Uniform Convergence Through Further Operations
The previous tutorial used a finite mesh to turn pointwise convergence into uniform convergence when the functions shared a common continuity estimate. Here we take uniform convergence as our starting point and ask what happens when the functions are transformed. This is useful whenever a limit is part of a larger expression: a product, a reciprocal, or the output of another function.
The central issue is that different operations need different kinds of control. A uniformly continuous outer function can transfer uniform convergence directly. Products need boundedness, and reciprocals need denominators that stay away from zero. In each case, a useful proof does more than establish convergence: it gives an error estimate that is uniform over the domain.
Recall the Supremum Criterion for Uniform Convergence: \(f_n\to f\) uniformly on a nonempty set \(E\) when $$ \sup_{x\in E}|f_n(x)-f(x)|\longrightarrow 0. $$ We will often write \(e_n=\sup_{x\in E}|f_n(x)-f(x)|\) for this uniform error.
Composition with a Uniformly Continuous Function
Continuity of a function at each point is not always enough to control its effect on a uniform error across an entire domain. Uniform continuity supplies exactly the needed control: one input tolerance works at every point in the outer function’s domain.
Proof. Let \(\varepsilon>0\). By uniform continuity of \(\phi\), there is a \(\delta>0\) such that for all \(u,v\in\mathbb{R}\), $$ |u-v|<\delta\quad\Longrightarrow\quad|\phi(u)-\phi(v)|<\varepsilon. $$ Since \(f_n\to f\) uniformly, there is an \(N\) such that \(n\geq N\) implies \(|f_n(x)-f(x)|<\delta\) for every \(x\in E\). Therefore, for every \(n\geq N\) and every \(x\in E\), $$ |\phi(f_n(x))-\phi(f(x))|<\varepsilon. $$ The same \(N\) works throughout \(E\), so the convergence of the compositions is uniform. \(\square\)
This theorem does not require the functions \(f_n\) to be bounded. Instead, the outer function must be uniformly continuous on all possible input values. When \(\phi\) is only known to be uniformly continuous on a set containing the ranges of \(f_n\) and \(f\), the same proof applies using that set’s uniform-continuity estimate.
Worked Example: Applying Sine to a Uniformly Convergent Sequence
Take \(E=\mathbb{R}\), \(f_n(x)=x+1/n\), and \(f(x)=x\). For every \(x\in\mathbb{R}\), $$ |f_n(x)-f(x)|=\frac{1}{n}, $$ so the convergence is uniform. The sine function is uniformly continuous because $$ |\sin u-\sin v|\leq |u-v| $$ for all real \(u,v\), by the Mean Value Theorem and the bound \(|\cos t|\leq1\). Consequently, $$ |\sin(f_n(x))-\sin(f(x))| =|\sin(x+1/n)-\sin x| \leq\frac{1}{n} $$ for every \(x\). Taking the supremum over \(x\in\mathbb{R}\) gives a uniform error at most \(1/n\), which tends to zero.
Products: Uniform Errors Need Bounds
For products, subtract and add an intermediate term so that each difference contains one known convergence error. If \(f_n\) and \(g_n\) approximate \(f\) and \(g\), respectively, then $$ f_ng_n-fg=f_n(g_n-g)+g(f_n-f). $$ This identity separates the two errors, but also shows why boundedness matters: the errors are multiplied by function values.
Proof. For every \(n\) and \(x\in E\), the displayed identity and the triangle inequality give $$ |f_n(x)g_n(x)-f(x)g(x)| \leq |f_n(x)|\,|g_n(x)-g(x)|+|g(x)|\,|f_n(x)-f(x)|. $$ Using the assumed bounds, $$ |f_n(x)g_n(x)-f(x)g(x)| \leq A|g_n(x)-g(x)|+B|f_n(x)-f(x)|. $$ Taking suprema over \(E\) yields $$ \sup_{x\in E}|f_n(x)g_n(x)-f(x)g(x)| \leq A\sup_{x\in E}|g_n(x)-g(x)| +B\sup_{x\in E}|f_n(x)-f(x)|. $$ Both terms on the right tend to zero by uniform convergence. Hence the supremum error for the products tends to zero, which proves uniform convergence. \(\square\)
A convenient way to verify the required bounds is to use the Uniform Limits of Bounded Functions Are Bounded Theorem and the Uniform Boundedness of a Uniformly Convergent Sequence Theorem from earlier in this course. In particular, if \(f\) and \(g\) are bounded and both sequences converge uniformly, then the hypotheses above hold: the sequence \((f_n)\) is uniformly bounded, and \(g\) is bounded.
Worked Example: A Product with an Explicit Uniform Error
On \([0,1]\), define $$ f_n(x)=x+\frac{1}{n}, \qquad g_n(x)=1-x+\frac{1}{n}. $$ The limits are \(f(x)=x\) and \(g(x)=1-x\), since each approximation has error \(1/n\) at every \(x\). For \(n\geq1\), both functions take values between \(0\) and \(2\), so they are uniformly bounded. Direct multiplication gives $$ f_n(x)g_n(x) =\left(x+\frac1n\right)\left(1-x+\frac1n\right) =x(1-x)+\frac{x+1-x}{n}+\frac{1}{n^2} =x(1-x)+\frac1n+\frac{1}{n^2}. $$ Thus $$ \sup_{x\in[0,1]}|f_n(x)g_n(x)-f(x)g(x)| =\frac1n+\frac{1}{n^2}\longrightarrow0. $$ The error is independent of \(x\), as uniform convergence requires.
Reciprocals: Keep the Limit Away from Zero
Taking reciprocals is more delicate than taking products. A small change in a denominator can produce a large change in its reciprocal when the denominator is close to zero. The condition that prevents this is a fixed positive lower bound for the absolute value of the limit.
Proof. Uniform convergence gives an \(N\) such that for \(n\geq N\) and every \(x\in E\), $$ |f_n(x)-f(x)|<\frac{c}{2}. $$ The reverse triangle inequality implies $$ |f_n(x)|\geq |f(x)|-|f_n(x)-f(x)|>\frac{c}{2}. $$ Thus \(f_n(x)\neq0\) for all \(x\in E\) when \(n\geq N\). For such \(n\), $$ \left|\frac{1}{f_n(x)}-\frac{1}{f(x)}\right| =\frac{|f_n(x)-f(x)|}{|f_n(x)|\,|f(x)|} \leq\frac{2}{c^2}|f_n(x)-f(x)|. $$ Taking suprema over \(x\in E\) gives $$ \sup_{x\in E}\left|\frac{1}{f_n(x)}-\frac{1}{f(x)}\right| \leq \frac{2}{c^2}\sup_{x\in E}|f_n(x)-f(x)|. $$ The right-hand side tends to zero, proving uniform convergence of the reciprocals. \(\square\)
Worked Example: Reciprocals of a Uniformly Convergent Sequence
For \(x\in\mathbb{R}\) and \(n\geq1\), let $$ f_n(x)=2+\frac{\sin x}{n}, \qquad f(x)=2. $$ Since \(|\sin x|\leq1\), $$ |f_n(x)-f(x)|=\frac{|\sin x|}{n}\leq\frac1n, $$ so \(f_n\to f\) uniformly. Also, $$ f_n(x)\geq2-\frac1n\geq1 $$ for every \(x\) and \(n\geq1\). In particular, the denominators are nonzero. A direct calculation gives $$ \left|\frac1{f_n(x)}-\frac12\right| =\frac{|f_n(x)-2|}{2|f_n(x)|} \leq\frac{1/n}{2} =\frac{1}{2n}. $$ The bound is uniform in \(x\), so \(1/f_n\) converges uniformly to \(1/2\).
Why Continuity Alone Can Fail
Uniform continuity in the composition theorem and boundedness in the product theorem are substantive hypotheses, not technical decorations. For instance, the exponential function is continuous everywhere but is not uniformly continuous on \(\mathbb{R}\). A uniformly small change in its input need not produce a uniformly small change in its output when the input can be arbitrarily large.
Worked Example: Uniform Convergence Lost Under Exponentiation
Again let \(f_n(x)=x+1/n\) and \(f(x)=x\) on \(\mathbb{R}\). We have already checked that \(f_n\to f\) uniformly, with error exactly \(1/n\). After applying the exponential function, however, $$ |e^{f_n(x)}-e^{f(x)}| =e^x(e^{1/n}-1). $$ For each fixed \(n\), \(e^{1/n}-1>0\). Letting \(x\) increase without bound makes \(e^x(e^{1/n}-1)\) unbounded. Therefore $$ \sup_{x\in\mathbb{R}}|e^{f_n(x)}-e^{f(x)}|=+\infty $$ for every \(n\), so the transformed sequence does not converge uniformly. There is no contradiction with the composition theorem: exponentiation is not uniformly continuous on the full real line.
The domain and range conditions should therefore be checked before transferring uniform convergence through an operation. On a closed bounded interval, a continuous function is uniformly continuous by the Heine–Cantor Theorem, so composition may be justified if all relevant function values lie in such an interval. For reciprocals, continuity of \(t\mapsto1/t\) near each nonzero value is not by itself the needed uniform estimate; a common lower bound \(c\) ensures that the denominators cannot approach zero anywhere in the domain.
A Proof Plan for Transformed Sequences
When a problem asks whether a transformed sequence converges uniformly, identify the error that must be controlled and the extra condition that controls its multiplier or sensitivity. The following sequence of checks keeps the argument quantitative.
For products, add and subtract an intermediate product; for reciprocals, combine the fractions; for composition, compare the two inputs to the outer function.
Look for bounded factors in a product, a positive lower bound for denominators, or a uniform-continuity estimate for an outer function.
Bound the transformed error by constants times the original uniform errors, or apply one uniform-continuity tolerance throughout the domain.
Take the supremum over the domain and verify that the resulting bound tends to zero.
The main distinction is between controlling an error at each fixed input and controlling it with a single estimate over the whole domain. Uniform convergence provides the latter for the original functions. The additional hypotheses determine whether a transformation preserves that control.
Check Your Understanding
Use the hypotheses and error estimates above to answer the following questions.
- Why does the composition theorem require uniform continuity rather than merely continuity of the outer function?
- In the product estimate, where do the bounds on the function values enter?
- If \(|f(x)|\geq c>0\), what lower bound on \(|f_n(x)|\) follows once \(|f_n(x)-f(x)|<c/2\)?
- Why does exponentiation fail to preserve uniform convergence in the example on the whole real line?
- Which additional condition would let you apply the composition theorem to a continuous outer function on a bounded range?