Examples of Pointwise Convergence

Pointwise Convergence Is Checked One Input at a Time

In the previous tutorial, the challenge was to choose one index that works for every input. Pointwise convergence asks for less: fix an input first, and then check whether the resulting sequence of real numbers converges. The index may depend on the input. That order of choices makes many examples straightforward, but it also means that behavior at exceptional points must be checked separately.

The definition and basic limit laws from “Pointwise Convergence” apply throughout. In practice, the central calculation is to fix an arbitrary \(x\) in the domain and evaluate \(\lim_{n\to\infty} f_n(x)\). If a formula behaves differently at different inputs, divide the domain into cases and find the scalar limit in each case. The result of those calculations is a function: the proposed pointwise limit.

Geometric Powers and an Exceptional Endpoint

Theorem (Pointwise Limit of Geometric Powers on \([0,1]\)): Define \(f_n:[0,1]\to\mathbb{R}\) by \(f_n(x)=x^n\). Then \(f_n\) converges pointwise on \([0,1]\) to the function \(f\) given by \(f(x)=0\) for \(0\leq x<1\) and \(f(1)=1\).

Proof. Fix \(x\in[0,1]\). If \(x=0\), then \(f_n(0)=0^n=0\) for every positive integer \(n\), so the sequence converges to \(0\). If \(0<x<1\), the geometric sequence \(x^n\) tends to \(0\), since its fixed ratio \(x\) lies strictly between \(0\) and \(1\). Finally, if \(x=1\), then \(f_n(1)=1^n=1\) for every \(n\), so the sequence converges to \(1\). These cases cover every point of \([0,1]\) and give exactly the stated function \(f\). Therefore \(f_n\to f\) pointwise on \([0,1]\). \(\square\)

The endpoint is decisive. For any fixed \(x<1\), powers eventually become small, but at \(x=1\) they remain equal to \(1\). Consequently, the limit is not the zero function on the closed interval. This is a useful model for pointwise proofs: first handle points where a standard limit applies, then inspect the boundary and any other exceptional points directly.

Worked Example: The Same Powers on a Half-Open Interval

Let \(g_n:[0,1)\to\mathbb{R}\) be given by \(g_n(x)=x^n\). To determine the limit, fix \(x\in[0,1)\). If \(x=0\), then \(g_n(0)=0\) for all \(n\). If \(0<x<1\), then the geometric sequence \(x^n\) tends to \(0\). Hence, in either case, \(\lim_{n\to\infty}g_n(x)=0\). The pointwise limit on this domain is therefore \(g(x)=0\) for every \(x\in[0,1)\).

This conclusion differs from the one on \([0,1]\) only because the point \(1\) is absent. The two domains contain the same points below \(1\), and the limit is \(0\) at each of them. On \([0,1]\), the additional point contributes the separate limit \(1\). A pointwise limit is defined on the domain of the sequence, so changing the domain can change the resulting limit function.

The limit in the theorem is discontinuous at \(1\): its value there is \(1\), whereas its values at points below \(1\) are \(0\). Each \(f_n(x)=x^n\) is a polynomial and hence continuous. Thus continuity of every function in a sequence does not, by itself, ensure continuity of its pointwise limit. The distinction between pointwise and uniform convergence will explain why continuity can fail in this way.

A Pointwise Squeeze Principle

A useful technique is to trap each function between two other sequences whose limits are known. The bounds need only work at each fixed input; they do not have to be chosen using one index that controls the whole domain. This is the pointwise form of the squeeze principle.

Theorem (Pointwise Squeeze Theorem): Let \(E\) be a set, and suppose \(a_n,b_n,f_n:E\to\mathbb{R}\) satisfy \(a_n(x)\leq f_n(x)\leq b_n(x)\) for every \(n\) and every \(x\in E\). Suppose also that \(a_n\to f\) and \(b_n\to f\) pointwise on \(E\). Then \(f_n\to f\) pointwise on \(E\).

Proof. Fix an arbitrary \(x\in E\). The hypotheses give the scalar inequalities \(a_n(x)\leq f_n(x)\leq b_n(x)\) for every \(n\). Since \(a_n(x)\to f(x)\) and \(b_n(x)\to f(x)\), the squeeze theorem for real sequences implies \(f_n(x)\to f(x)\). This argument applies to every \(x\in E\). Therefore \(f_n\to f\) pointwise on \(E\). \(\square\)

Worked Example: Squeezing an Oscillating Sequence

For \(x\in\mathbb{R}\), define \(h_n(x)=\sin(nx)/n\), where sine is the function defined by its series earlier in this course. The established bound \(|\sin(t)|\leq1\) for every real \(t\) gives, for every \(x\in\mathbb{R}\) and every positive integer \(n\),

-\frac{1}{n}\leq\frac{\sin(nx)}{n}\leq\frac{1}{n}.

Both outer expressions tend to \(0\) as \(n\to\infty\). By the pointwise squeeze theorem, \(h_n(x)\to0\) at every real \(x\). In this example the same bounds even hold across the entire domain at once, but pointwise convergence requires only that the squeeze argument succeed at each fixed input.

The proof of the squeeze theorem emphasizes the order of reasoning. First fix \(x\); then apply a theorem about ordinary real sequences. The pointwise limit of each bounding sequence must agree with the same function \(f\). If the two bounds instead tend to different values at a point, this theorem gives no conclusion there.

Inputs That Vary with the Index

Pointwise convergence concerns a fixed input. It does not say that values remain small when the input is allowed to change as \(n\) changes. A sequence can converge to zero at every fixed point while retaining a substantial value at a point that moves with the index. The next example makes this distinction explicit.

Worked Example: A Narrow Peak Moving Toward Zero

For \(n\geq1\), define \(p_n:\mathbb{R}\to\mathbb{R}\) by

p_n(x)=\frac{nx}{1+n^2x^2}.

Fix \(x\in\mathbb{R}\). If \(x=0\), then \(p_n(0)=0\) for every \(n\). Now suppose \(x\neq0\). Because \(1+n^2x^2\geq n^2x^2>0\), we obtain

|p_n(x)| =\frac{n|x|}{1+n^2x^2} \leq\frac{n|x|}{n^2x^2} =\frac{1}{n|x|}.

Here \(x\) is fixed and nonzero, so \(1/(n|x|)\to0\). The displayed bound therefore implies \(p_n(x)\to0\). Together with the calculation at \(x=0\), this proves that \(p_n\) converges pointwise to zero on \(\mathbb{R}\).

The estimate depends on the fixed nonzero input through \(1/|x|\). It cannot be used unchanged when \(x\) is allowed to vary with \(n\). Indeed, set \(x_n=1/n\). Substitution gives

p_n(x_n) =\frac{n(1/n)}{1+n^2(1/n)^2} =\frac{1}{1+1} =\frac{1}{2}.

Thus the values at these moving inputs do not tend to zero. There is no contradiction: the pointwise calculation fixes \(x\) before letting \(n\) grow, whereas \(x_n\) changes at every step. Keeping these two kinds of statements separate is essential when reading or writing a convergence proof.

A Limit That Keeps One Exceptional Value

A sequence may also have a pointwise limit that is nonzero at just one input. To establish such a limit, identify which inputs eventually fall outside the part of the domain where the functions are nonzero, and check separately any point that remains exceptional.

Worked Example: A Triangular Spike at the Origin

For \(x\in\mathbb{R}\), define

q_n(x)=\max\{1-n|x|,0\}.

At \(x=0\), the value is \(q_n(0)=\max\{1,0\}=1\) for every \(n\), so the pointwise limit at the origin is \(1\). Now fix \(x\neq0\). Choose a positive integer \(N\) with \(N\geq1/|x|\). For every \(n\geq N\), we have \(n|x|\geq1\), and therefore \(1-n|x|\leq0\). It follows that

q_n(x)=\max\{1-n|x|,0\}=0\qquad(n\geq N).

So the sequence is eventually zero at each fixed nonzero \(x\). Its pointwise limit is the function \(q\) defined by \(q(0)=1\) and \(q(x)=0\) for \(x\neq0\). Notice that the index \(N\) chosen for a nonzero \(x\) depends on \(|x|\); the argument establishes pointwise convergence without claiming that one index works for all real inputs.

This example and the geometric-powers example share a practical feature: the limit is determined by separating an exceptional set from the remaining inputs. For \(q_n\), the exceptional point is the origin; for \(x^n\) on \([0,1]\), it is the endpoint \(1\). Writing down the exceptional value explicitly prevents a common error: extending a formula valid away from that point to the entire domain without checking it.

A Reliable Procedure for Examples

For a new sequence of functions, use the following sequence of checks. The first step is the defining one: fix an arbitrary input before taking a limit. The remaining steps help prevent errors about domains, exceptional points, and changing inputs.

Fix a point in the stated domain.
Write \(f_n(x)\) with \(x\) held constant. Do not replace \(x\) by a sequence depending on \(n\) when calculating a pointwise limit.

Identify cases that behave differently.
Check endpoints, zeros of denominators, and points where a formula changes. Each case must be included in the domain.

Find the scalar limit in each case.
Use a known sequence limit, a valid bound, or the pointwise squeeze theorem. If a bound depends on the fixed input, make that dependence explicit.

Assemble the limit function.
State its value at every point, including exceptional points. Then check that the cases cover the full domain.

A pointwise limit is not automatically continuous, and pointwise convergence does not control the sequence at inputs that move with the index. Those are not defects in the definition; they are consequences of checking one fixed input at a time. In the next tutorial, examples of uniform convergence will sharpen the contrast by requiring a single error control across the domain.

Takeaway: To establish pointwise convergence, fix an arbitrary input and compute the resulting numerical limit. Separate exceptional points carefully, and do not confuse behavior at fixed inputs with behavior along a changing sequence of inputs.

Check Your Understanding

Use the definitions and examples above to answer the following questions.

What is the pointwise limit of \(x^n\) on \([0,1]\), and why must the point \(x=1\) be treated separately?
In the pointwise squeeze theorem, why do the two bounding sequences need to converge to the same function?
For \(p_n(x)=nx/(1+n^2x^2)\), what estimate proves convergence when a nonzero \(x\) is fixed?
Why does the calculation \(p_n(1/n)=1/2\) not contradict pointwise convergence to zero?
For the triangular spike \(q_n\), what is its pointwise limit at the origin and at a fixed nonzero input?