Continuity of Sums

Adding Continuous Functions

The previous tutorial showed that taking absolute values preserves continuity. Addition is another basic operation: if two functions are continuous at the same point, their sum is continuous there as well. That two-function result was established in “Continuity of Polynomial Functions.” Here we use it as a starting point and extend the discussion to finite sums, uniform continuity, and useful distance estimates.

The functions being added must have a common domain. At each input in that domain, their values are real numbers, so ordinary addition defines a new function.

For two functions, “Continuity of Polynomial Functions” established that if \(f\) and \(g\) are continuous at \(a\in E\), then \(f+g\) is continuous at \(a\). The same result applies repeatedly when more terms are added. The next theorem makes that finite extension precise.

Continuity of a Finite Sum

Proof. We use induction on \(n\). For \(n=1\), the sum is \(f_1\), which is continuous at \(a\) by hypothesis. Suppose the result holds for sums of \(n\) functions, and consider \(n+1\) functions, each continuous at \(a\). By the induction hypothesis, \(S_n=\sum_{j=1}^{n}f_j\) is continuous at \(a\). The two-function sum result from “Continuity of Polynomial Functions,” applied to \(S_n\) and \(f_{n+1}\), then shows that \(S_n+f_{n+1}\) is continuous at \(a\). Since \(S_n(x)+f_{n+1}(x)=\sum_{j=1}^{n+1}f_j(x)\) for every \(x\in E\), this is the desired sum. The induction proves the claim for every finite \(n\). \(\square\)

The common domain and common point matter in this statement: every summand must be defined at \(a\), and every summand must be continuous there. The result is local. It says that if the component functions are continuous at each point of \(E\), then their finite sum is continuous at each point of \(E\).

Uniform Continuity of Finite Sums

Pointwise continuity is a statement about a chosen point. Uniform continuity is stronger: for a given \(\varepsilon\), one \(\delta\) must work for every pair of inputs in the domain. When several functions are added, their individual choices of \(\delta\) may differ. A finite minimum lets us choose one that works for all the summands.

Proof. Fix \(\varepsilon>0\). For each \(j\in\{1,\ldots,n\}\), uniform continuity of \(f_j\) gives a number \(\delta_j>0\) such that, for all \(x,y\in E\), \(|x-y|<\delta_j\) implies \(|f_j(x)-f_j(y)|<\varepsilon/n\). Because there are only finitely many \(\delta_j\), their minimum \(\delta=\min\{\delta_1,\ldots,\delta_n\}\) is positive. Take any \(x,y\in E\) with \(|x-y|<\delta\). Then \(|x-y|<\delta_j\) for every \(j\), so each difference satisfies \(|f_j(x)-f_j(y)|<\varepsilon/n\). The triangle inequality gives \[ \left|\sum_{j=1}^{n}f_j(x)-\sum_{j=1}^{n}f_j(y)\right| \leq \sum_{j=1}^{n}|f_j(x)-f_j(y)| <\sum_{j=1}^{n}\frac{\varepsilon}{n} =\varepsilon. \] The same \(\delta\) works for every \(x,y\in E\), which is precisely uniform continuity of the sum. \(\square\)

Adding Lipschitz Bounds

Uniform continuity is qualitative: sufficiently close inputs produce close outputs, with a suitable threshold. Sometimes a more precise estimate is available. A function \(f:E\to\mathbb{R}\) is Lipschitz with constant \(L\geq0\) if \(|f(x)-f(y)|\leq L|x-y|\) for every \(x,y\in E\). Such a bound immediately implies uniform continuity when \(L>0\); when \(L=0\), the function is constant and is uniformly continuous as well.

Proof. For any \(x,y\in E\), the triangle inequality and the Lipschitz hypotheses give \[ \left|\sum_{j=1}^{n}f_j(x)-\sum_{j=1}^{n}f_j(y)\right| \leq\sum_{j=1}^{n}|f_j(x)-f_j(y)| \leq\sum_{j=1}^{n}L_j|x-y| =\left(\sum_{j=1}^{n}L_j\right)|x-y|. \] This is the defining Lipschitz inequality for the sum, with constant \(\sum_{j=1}^{n}L_j\). \(\square\)

This result is useful when the goal is not only to know that a sum is continuous, but also to control the size of its changes. If all the summands have a common Lipschitz constant \(L\), this theorem gives the bound \(nL\) for a sum of \(n\) terms. The bound may not be the smallest possible one, but it is always valid.

Domains and a Common Pitfall

For a sum \(f+g\), both functions must be defined on the inputs under discussion. If their original domains differ, the natural domain of their sum is the intersection of those domains. Continuity of each summand on that common domain then gives continuity of the sum there. For example, a reciprocal term restricts the domain to points where its denominator is nonzero; adding another function does not fill in a point where the reciprocal is undefined.

A different pitfall is to reverse the sum theorem. Continuity of \(f\) and \(g\) guarantees continuity of \(f+g\), but continuity of the sum alone does not guarantee continuity of either summand. For instance, on \(\mathbb{R}\), define \(f(0)=1\), \(f(x)=0\) for \(x\ne0\), and \(g(x)=-f(x)\). The function \(f\) is not continuous at \(0\): values at nonzero points are \(0\), while \(f(0)=1\). Yet \(f(x)+g(x)=0\) for every \(x\), so the sum is continuous. Addition can cancel discontinuities; it does not generally allow them to be recovered from the sum.

The main practical choices are therefore clear. Use the finite-sum theorem when only continuity at a point is needed, the uniform-continuity theorem when one \(\delta\) must work across the whole domain, and a Lipschitz estimate when a numerical bound on output changes is useful. In each case, the triangle inequality is what allows the separate controls to combine.