Continuity of Products

Products: From Pointwise to Global Control

The pointwise product theorem established earlier in “Continuity of Polynomial Functions” says that if \(f\) and \(g\) are continuous at \(a\), then their product is continuous at \(a\). The same conclusion applies on any common domain: if both functions are continuous at every point of that domain, their product is continuous there. This tutorial develops stronger conclusions. Uniform continuity and Lipschitz continuity control changes across an entire domain, and for products these properties require attention to the sizes of the factors as well as to how quickly they change.

A useful algebraic identity separates a change in a product into changes in its factors:

The identity follows by expanding the right-hand side: it equals \(f(x)g(x)-f(x)g(y)+g(y)f(x)-g(y)f(y)\), and the middle terms cancel. Taking absolute values gives

This estimate is the main tool in the global results below. Each factor’s change is multiplied by a value of the other factor. Bounds on those values therefore let us turn control of the individual changes into control of the product.

Uniform Continuity of Products

Uniform continuity of each factor controls its change using a distance threshold that works throughout the domain. But the product estimate also contains the values \(|f(x)|\) and \(|g(y)|\). If these values can grow without bound, small changes in one factor may be magnified by the other. Boundedness supplies the missing control.

Proof. Choose finite constants \(M_f,M_g\geq0\) such that \(|f(x)|\leq M_f\) and \(|g(x)|\leq M_g\) for every \(x\in E\). Fix \(\varepsilon>0\), and set

This number is positive, including when both bounds are zero. Uniform continuity of \(f\) gives \(\delta_f>0\) such that \(x,y\in E\) and \(|x-y|<\delta_f\) imply \(|f(x)-f(y)|<\eta\). Uniform continuity of \(g\) gives \(\delta_g>0\) with the corresponding property for \(g\). Let \(\delta=\min\{\delta_f,\delta_g\}>0\). For any \(x,y\in E\) with \(|x-y|<\delta\), both changes are less than \(\eta\). The product estimate then gives

The second line uses non-strict bounds on the coefficients and the final line is strict because \(M_f+M_g<M_f+M_g+1\). This also covers \(M_f=M_g=0\): then the bound on the second line is \(0\), and \(0<\varepsilon\). Thus the chosen \(\delta\) verifies uniform continuity of \(fg\). \(\square\)

Why Boundedness Matters

The boundedness condition is sufficient, not a claim that every product of uniformly continuous functions must fail without it. In fact, the identity function is uniformly continuous on \(\mathbb{R}\), but its product with itself is not. This shows that uniform continuity of the two factors alone cannot guarantee uniform continuity of their product.

Lipschitz Bounds for Products

A Lipschitz estimate gives a numerical rate of control: a function \(h\) is Lipschitz with constant \(L\geq0\) on \(E\) if \(|h(x)-h(y)|\leq L|x-y|\) for all \(x,y\in E\). A product of Lipschitz functions has a useful Lipschitz bound when the functions are bounded. This is a quantitative counterpart of the uniform-continuity result.

Proof. For any \(x,y\in E\), apply the product estimate, then use the bounds on the factor values and the Lipschitz inequalities:

This is the Lipschitz inequality for the product with the stated constant. The constant need not be the smallest possible one, but the estimate is valid even if one or more of the bounds or Lipschitz constants are zero. \(\square\)

Local Continuity and Cancellation

For continuity at a single point, boundedness does not need to hold throughout the domain. The pointwise product theorem gives continuity of the product at \(a\) whenever both factors are continuous there, without any global boundedness hypothesis. Uniform continuity is different because its single distance threshold must work over the entire domain.

A product being continuous also does not imply that each factor is continuous. For example, define \(h(0)=1\) and \(h(x)=0\) for \(x\ne0\), and let \(k(x)=0\) for every real \(x\). The function \(h\) is discontinuous at \(0\): for every \(\delta>0\), \(x=\delta/2\) satisfies \(|x|<\delta\), \(x\ne0\), and \(|h(x)-h(0)|=1\). But \(hk\) is the constant zero function, hence continuous everywhere. The product can conceal discontinuity through a zero factor.

When applying a product theorem, check the hypotheses suited to the conclusion. Pointwise continuity uses continuity of both factors at the point. The uniform-continuity theorem proved here also requires both factors to be bounded on the full domain. The Lipschitz estimate likewise uses global bounds on the factors. A common error is to infer uniform continuity of a product from uniform continuity of its factors without checking whether their values can become arbitrarily large.