Uniform Convergence and Boundedness

Individual Bounds and a Common Bound

A function can be bounded even when its bound is very large, and a sequence can consist entirely of bounded functions while their bounds grow without limit. Uniform convergence gives a useful way to rule out that growth: once two sufficiently late functions are uniformly close, one bounded function controls all the others in the tail. The finitely many earlier functions can then be handled separately.

Throughout, \(E\) is a nonempty set and each \(f_n:E\to\mathbb{R}\). A function \(f:E\to\mathbb{R}\) is bounded if there is a finite \(M\geq0\) such that \(|f(x)|\leq M\) for every \(x\in E\). The bound may depend on the function.

Uniform boundedness is stronger than saying that each \(f_n\) is bounded. In the latter statement, a different bound \(M_n\) may be used for each index, and those bounds might grow without limit. The distinction is similar to the distinction between having a bound at each fixed input and having a bound that works simultaneously for every input.

Boundedness Under a Uniform Error

The basic estimate behind the topic is the triangle inequality. If two functions differ by at most a fixed finite amount everywhere, then one is bounded exactly when the other is bounded. This fact is useful even when the functions are not limits of a sequence.

Proof. Suppose first that \(f\) is bounded, with \(|f(x)|\leq M\) for every \(x\in E\). For each \(x\in E\),

Thus \(g\) is bounded. Interchanging \(f\) and \(g\) gives \(|f(x)|\leq |f(x)-g(x)|+|g(x)|\leq C+|g(x)|\), so boundedness of \(g\) implies boundedness of \(f\). This proves both directions. \(\square\)

In terms of the supremum norm from earlier in the course, the hypothesis holds whenever \(\|f-g\|_\infty\) is finite. Uniform convergence gives just this kind of control between a sufficiently late term and the limit: for example, if \(f_n\to f\) uniformly, then for all sufficiently large \(n\), \(|f_n(x)-f(x)|<1\) at every \(x\). Consequently, if the limit is bounded, all sufficiently late terms are bounded by a common bound. The theorem also explains why finite uniform error alone cannot make an unbounded function bounded.

Uniform Convergence Gives a Common Bound for Bounded Terms

The next theorem does not require first identifying or bounding the limit. Instead, it uses the Uniform Cauchy Criterion established earlier in the course. Far enough along the sequence, any two terms are uniformly close. Fixing one bounded term in that tail provides a bound for every later term.

Proof. Uniform convergence implies that \((f_n)\) is uniformly Cauchy, by the Uniform Cauchy Criterion. Choose an index \(N\) such that whenever \(m,n\geq N\),

In particular, take \(m=N\). Since \(f_N\) is bounded, choose \(M_N\geq0\) such that \(|f_N(x)|\leq M_N\) for every \(x\in E\). For \(n\geq N\) and \(x\in E\),

The terms with indices \(1,\ldots,N-1\) are also bounded individually. For each such \(j\), choose \(M_j\geq0\) with \(|f_j(x)|\leq M_j\) for all \(x\in E\). Then the finite number

bounds every term: it is at least \(1+M_N\) for \(n\geq N\), and at least \(M_j\) for each \(j<N\). (When \(N=1\), the sum is zero.) Hence \(|f_n(x)|\leq M\) for every \(n\) and every \(x\in E\). The sequence is uniformly bounded. \(\square\)

The proof separates the sequence into a tail and a finite beginning. Uniform Cauchy behavior controls the whole tail using one fixed function, \(f_N\). Individual boundedness then supplies bounds for the finite number of terms before that tail. The finite beginning matters: a tail bound by itself says nothing about whether an earlier function is bounded.

When a Common Bound Passes to a Limit

There is a useful converse-looking observation. If a single bound already works for every function in a sequence, then any pointwise limit inherits that bound. Uniform convergence is not needed for this particular conclusion; pointwise convergence at each input is enough.

Proof. Fix any \(x\in E\). Pointwise convergence gives \(f_n(x)\to f(x)\). For every \(n\), the triangle inequality and the common bound give

Let \(\varepsilon>0\). Choose \(n\) large enough that \(|f(x)-f_n(x)|<\varepsilon\). Then \(|f(x)|\leq M+\varepsilon\). Since this holds for every \(\varepsilon>0\), it follows that \(|f(x)|\leq M\). The point \(x\) was arbitrary, so the inequality holds throughout \(E\). \(\square\)

The crucial hypothesis is the existence of one \(M\) that works for every index and every input. Merely knowing that each function is bounded does not provide such an \(M\). The following example makes that distinction visible even when the sequence has a pointwise limit.

What the Bounds Do—and Do Not—Say

The results above answer different questions. If each term is bounded and the sequence converges uniformly, the Uniform Cauchy Criterion supplies a common bound for the entire sequence. If the sequence already has a common bound and converges pointwise, that bound passes to the limit. In contrast, a uniform limit of bounded functions is itself bounded by the earlier theorem Uniform Limits of Bounded Functions Are Bounded; that result concerns the limit, whereas the theorem proved here controls all terms of the sequence at once.

A common error is to treat “every function is bounded” as if it meant “the sequence is uniformly bounded.” The moving-spike example shows why that inference fails: each individual function has a finite bound, but the bounds increase with the index. Another error is to infer boundedness just from uniform convergence. The sequence \(x+1/n\) converges uniformly on \(\mathbb{R}\), yet every term is unbounded. Always check which function or family the bound is supposed to control, and whether the bound may depend on the index or the input.