_V. $$ This estimate holds for every $x$, so the composition is bounded. The Operator Norm Is the Least Uniform Bound theorem, now applied to $S\circ T$, gives $\|S\circ T\|\leq\|S\|\|T\|$. The argument also covers a zero domain: in that case the composition is the zero map and its norm is zero. $\square$

The order matters: $S\circ T$ means that $T$ acts first and $S$ acts second. The product estimate need not be an equality. Inputs that nearly maximize $T$ need not be sent to directions on which $S$ produces its largest output.

Worked Example: A Strict Composition Estimate

Let the domain be $\mathbb{R}^2$ with the norm $\|(x,y)\|_1=|x|+|y|$, let the intermediate space be $\mathbb{R}^2$ with the norm $\|(u,v)\|_\infty=\max\{|u|,|v|\}$, and let the codomain be $\mathbb{R}$ with absolute value. Define $$ T(x,y)=(x,y),\qquad S(u,v)=u-v. $$ For every $(x,y)$, $$ \|T(x,y)\|_\infty=\max\{|x|,|y|\}\leq |x|+|y|=\|(x,y)\|_1. $$ At $(1,0)$, both norms equal $1$, so $\|T\|=1$. Also, $$ |S(u,v)|=|u-v|\leq |u|+|v|\leq2\max\{|u|,|v|\}. $$ For $(u,v)=(1,-1)$, the input norm is $1$ and $|S(1,-1)|=2$, proving $\|S\|=2$.

The composition is $(S\circ T)(x,y)=x-y$. Its output satisfies $$ |x-y|\leq |x|+|y|=\|(x,y)\|_1, $$ and equality holds at $(1,0)$. Hence $\|S\circ T\|=1$, whereas $\|S\|\|T\|=2$. This example verifies that the composition estimate can be strict.

Comparing Nearby Maps

The sum estimate also controls how much the operator norm can change when a map is replaced by another one. This is a stability property: a small operator-norm difference guarantees a small difference between the two operator norms.

Theorem (Reverse Triangle Inequality for Operator Norms): If $S,T:V\to W$ are bounded linear maps, then $$ \big|\|S\|-\|T\|\big|\leq\|S-T\|. $$

Proof. Since $S=T+(S-T)$, the sum estimate gives $$ \|S\|\leq\|T\|+\|S-T\|. $$ Subtracting $\|T\|$ yields $\|S\|-\|T\|\leq\|S-T\|$. Interchanging $S$ and $T$ gives $$ \|T\|-\|S\|\leq\|T-S\|. $$ Because $T-S=-(S-T)$, the scalar-multiple formula shows $\|T-S\|=\|S-T\|$. The two inequalities together are exactly $$ \big|\|S\|-\|T\|\big|\leq\|S-T\|. $$ $\square$

For example, if $\|S-T\|\leq\varepsilon$, then the theorem immediately gives $\|S\|-\varepsilon\leq\|T\|\leq\|S\|+\varepsilon$. The estimate requires the same domain and codomain norms for $S$, $T$, and $S-T$, so that the difference is measured in one operator norm.

Worked Example: A Multiplication Operator on Continuous Functions

On $C[0,1]$ with the supremum norm, define $M:C[0,1]\to C[0,1]$ by $$ (Mf)(x)=x^2f(x). $$ The product is continuous whenever $f$ is continuous, and the definition is linear. For every $f\in C[0,1]$, $$ \|Mf\|_\infty =\sup_{x\in[0,1]}|x^2f(x)| \leq\left(\sup_{x\in[0,1]}x^2\right)\|f\|_\infty =\|f\|_\infty. $$ Thus $\|M\|\leq1$. To see that this bound is sharp, use the constant function $f(x)=1$. Its norm is $1$, and $Mf(x)=x^2$, whose supremum on $[0,1]$ is $1$. Therefore $\|M\|=1$.

This calculation illustrates a useful method: a pointwise bound gives an upper bound for the operator norm, and a suitable test function proves the matching lower bound. The test function need not make the output constant; it only needs to produce the claimed supremum norm.

Identity Maps, Isometries, and the Zero Space

The identity map and norm-preserving maps provide basic reference values for operator norms. The zero space requires separate attention: it has no unit vectors, and its only linear map into any normed space is the zero map.

Proposition (Norm of the Identity and of a Norm-Preserving Map): Let $I_V:V\to V$ be the identity map. If $V\neq\{0_V\}$, then $\|I_V\|=1$. If $V=\{0_V\}$, then $\|I_V\|=0$. More generally, suppose $U:V\to W$ is linear and satisfies $\|U(x)\|_W=\|x\|_V$ for every $x\in V$. If $V\neq\{0_V\}$, then $\|U\|=1$; if $V=\{0_V\}$, then $\|U\|=0$.

Proof. The identity satisfies $\|I_V(x)\|_V=\|x\|_V$, so it is bounded with bound $1$. If $V\neq\{0_V\}$, choose any nonzero $x\in V$ and set $u=x/\|x\|_V$. Then $\|u\|_V=1$ and $\|I_V(u)\|_V=1$. The unit-ball definition therefore gives $\|I_V\|=1$. If $V=\{0_V\}$, the identity sends the only vector, $0_V$, to itself, and the only output norm in the unit ball is zero. Hence $\|I_V\|=0$.

For $U$, the assumed equality gives $\|U(x)\|_W=\|x\|_V$ for every $x$, so $U$ is bounded with bound $1$. When $V\neq\{0_V\}$, the same normalization $u=x/\|x\|_V$ gives $\|u\|_V=1$ and $\|U(u)\|_W=1$. Thus the supremum defining $\|U\|$ is both at most $1$ and at least $1$, so it equals $1$. When $V=\{0_V\}$, $U$ is the zero map on its one-element domain, so its operator norm is $0$. $\square$

The nonzero-domain hypothesis in the norm-one conclusion is essential. The supremum over the unit ball of the zero space is a supremum of the single value $0$, not $1$. This is a general reminder to check whether a normalization argument requires a nonzero input.

Worked Example: The Identity on a Zero-Dimensional Space

Let $V=\{0_V\}$ and let $I_V$ be its identity map. The closed unit ball is $\{0_V\}$, and $$ \|I_V(0_V)\|_V=\|0_V\|_V=0. $$ Consequently, $\|I_V\|=0$. This agrees with the fact that $I_V$ is the zero map as a map whose domain contains only the zero vector. In contrast, on any nonzero normed space, a unit vector exists, and the identity sends that vector to an output of norm $1$, giving operator norm $1$.

Using the Properties Carefully

These estimates are especially useful when a complicated map can be built from simpler ones. The sum estimate controls a sum, the scalar formula handles rescaling, and the composition estimate controls a sequence of operations. None of the upper-bound inequalities for sums or compositions should automatically be read as equalities; cancellation and the directions of intermediate outputs can make the actual norm smaller.

The zero-space case is another worthwhile check. Any linear map with zero domain has operator norm zero, because its only input is zero. Thus a claim that a map has norm one by testing a unit vector is valid only when the domain contains a nonzero vector. The proposition above states the identity and norm-preserving cases with this distinction explicit.

Takeaway: Operator norms respect scalar multiplication exactly, are subadditive under addition, and are submultiplicative under composition. They also satisfy a reverse triangle inequality, so the norm changes by no more than the operator-norm size of a perturbation. Check separately whether the domain is the zero space before claiming a norm-one conclusion.

Check Your Understanding

Use the estimates and examples in this tutorial to answer the following questions.

Why does the scalar formula $\|aT\|=|a|\|T\|$ require a separate argument when $a=0$?
For bounded maps $T:V\to W$ and $S:W\to Z$, which map acts first in $S\circ T$, and what is the operator-norm estimate for the composition?
In the strict-composition example, what are $\|S\|$, $\|T\|$, and $\|S\circ T\|$?
If $\|S-T\|\leq\varepsilon$, what interval must contain $\|T\|$ in terms of $\|S\|$ and $\varepsilon$?
Why does the identity on the zero space have operator norm zero rather than one?