Pointwise Convergence of Functions

Convergence One Input at a Time

A sequence of functions is a sequence of objects, but pointwise convergence asks a separate question at each input: does the sequence of output values converge in the target space? The definition keeps the domain and target in different roles. The domain determines which inputs must be tested; the topology on the target determines what it means for the outputs to converge.

Let \(X\) be a set, let \((Y,\mathcal{T})\) be a topological space, and let \(f_n:X\to Y\) for each \(n\in\mathbb{N}\). Write \(Y^X\) for the set of all maps from \(X\) to \(Y\). No topology on \(Y^X\) is needed to define the convergence considered here.

Definition: The sequence \((f_n)\) converges pointwise to a map \(f:X\to Y\) if, for every \(x\in X\), the sequence of values \((f_n(x))\) converges to \(f(x)\) in \(Y\). Explicitly, for every \(x\in X\) and every open neighborhood \(U\) of \(f(x)\), there is an \(N\in\mathbb{N}\) such that \(f_n(x)\in U\) for all \(n\ge N\).

The order of the quantifiers is essential: for each fixed \(x\), a threshold \(N\) may be chosen. The threshold can depend on \(x\), on the neighborhood \(U\), and on the sequence. If \(X=\varnothing\), there is exactly one map \(X\to Y\), and every sequence of such maps converges to it: the condition has no points \(x\) to test.

When \(Y=\mathbb{R}\) with its standard topology, the neighborhood definition is equivalent to the familiar epsilon condition: for each fixed \(x\in X\) and each \(\varepsilon>0\), there is \(N\) such that \(n\ge N\) implies \(|f_n(x)-f(x)|<\varepsilon\). Indeed, open intervals centered at \(f(x)\) form a neighborhood base in \(\mathbb{R}\): applying convergence to \((f(x)-\varepsilon,f(x)+\varepsilon)\) gives the inequality, and every open neighborhood of \(f(x)\) contains such an interval for some \(\varepsilon>0\).

Pointwise Limits on an Interval

In the following examples, the domain is \(I=[0,1]\) with the subspace topology inherited from the standard topology on \(\mathbb{R}\), and the target is \(\mathbb{R}\) with its standard topology. The formulas define maps on every point of \(I\), including both endpoints.

Worked Example: Powers on the Unit Interval

For \(n\in\mathbb{N}\), define \(f_n:I\to\mathbb{R}\) by \(f_n(x)=x^n\). Find the pointwise limit.

Fix \(x\in I\). If \(x=1\), then \(f_n(1)=1^n=1\) for every \(n\), so the value sequence converges to \(1\). If \(0\le x<1\), then \(x^n\to0\). For \(x=0\), this is immediate because \(0^n=0\) for all \(n\in\mathbb{N}\). For \(0<x<1\), put \(c=1/x-1\), so \(c>0\) and \(x=1/(1+c)\). The binomial inequality \((1+c)^n\ge 1+nc\) gives \[ 0\le x^n=\frac{1}{(1+c)^n}\le\frac{1}{1+nc}. \] The right-hand side tends to \(0\), so \(x^n\to0\).

Therefore the pointwise limit \(f\) is \[ f(x)= \begin{cases} 0,&0\le x<1,\\ 1,&x=1. \end{cases} \]

Check: At \(x=1/2\), the first four values are \(1/2,\ 1/4,\ 1/8,\ 1/16\), and they tend to the stated value \(0\). At \(x=1\), every value is \(1\), matching \(f(1)=1\). These checks also show why the endpoint \(1\) must be handled separately.

Worked Example: A Reciprocal Family

Define \(g_n:I\to\mathbb{R}\) by \(g_n(x)=1/(1+nx)\). Find its pointwise limit, paying particular attention to \(x=0\).

At \(x=0\), \(g_n(0)=1/(1+n\cdot0)=1\) for every \(n\), so the limit is \(1\). Fix instead \(x>0\). Then \[ 0<g_n(x)=\frac{1}{1+nx}\le\frac{1}{nx}. \] For this fixed \(x\), the right-hand side tends to \(0\) as \(n\to\infty\), and hence \(g_n(x)\to0\). Thus the pointwise limit is \(g(0)=1\) and \(g(x)=0\) for \(0<x\le1\).

Check: At \(x=1/2\), direct substitution gives \(g_n(1/2)=1/(1+n/2)=2/(n+2)\), which tends to \(0\). At \(x=0\), direct substitution gives \(1/(1+0)=1\), not \(0\). The calculation confirms both parts of the stated limit.

In both examples, each input is fixed before the values are tested. In particular, the estimate for \(g_n(x)\) uses a fixed positive \(x\); it does not apply at \(x=0\). Such endpoint and zero cases are not optional details: a pointwise limit is specified at every point in the domain.

Sequences on Discrete Domains

Let \(\mathbb{N}\) have the discrete topology, in which every subset is open, and let the target be \(\mathbb{R}\) with its standard topology. The definition still tests one input \(k\in\mathbb{N}\) at a time. The discreteness of the domain does not change convergence of the real value sequence \((f_n(k))\); it means only that arbitrary functions on \(\mathbb{N}\) are continuous, a fact that is not needed for the computations below.

Worked Example: Indicators of Growing Initial Segments

For \(n,k\in\mathbb{N}\), define \(u_n(k)=1\) if \(k\le n\), and \(u_n(k)=0\) if \(k>n\). Determine the pointwise limit.

Fix an input \(k\in\mathbb{N}\). If \(n\ge k\), then \(k\le n\), so \(u_n(k)=1\). Thus the value sequence at this fixed input is eventually constant at \(1\). More precisely, for every \(\varepsilon>0\), choosing \(N=k\) gives \(|u_n(k)-1|=0<\varepsilon\) whenever \(n\ge N\). The pointwise limit is therefore the constant function \(u(k)=1\).

Check: At input \(k=3\), the values for \(n=1,2,3,4\) are \(0,0,1,1\), respectively. They are \(1\) from \(n=3\) onward, as the general calculation predicts. The inequality defining the function is \(k\le n\), so equality at \(n=k\) gives value \(1\).

Worked Example: A Moving Spike

For \(n,k\in\mathbb{N}\), define \(v_n(k)=1\) if \(k=n\), and \(v_n(k)=0\) if \(k\ne n\). Determine the pointwise limit.

Fix \(k\in\mathbb{N}\). At the single index \(n=k\), the value is \(v_k(k)=1\). For every \(n\ge k+1\), we have \(n\ne k\), so \(v_n(k)=0\). Hence the value sequence at \(k\) is eventually \(0\), and \(v_n(k)\to0\). Since this holds for every fixed \(k\), the pointwise limit is the constant zero function.

Check: At input \(k=2\), the values for \(n=1,2,3,4\) are \(0,1,0,0\); from \(n=3\) onward they are \(0\). At input \(k=4\), the value at \(n=4\) is \(1\), but all values for \(n\ge5\) are \(0\). Both calculations agree with the pointwise limit \(0\).

Two Consequences of the Definition

Theorem: If \(Y\) is Hausdorff, a sequence of maps \(f_n:X\to Y\) has at most one pointwise limit.

Proof. Suppose \((f_n)\) converges pointwise both to \(f:X\to Y\) and to \(g:X\to Y\). Fix \(x\in X\). If \(f(x)\ne g(x)\), the Hausdorff property gives disjoint open neighborhoods \(U\) of \(f(x)\) and \(V\) of \(g(x)\). Convergence to \(f(x)\) gives an index \(N_1\) such that \(f_n(x)\in U\) for every \(n\ge N_1\); convergence to \(g(x)\) gives an index \(N_2\) such that \(f_n(x)\in V\) for every \(n\ge N_2\). Choose \(n\ge\max(N_1,N_2)\). Then \(f_n(x)\in U\cap V\), contradicting \(U\cap V=\varnothing\). So \(f(x)=g(x)\). Since this holds for every \(x\in X\), \(f=g\). If \(X\) is empty, there is only one map \(X\to Y\), and the conclusion holds as well. \(\square\)

The Hausdorff hypothesis matters. For example, let \(Y=\{a,b\}\) have the indiscrete topology \(\{\varnothing,Y\}\), and let \(X=\{p\}\). The constant sequence of maps \(f_n(p)=a\) converges to the map with value \(a\): the only open neighborhood of \(a\) is \(Y\), which contains every term. It also converges to the map with value \(b\), because the only open neighborhood of \(b\) is \(Y\) as well. The two limits differ. This example verifies that uniqueness need not hold for a non-Hausdorff target.

Proposition: If \(X\) is finite and \(Y=\mathbb{R}\), pointwise convergence gives one threshold that works for all inputs at once for any fixed \(\varepsilon>0\).

Proof. Let \(f_n:X\to\mathbb{R}\) converge pointwise to \(f\), and fix \(\varepsilon>0\). For each \(x\in X\), pointwise convergence gives an index \(N_x\) such that \(|f_n(x)-f(x)|<\varepsilon\) for all \(n\ge N_x\). If \(X\) is finite and nonempty, the finite set of integers \(\{N_x:x\in X\}\) has a maximum \(N\). For every \(x\in X\) and \(n\ge N\), we have \(n\ge N_x\), so \(|f_n(x)-f(x)|<\varepsilon\). If \(X=\varnothing\), the assertion has no inputs to check and any \(N\in\mathbb{N}\) works. \(\square\)

Finiteness is essential to this argument: for an infinite domain, the individual thresholds need not have a finite maximum. In the moving-spike example, take \(\varepsilon=1/2\). At any proposed index \(N\), use the input \(k=N\). Then \(v_N(N)=1\) and the pointwise limit at \(N\) is \(0\), so \(|v_N(N)-0|=1\), which is not less than \(1/2\). This does not contradict pointwise convergence: for each fixed \(k\), all terms with \(n\ge k+1\) have value \(0\) at that input. The failed calculation would be to use the pointwise threshold \(N=k\) as though it worked for every \(k\); it depends on \(k\).

Check Your Understanding

For each question, keep the input fixed before testing the sequence of values.

On \(I=[0,1]\), find the pointwise limit of \(h_n(x)=(1-x)^n\), treating \(x=0\) and \(x>0\) separately.
On \(\mathbb{N}\), define \(w_n(k)=1\) when \(k>n\) and \(w_n(k)=0\) when \(k\le n\). Find the pointwise limit and state a valid threshold for a fixed input \(k\).
Which part of the uniqueness proof uses the Hausdorff property of the target?
In the moving-spike example, why does the value \(v_k(k)=1\) not prevent convergence to \(0\) at the fixed input \(k\)?
Explain why pointwise convergence on a finite nonempty domain gives a maximum of the finitely many thresholds, while that proof does not extend directly to an infinite domain.