Continuity and Linear Structure
Boundedness gives a global estimate for a linear map: one constant controls the output for every input. Continuity is a local condition, asking whether inputs sufficiently close to a given point have outputs close to its image. Linearity connects these ideas. Because differences of outputs can be written as the map applied to a difference of inputs, control near zero becomes control everywhere.
Throughout this tutorial, \(V\) and \(W\) are normed vector spaces over \(\mathbb{R}\), with norms \(\|\cdot\|_V\) and \(\|\cdot\|_W\), and \(T:V\to W\) is linear. Continuity between normed spaces is defined using their norm metrics.
For a general function, continuity is a local property that may hold at some points and fail at others. For a linear map, the relation $$ T(x)-T(a)=T(x-a) $$ shows that continuity at \(a\) is governed by the behavior of \(T\) at zero. The next theorem makes the resulting global conclusion precise.
Continuity at Zero Is Equivalent to Boundedness
- \(T\) is continuous at \(0_V\);
- \(T\) is continuous at every point of \(V\);
- \(T\) is bounded.
Proof. First suppose that \(T\) is bounded. By the Boundedness Criterion for Linear Maps from the previous tutorial, there is a finite \(M\geq0\) such that $$ \|T(x)\|_W\leq M\|x\|_V\qquad\text{for every }x\in V. $$ Given \(\varepsilon>0\), choose \(\delta=\varepsilon/(M+1)\), which is positive. If \(\|x-0_V\|_V<\delta\), then $$ \|T(x)-T(0_V)\|_W =\|T(x)\|_W \leq M\|x\|_V \leq\frac{M\varepsilon}{M+1} <\varepsilon. $$ Here \(T(0_V)=0_W\) by linearity. Thus \(T\) is continuous at zero.
Next suppose that \(T\) is continuous at zero. Apply the definition with \(\varepsilon=1\). There is a \(\delta>0\) such that $$ \|u\|_V<\delta\quad\Longrightarrow\quad\|T(u)\|_W<1. $$ If \(\|x\|_V\leq1\), put \(u=(\delta/2)x\). Then $$ \|u\|_V=\frac{\delta}{2}\|x\|_V\leq\frac{\delta}{2}<\delta, $$ so \(\|T(u)\|_W<1\). By linearity, $$ \|T(u)\|_W=\frac{\delta}{2}\|T(x)\|_W. $$ Consequently \(\|T(x)\|_W<2/\delta\) for every \(x\) in the closed unit ball. The image of that ball is bounded, so the Boundedness Criterion for Linear Maps shows that \(T\) is bounded. This argument also works if \(V\) contains only its zero vector.
Finally, suppose \(T\) is continuous at some \(a\in V\). Given \(\varepsilon>0\), take \(\delta>0\) from continuity at \(a\). If \(\|h\|_V<\delta\), set \(x=a+h\). Then \(\|x-a\|_V<\delta\), and linearity gives $$ \|T(h)\|_W =\|T(a+h)-T(a)\|_W <\varepsilon. $$ Thus \(T\) is continuous at zero. The implications already proved show that it is bounded and continuous at zero. For any \(b\in V\), whenever \(\|x-b\|_V<\delta\), we have $$ \|T(x)-T(b)\|_W=\|T(x-b)\|_W<\varepsilon, $$ so \(T\) is continuous at \(b\). Since \(b\) was arbitrary, \(T\) is continuous everywhere. This proves all the equivalences. \(\square\)
The rescaling in this proof is important. Continuity at zero only gives information about inputs inside one small ball. Linearity lets us shrink any vector into that ball and then scale the resulting estimate back to the original vector. The use of \(\delta/2\), rather than \(\delta\), ensures the rescaled vector lies strictly inside the neighborhood required by the definition of continuity.
A Global Lipschitz Estimate
Proof. If \(T\) is continuous, the theorem shows that it is bounded. The Boundedness Criterion gives a finite \(L\geq0\) such that \(\|T(z)\|_W\leq L\|z\|_V\) for every \(z\in V\). Apply this inequality to \(z=x-y\). Linearity yields $$ \|T(x)-T(y)\|_W =\|T(x-y)\|_W \leq L\|x-y\|_V. $$ Conversely, suppose the displayed Lipschitz estimate holds. For any \(a\in V\) and any \(\varepsilon>0\), if \(L>0\), choose \(\delta=\varepsilon/L\). Then \(\|x-a\|_V<\delta\) implies $$ \|T(x)-T(a)\|_W\leq L\|x-a\|_V<\varepsilon. $$ If \(L=0\), the estimate says \(\|T(x)-T(y)\|_W=0\) for all \(x,y\), so \(T\) is constant and therefore continuous. Thus the estimate implies continuity in every case. \(\square\)
The estimate compares outputs at any two inputs, not just at the origin. The same constant works throughout the domain, which is stronger than merely having a separate local estimate at each point. For a bounded linear map, the previous tutorial’s least-uniform-bound property allows the operator norm itself to serve as such a constant.
Worked Examples
Worked Example: A Linear Functional on \(\mathbb{R}^2\)
Give \(\mathbb{R}^2\) the \(1\)-norm \(\|(x,y)\|_1=|x|+|y|\), and give \(\mathbb{R}\) its usual absolute-value norm. Define \(T(x,y)=2x-y\). The map is linear, since for any scalars \(\alpha,\beta\) and vectors \((x,y),(u,v)\), $$ T\bigl(\alpha(x,y)+\beta(u,v)\bigr) =2(\alpha x+\beta u)-(\alpha y+\beta v) =\alpha T(x,y)+\beta T(u,v). $$ For every \((x,y)\), $$ |T(x,y)|=|2x-y| \leq2|x|+|y| \leq2(|x|+|y|) =2\|(x,y)\|_1. $$ Thus \(T\) is bounded and hence continuous. In fact, the displayed estimate gives, for any two input vectors, $$ |T(x,y)-T(u,v)| =|T(x-u,y-v)| \leq2\|(x-u,y-v)\|_1. $$ This directly exhibits a global Lipschitz bound.
Worked Example: A Weighted Integral Functional
On \(C[0,1]\) with the supremum norm, define \(T:C[0,1]\to\mathbb{R}\) by $$ T(f)=\int_0^1 x f(x)\,dx. $$ Linearity follows from the linearity of the Riemann integral: $$ T(\alpha f+\beta g) =\int_0^1 x\bigl(\alpha f(x)+\beta g(x)\bigr)\,dx =\alpha T(f)+\beta T(g). $$ Since \(|f(x)|\leq\|f\|_\infty\) on \([0,1]\), $$ |T(f)| \leq\int_0^1 x|f(x)|\,dx \leq\|f\|_\infty\int_0^1 x\,dx =\frac12\|f\|_\infty. $$ Therefore \(T\) is bounded and continuous. The bound is attained for the constant function \(f(x)=1\): its supremum norm is \(1\), and $$ T(f)=\int_0^1 x\,dx=\frac12. $$ The calculation illustrates how an integral estimate can establish continuity without first analyzing the definition point by point.
Worked Example: Multiplication by a Continuous Function
Define \(M:C[0,1]\to C[0,1]\) by $$ (Mf)(x)=(1+x)f(x). $$ The product of the continuous functions \(1+x\) and \(f(x)\) is continuous, so the output belongs to \(C[0,1]\). Also, for \(f,g\in C[0,1]\) and \(\alpha,\beta\in\mathbb{R}\), $$ M(\alpha f+\beta g)(x) =(1+x)\bigl(\alpha f(x)+\beta g(x)\bigr) =\alpha(Mf)(x)+\beta(Mg)(x). $$ Thus \(M\) is linear. Because \(0\leq x\leq1\), we have \(|1+x|\leq2\), and hence $$ \|Mf\|_\infty =\sup_{x\in[0,1]}|(1+x)f(x)| \leq2\|f\|_\infty. $$ It follows that \(M\) is bounded and continuous. The constant function \(f(x)=1\) also shows that the factor \(2\) cannot be decreased in this estimate: \(\|f\|_\infty=1\) and \(\|Mf\|_\infty=2\).
Worked Example: The Identity Map Can Fail to Be Continuous
Consider the identity map from \(C[0,1]\) with the \(1\)-norm \(\|f\|_1=\int_0^1|f(x)|\,dx\) to the same space with the supremum norm. For each integer \(n\geq2\), define $$ f_n(x)=\max\bigl(1-n|x-\tfrac12|,0\bigr). $$ Each \(f_n\) is continuous, is nonnegative, and has height \(1\) at \(x=1/2\). Its support is contained in \([1/2-1/n,1/2+1/n]\), which lies in \([0,1]\) for \(n\geq2\). Symmetry around \(1/2\) gives $$ \|f_n\|_1 =2\int_0^{1/n}(1-nt)\,dt =2\left(\frac1n-\frac{1}{2n}\right) =\frac1n, \qquad \|f_n\|_\infty=1. $$ Suppose the identity map \(I:(C[0,1],\|\cdot\|_1)\to(C[0,1],\|\cdot\|_\infty)\) were continuous at zero. With \(\varepsilon=1/2\), there would be a \(\delta>0\) such that \(\|f\|_1<\delta\) implies \(\|I(f)\|_\infty<1/2\). Choose an integer \(n\geq2\) with \(n>1/\delta\). Then \(\|f_n\|_1=1/n<\delta\), but $$ \|I(f_n)\|_\infty=\|f_n\|_\infty=1, $$ a contradiction. The identity map is linear but not continuous for these choices of norms. This does not contradict continuity of the identity map when the same norm is used on its domain and codomain: the two norms here measure function size differently.
Why the Linear Hypothesis Matters
The theorem depends essentially on linearity. For a linear map, subtracting the output at a point is the same as applying the map to the difference of the inputs. That relation transfers continuity from one point to every point. A general function need not have this structure: for example, the function \(F:\mathbb{R}\to\mathbb{R}\) defined by \(F(x)=0\) for \(x\leq0\) and \(F(x)=1\) for \(x>0\) is continuous at \(-1\), but discontinuous at \(0\).
Another common pitfall is to treat boundedness as a pointwise claim. Each individual value \(T(x)\) is a vector, but boundedness of a linear map requires a single finite constant controlling all inputs in proportion to their norms. The identity-map example shows why the norms must also be specified: the same algebraic map can be continuous for one pair of norms and discontinuous for another.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why does linearity allow continuity at one point to imply continuity at zero?
- In the proof that continuity at zero implies boundedness, why is the input rescaled by \(\delta/2\)?
- State the global estimate satisfied by a bounded linear map at two inputs \(x\) and \(y\).
- Why is the weighted integral functional continuous in the supremum norm?
- What property of the triangular functions shows that the identity map in the final worked example is not continuous?