Measuring the Size of a Linear Map
The previous tutorial treated linearity as an algebraic property: a linear map preserves linear combinations. In normed spaces, we can ask an additional question: how large can the output be relative to the input? A map may preserve all vector-space operations and still have outputs that grow too rapidly for any single constant to control them.
A uniform bound on that growth is useful because it controls the map on every input at once. It also gives a natural way to measure the map itself. Throughout this tutorial, \(V\) and \(W\) are normed vector spaces over \(\mathbb{R}\), with norms denoted by \(\|\cdot\|_V\) and \(\|\cdot\|_W\), and \(T:V\to W\) is linear.
The definition does not say that all outputs \(T(x)\) are bounded independently of \(x\). A nonzero linear map on a nonzero vector space typically has arbitrarily large outputs when its inputs are scaled. Boundedness instead requires output size to be controlled in proportion to input size, with one constant that works for every input.
Why the Two Definitions of Boundedness Agree
Proof. Suppose first that \(T\) maps bounded sets to bounded sets. The closed unit ball $$ \overline{B}_V(0,1)=\{x\in V:\|x\|_V\leq1\} $$ is bounded, so its image under \(T\) is bounded. Therefore there is a finite \(C\geq0\) such that \(\|T(u)\|_W\leq C\) whenever \(\|u\|_V\leq1\).
If \(x\neq0_V\), then \(u=x/\|x\|_V\) has norm \(1\). Linearity gives \(T(u)=T(x)/\|x\|_V\), and hence $$ \|T(x)\|_W=\|x\|_V\|T(u)\|_W\leq C\|x\|_V. $$ For \(x=0_V\), linearity gives \(T(x)=0_W\), so the same inequality holds. Thus the required estimate holds with \(M=C\).
Conversely, suppose such an \(M\) exists, and let \(A\subseteq V\) be bounded. There is a finite \(R\geq0\) such that \(\|x\|_V\leq R\) for all \(x\in A\). For each \(x\in A\), $$ \|T(x)\|_W\leq M\|x\|_V\leq MR. $$ Thus \(T(A)\) is bounded. Both implications hold, proving the criterion. \(\square\)
This proof also explains why the unit ball is the right place to test a linear map. Every nonzero vector can be rescaled to have norm \(1\), and linearity then transfers any bound on the unit ball to a bound on the whole space. The zero vector causes no exception because every linear map sends it to zero.
The Operator Norm
The operator norm is the largest output size on the unit ball, understood as a supremum rather than necessarily an attained maximum. Scaling shows that it controls the map on all of \(V\). In fact, it is the smallest possible constant in the boundedness estimate.
Proof. For \(x=0_V\), the claimed estimate holds because \(T(0_V)=0_W\). For \(x\neq0_V\), the vector \(u=x/\|x\|_V\) has norm \(1\). By the definition of the supremum, $$ \|T(u)\|_W\leq\|T\|. $$ Linearity and homogeneity of the norm give $$ \|T(x)\|_W=\|x\|_V\left\|T\left(\frac{x}{\|x\|_V}\right)\right\|_W \leq\|T\|\|x\|_V. $$ Now suppose \(M\) satisfies the stated bound. For each \(u\) with \(\|u\|_V\leq1\), we have $$ \|T(u)\|_W\leq M\|u\|_V\leq M. $$ Thus \(M\) is an upper bound for the set whose supremum defines \(\|T\|\), so \(\|T\|\leq M\). \(\square\)
One can also write the operator norm as the supremum of \(\|T(x)\|_W/\|x\|_V\) over nonzero \(x\in V\), provided \(V\) contains a nonzero vector. The unit-ball definition is convenient even when \(V=\{0_V\}\): in that case the only linear map sends \(0_V\) to \(0_W\), and its operator norm is \(0\).
Worked Examples: Computing and Testing Operator Norms
Worked Example: A Linear Functional on Euclidean Space
Let \(T:\mathbb{R}^2\to\mathbb{R}\) be given by \(T(x,y)=3x-4y\), with the Euclidean norm on \(\mathbb{R}^2\) and absolute value on \(\mathbb{R}\). This map is linear because $$ T(\alpha(x,y)+\beta(u,v)) =3(\alpha x+\beta u)-4(\alpha y+\beta v) =\alpha T(x,y)+\beta T(u,v). $$
The Cauchy–Schwarz inequality gives $$ |T(x,y)|=|(3,-4)\mathbin{\cdot}(x,y)| \leq\sqrt{3^2+(-4)^2}\sqrt{x^2+y^2} =5\|(x,y)\|_2. $$ Therefore \(T\) is bounded and \(\|T\|\leq5\). To see that this bound is sharp, take \((x,y)=(3/5,-4/5)\). Its Euclidean norm is \(1\), and $$ T(3/5,-4/5)=9/5+16/5=5. $$ So the supremum on the unit ball is at least \(5\). Combining the two inequalities gives \(\|T\|=5\).
Worked Example: Inclusion from the Supremum Norm to the Integral Norm
On \(C[0,1]\), consider the identity map \(I\) from the supremum norm to the integral norm: $$ I:(C[0,1],\|\cdot\|_\infty)\longrightarrow(C[0,1],\|\cdot\|_1), \qquad I(f)=f, $$ where \(\|f\|_1=\int_0^1|f(x)|\,dx\). The map is linear because it leaves each function unchanged, so it preserves sums and scalar multiples. For every \(f\in C[0,1]\), $$ \|I(f)\|_1=\int_0^1|f(x)|\,dx \leq\int_0^1\|f\|_\infty\,dx =\|f\|_\infty. $$ Thus \(I\) is bounded and \(\|I\|\leq1\). For the constant function \(f(x)=1\), the input norm is \(1\) and the output norm is also \(1\), since \(\int_0^1 1\,dx=1\). Hence \(\|I\|\geq1\), and therefore \(\|I\|=1\).
This example illustrates that the domain and codomain can be the same vector space with different norms. Boundedness compares the chosen input and output norms; the formula for the map alone does not determine its operator norm.
Worked Example: An Unbounded Linear Differentiation Map
Let \(V=C^1[0,1]\), equipped only with the supremum norm, and let \(W=C[0,1]\), also equipped with the supremum norm. Differentiation defines a linear map \(D:V\to W\) by \(D(f)=f'\). For each positive integer \(n\), set \(f_n(x)=x^n\). Then \(f_n\in C^1[0,1]\), and direct calculation gives $$ \|f_n\|_\infty=\sup_{0\leq x\leq1}x^n=1,\qquad \|D(f_n)\|_\infty=\sup_{0\leq x\leq1}nx^{n-1}=n. $$ The second supremum is \(n\), since \(x^{n-1}\leq1\) on this interval and equals \(1\) at \(x=1\); this also holds when \(n=1\).
If \(D\) were bounded, some finite \(M\) would satisfy \(\|D(f)\|_\infty\leq M\|f\|_\infty\) for every \(f\in V\). Applying this to \(f_n\) would give \(n\leq M\) for every positive integer \(n\), which is impossible. Thus \(D\) is linear but unbounded for these norms. The choice of norm matters: this conclusion concerns the supremum norm on the domain, which does not control the size of a derivative.
Bounded Linear Maps Form a Normed Space
Suppose now that \(V\) and \(W\) are fixed normed spaces. The bounded linear maps from \(V\) to \(W\) can themselves be added and multiplied by real scalars pointwise. Their operator norms measure the sizes of those maps.
Proof. If \(S,T:V\to W\) are bounded and linear, then \(S+T\) and \(cT\) are linear. For every \(x\in V\), $$ \|(S+T)(x)\|_W \leq\|S(x)\|_W+\|T(x)\|_W \leq(\|S\|+\|T\|)\|x\|_V, $$ and $$ \|(cT)(x)\|_W=|c|\|T(x)\|_W \leq |c|\|T\|\|x\|_V. $$ The boundedness criterion shows that both maps are bounded. The zero map is bounded, and the vector-space operations are inherited from all linear maps \(V\to W\).
The operator norm is nonnegative. If \(\|T\|=0\), the least-uniform-bound theorem gives \(\|T(x)\|_W\leq0\) for every \(x\), so \(T(x)=0_W\) for every \(x\); hence \(T\) is the zero map. Conversely, the zero map has operator norm \(0\). Homogeneity follows from $$ \|cT\|=\sup_{\|x\|_V\leq1}\|cT(x)\|_W =|c|\sup_{\|x\|_V\leq1}\|T(x)\|_W =|c|\|T\|. $$ Finally, the triangle inequality for the norm on \(W\) gives, for every \(\|x\|_V\leq1\), $$ \|(S+T)(x)\|_W\leq\|S(x)\|_W+\|T(x)\|_W\leq\|S\|+\|T\|. $$ Taking the supremum over the unit ball yields \(\|S+T\|\leq\|S\|+\|T\|\). These are the norm axioms, so the bounded linear maps form a normed space under the operator norm. \(\square\)
There is also a useful estimate for successive bounded maps. If \(T:V\to W\) and \(S:W\to Z\) are bounded and linear, their composition is linear by the composition result from the previous tutorial. Moreover, for every \(x\in V\), $$ \|S(T(x))\|_Z\leq\|S\|\|T(x)\|_W \leq\|S\|\|T\|\|x\|_V. $$ The boundedness criterion and least-bound property therefore give \(\|S\circ T\|\leq\|S\|\|T\|\). This estimate allows bounds to be combined across multiple stages of a transformation.
What Boundedness Tells Us
A bounded linear map has one finite constant that controls every input, not merely a separate bound for each vector or each bounded set. The operator norm records the best such constant. When a proposed bound is not sharp, testing vectors of norm \(1\) can reveal whether the estimate can be improved, as in the Euclidean functional example.
A common mistake is to infer boundedness from linearity. Linearity governs how outputs respond to sums and scalar multiples, but it does not prevent the output from growing faster than the input norm. The differentiation example shows this directly: the functions \(x^n\) all have supremum norm \(1\), while their derivatives have supremum norms \(n\). Another important point is that boundedness depends on the norms assigned to the domain and codomain. Changing either norm can change whether the same algebraic map is bounded and can change its operator norm.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why is it enough to test a linear map on the closed unit ball when deciding whether it is bounded?
- What does the operator norm measure, and why is it the least constant in the global norm estimate?
- For the map \(T(x,y)=3x-4y\) with Euclidean input norm, which unit vector shows that the upper bound \(5\) is sharp?
- Why does the differentiation map in the worked example fail to be bounded for the stated norms?
- State the operator-norm estimate for the composition of two bounded linear maps.