Uniform Convergence and the Supremum Norm

Uniform Error as a Distance Between Functions

The supremum norm turns the idea of “largest error across the domain” into a measurement: for bounded functions \(f_n\) and \(f\), the error at stage \(n\) is measured by \(\|f_n-f\|_\infty\). The Supremum Criterion for Uniform Convergence, established earlier in the course, says that uniform convergence is equivalent to these supremum errors tending to zero. This tutorial develops the consequences of that connection. In particular, we will see what uniform convergence implies about boundedness and about the sizes of the functions themselves.

The domain matters. If \(f_n,f\in B(E)\), then \(f_n-f\in B(E)\), so its supremum norm is defined. More generally, the difference may be bounded even when \(f_n\) and \(f\) are not individually bounded; in that case, the norm of the difference is still meaningful. The key point is that a supremum norm measures error over all inputs at once, not at one input chosen at a time.

Definition: When \(f_n-f\in B(E)\), the supremum-norm error at stage \(n\) is \(\|f_n-f\|_\infty=\sup_{x\in E}|f_n(x)-f(x)|\). Uniform convergence \(f_n\to f\) means precisely that these errors tend to zero, as given by the Supremum Criterion for Uniform Convergence.

This gives a useful interpretation: uniform convergence is convergence in the metric \(d_\infty\) from “The Supremum Norm.” If the functions all belong to \(B(E)\), then \(d_\infty(f_n,f)=\|f_n-f\|_\infty\). The statement \(d_\infty(f_n,f)\to0\) says that the functions become close in the supremum metric.

Calculating the Supremum-Norm Error

The supremum in the error formula need not be attained. It is enough to find an upper bound and show that values of the error approach that bound. The following example illustrates why the supremum, rather than a maximum, remains the appropriate measure.

Worked Example: Uniform Error on an Unbounded Domain

For positive integers \(n\), define \(f_n:[0,\infty)\to\mathbb{R}\) by \(f_n(x)=x/(n(1+x))\), and let \(f(x)=0\). For every \(x\geq0\),

|f_n(x)-f(x)|=\frac{x}{n(1+x)}<\frac{1}{n},

because \(x<1+x\). Thus \(1/n\) is an upper bound for the error values. For positive integers \(k\), the values at \(x=k\) satisfy

|f_n(k)-f(k)|=\frac{k}{n(1+k)} =\frac{1}{n}\left(1-\frac{1}{1+k}\right)\longrightarrow\frac{1}{n} \qquad\text{as }k\to\infty.

No error value equals \(1/n\), but the values approach it arbitrarily closely. Therefore \(\|f_n-f\|_\infty=1/n\). Since \(1/n\to0\), the Supremum Criterion for Uniform Convergence gives \(f_n\to f\) uniformly on \([0,\infty)\).

A supremum-norm calculation often gives more information than a bare convergence statement: it provides an explicit bound on the error at every input. Here, for example, \(|f_n(x)-f(x)|\leq1/n\) holds simultaneously for all \(x\geq0\).

Uniform Limits of Bounded Functions Are Bounded

Uniform convergence does not, by itself, say that the limit is bounded. However, if every function in the sequence is bounded, uniform convergence prevents the limit from becoming unbounded. One stage of the sequence can be chosen close enough to the limit everywhere, and that single bounded function then supplies a global bound for the limit.

Theorem (Uniform Limits of Bounded Functions Are Bounded): Let \(E\) be nonempty. Suppose every \(f_n:E\to\mathbb{R}\) is bounded and \(f_n\to f\) uniformly on \(E\). Then \(f\) is bounded on \(E\).

Proof. By uniform convergence, there is an index \(N\) such that for every \(x\in E\),

$$ |f_N(x)-f(x)|<1. $$

Because \(f_N\) is bounded, there is a finite \(M\geq0\) such that \(|f_N(x)|\leq M\) for all \(x\in E\). The triangle inequality now gives, for every \(x\in E\),

|f(x)|\leq |f(x)-f_N(x)|+|f_N(x)|<1+M.

Thus \(1+M\) is a finite upper bound for \(|f(x)|\) on \(E\), so \(f\) is bounded. \(\square\)

Only one sufficiently late function is needed in this proof. It would be incorrect to infer boundedness of \(f\) merely from pointwise convergence of bounded functions: pointwise closeness is not necessarily uniform closeness over the domain.

Worked Example: Bounded Approximations with an Unbounded Pointwise Limit

On \([0,\infty)\), define \(g_n(x)=\min(x,n)\). For each fixed \(n\), \(0\leq g_n(x)\leq n\), so \(g_n\) is bounded. For each fixed \(x\), once \(n\geq x\), \(g_n(x)=x\). Hence \(g_n\) converges pointwise to \(g(x)=x\), which is unbounded.

The convergence cannot be uniform, in agreement with the theorem. In fact, for any fixed \(n\), if \(x\geq n\), then

$$ |g_n(x)-g(x)|=|n-x|=x-n. $$

As \(x\) increases without bound, \(x-n\) increases without bound. Thus the error is not even bounded on the domain for any \(n\), and it cannot tend uniformly to zero. This example shows why the uniformity hypothesis in the theorem matters.

Norms of Uniformly Convergent Functions

Uniform convergence controls the difference between functions. The reverse triangle inequality from “The Supremum Norm” converts that control into a statement about their individual sizes. If \(f_n\) is close to \(f\) in supremum norm, then their norms must be close as real numbers.

Theorem (Norms Converge Under Uniform Convergence): Let \(E\) be nonempty, and suppose \(f_n,f\in B(E)\) for every \(n\). If \(f_n\to f\) uniformly on \(E\), then \(\|f_n\|_\infty\to\|f\|_\infty\).

Proof. The reverse triangle inequality gives, for every \(n\),

\big|\|f_n\|_\infty-\|f\|_\infty\big| \leq\|f_n-f\|_\infty.

By the Supremum Criterion for Uniform Convergence, the right-hand side tends to zero. Therefore the nonnegative quantity on the left also tends to zero. This is exactly the assertion that \(\|f_n\|_\infty\to\|f\|_\infty\). \(\square\)

The boundedness assumptions ensure that the norms in the statement are finite. If the functions themselves are not bounded, their individual supremum norms need not be available, even though the differences \(f_n-f\) may have finite norms.

Worked Example: Norms Converge When the Supremum Is Not Attained

Let \(E=[0,1)\), and define \(h_n(x)=1+x/n\) and \(h(x)=1\). For each \(x\in E\), \(1\leq h_n(x)<1+1/n\). The values approach \(1+1/n\) as \(x\) approaches \(1\), so

\|h_n\|_\infty=\sup_{x\in[0,1)}\left(1+\frac{x}{n}\right)=1+\frac{1}{n}.

The supremum is not attained because \(x=1\) is not in the domain. The error is \[ |h_n(x)-h(x)|=\frac{x}{n}, \] whose supremum on \([0,1)\) is \(1/n\), also not attained. Thus \(\|h_n-h\|_\infty=1/n\to0\), so the convergence is uniform. Meanwhile,

\|h_n\|_\infty=1+\frac{1}{n}\longrightarrow1=\|h\|_\infty.

The reverse triangle inequality predicts that the difference of the norms is no greater than the norm of the error; here both quantities equal \(1/n\).

A related consequence is that a uniformly convergent sequence in \(B(E)\) has bounded norms. Indeed, if \(f_n\to f\) uniformly and \(f,f_n\in B(E)\), then for all sufficiently large \(n\), \(\|f_n-f\|_\infty<1\). The triangle inequality gives \(\|f_n\|_\infty\leq\|f\|_\infty+1\) for those indices. There are only finitely many earlier indices, and each has finite norm, so the entire sequence of real numbers \(\|f_n\|_\infty\) is bounded.

Pointwise Convergence Does Not Control the Supremum Error

Pointwise convergence allows the index needed for a chosen accuracy to depend on the input. The supremum norm instead measures the worst error over the whole domain at each stage. Consequently, pointwise convergence need not make the supremum errors small.

Worked Example: Pointwise Convergence with a Persistent Supremum Error

For \(n\geq1\), let \(p_n:[0,1]\to\mathbb{R}\) be \(p_n(x)=x^n\). At every \(x\in[0,1)\), \(x^n\to0\), while \(p_n(1)=1\) for all \(n\). The pointwise limit is therefore

p(x)= \begin{cases} 0,&0\leq x<1,\\ 1,&x=1. \end{cases}

For \(0\leq x<1\), \(|p_n(x)-p(x)|=x^n\); at \(x=1\), the error is zero. The supremum of the error values is \(1\): the values \(x^n\) approach \(1\) as \(x\) approaches \(1\) from below, but no \(x<1\) gives \(x^n=1\). Hence

\|p_n-p\|_\infty=1

for every \(n\), and the sequence does not converge uniformly. The moving-input criterion from “Pointwise Versus Uniform Convergence” expresses the same obstruction: inputs near \(1\) can be chosen to keep the error large as \(n\) changes.

Using the Supremum Norm Carefully

The supremum norm is especially useful when a proof needs a single error bound valid across the entire domain. When \(f_n\) and \(f\) are bounded, the norm of their difference is exactly the supremum error from the Supremum Criterion for Uniform Convergence. If the difference is bounded but the functions are not, the norm still measures uniform error, but the individual norms \(\|f_n\|_\infty\) and \(\|f\|_\infty\) may not exist as finite quantities.

Keep three distinctions in view. First, a supremum need not be attained. Second, pointwise convergence does not imply that supremum errors tend to zero. Third, the conclusion that a uniform limit is bounded requires that the approximating functions be bounded; it is uniform closeness to one bounded approximant that provides the bound. These facts make the supremum norm a precise tool for understanding both convergence and the size of functions.

Takeaway: For bounded functions, uniform convergence is convergence of the differences in supremum norm. It forces bounded approximating functions to have a bounded limit, and the norms of the functions converge to the norm of that limit.

Check Your Understanding

Use the supremum-norm interpretation and the results proved here to answer the following questions.

When does the supremum-norm error \(\|f_n-f\|_\infty\) express the uniform error between \(f_n\) and \(f\)?
In the sequence \(f_n(x)=x/(n(1+x))\) on \([0,\infty)\), is the supremum error attained? What is its value?
Why does a uniform limit of bounded functions have to be bounded?
Which earlier result gives the inequality used to prove that \(\|f_n\|_\infty\to\|f\|_\infty\)?
For \(p_n(x)=x^n\) on \([0,1]\), why do the supremum errors fail to tend to zero?