Sequences of Functions: Pointwise Convergence

Pointwise convergence is the most natural way to take a limit of functions, but it is the WRONG one for analysis. The problem is that the order of "for each $x$" and "for each $n$" matters enormously: if you fix $x$ first and then send $n\\to\\infty$, the limit $f(x)=\\lim f\_n(x)$ exists for every $x$, but the resulting function $f$ might be discontinuous even though every $f\_n$ is smooth. Worse, you cannot pass limits through integrals or derivatives. The standard example $f\_n(x)=x^n$ on \[0,1\] has continuous $f\_n$ but a discontinuous pointwise limit. Pointwise convergence is the warm-up; uniform convergence (next section) is what actually fixes things.

The definition

A sequence of functions $\\{f\_n\\}$ defined on a set $S$ converges pointwise to a function $f$ on $S$ if for every $x\\in S$, the numerical sequence $f\_1(x), f\_2(x), f\_3(x), \\ldots$ converges to $f(x)$. Formally: for every $x\\in S$ and every $\\epsilon>0$, there exists $N(x,\\epsilon)$ such that $|f\_n(x)-f(x)|<\\epsilon$ for all $n\\geq N$. Note crucially that $N$ may depend on $x$, that is the entire weakness of the concept.

The classic example: $f\_n(x)=x^n$ on \[0,1\]

For $0\\leq x<1$, the geometric $x^n\\to 0$. At $x=1$, $1^n=1$ for every $n$. So the pointwise limit is $f(x)=0$ for $x\\in\[0,1)$ and $f(1)=1$, a discontinuous step function, even though every $f\_n$ is a smooth polynomial. The convergence is "slow" near $x=1$: to get $x^n<\\epsilon$, you need $n>\\log\\epsilon/\\log x$, which blows up as $x\\to 1^-$. That is precisely why no single $N$ works for all $x$ at once.

Why integrals and limits need not commute

Consider $f\_n(x)=n$ on $(0,1/n)$ and $0$ elsewhere on \[0,1\]. Each $f\_n$ has $\\int\_0^1 f\_n=1$. But pointwise, for every fixed $x>0$ we have $f\_n(x)=0$ for all sufficiently large $n$, so the pointwise limit is $f\\equiv 0$. Therefore $\\lim\_n \\int f\_n=1\\neq 0=\\int\\lim\_n f\_n$. The mass "escapes to a point", a phenomenon unique to limits of unbounded sequences. This is the failure mode pointwise convergence cannot detect.

Use the grapher to plot $f\_n(x)=x^n$ on \[0,1\] for $n=1, 2, 5, 10, 50$. Watch how each curve hugs the $x$-axis for small $x$ and then sharply rises to 1 near $x=1$. As $n$ grows, the transition becomes a near-vertical wall just left of 1. The pointwise limit is the step function (0 for $x<1$, 1 at $x=1$), but no continuous curve "looks like" the step function for any finite $n$, that is the visual signature of failed uniform convergence.

• Numerical integration & convergence theory: the Lebesgue Dominated Convergence Theorem and the Monotone Convergence Theorem give exactly the conditions under which pointwise convergence of integrands implies convergence of integrals. Numerical analysts rely on these results when they pass to the limit in Monte Carlo, finite-element, or spectral methods.
• Approximation theory: a sequence of approximation operators (Bernstein polynomials, Fourier partial sums, splines) is typically constructed to converge pointwise to a target function; whether the convergence is uniform determines whether the approximation is acceptable for engineering applications.
• Probability & convergence of distributions: a sequence of random variables converges "in distribution" iff their CDFs converge pointwise at every continuity point. The Central Limit Theorem is fundamentally a pointwise-CDF-convergence statement.

Pause and think: Why does pointwise convergence "lose" continuity of the limit even when each $f\_n$ is continuous? Try to articulate, in your own words, what extra control we would need on $\\{f\_n\\}$ to guarantee the limit is continuous.

Try it

• Predict first: find the pointwise limit of $f\_n(x)=x/n$ on $\\mathbb{R}$. Is the limit continuous? Is the convergence uniform? (Spoiler: limit is 0, yes continuous, no uniform on all of $\\mathbb{R}$.)
• Compute the pointwise limit of $f\_n(x)=1/(1+x^{2n})$ on $\[0,\\infty)$. (Hint: split into $x<1$, $x=1$, $x>1$.) Is the limit continuous?
• Construct a sequence of continuous $f\_n$ on \[0,1\] with $\\int\_0^1 f\_n=1$ but $f\_n\\to 0$ pointwise. (Tent functions of height $n$ and base $2/n$ work.)
• True or false: if $f\_n\\to f$ pointwise on \[a,b\] and each $f\_n$ is bounded, then $f$ is bounded. Justify or give a counterexample.

A trap to watch for

Pointwise convergence is so weak that most "nice" theorems about limits and analysis ARE FALSE under it. Specifically: pointwise limit of continuous can be discontinuous; pointwise limit of integrable can be non-integrable; $\\lim\\int\\neq\\int\\lim$ in general; $\\lim f\_n'\\neq(\\lim f\_n)'$ in general. Whenever you want to swap a limit with another operation, you need a stronger hypothesis, usually uniform convergence (Section 2.7) or a dominated/monotone hypothesis (Lebesgue integration).

What you now know

You can compute pointwise limits, construct discontinuous-limit examples, build counterexamples to interchange-of-limits, and articulate why pointwise convergence alone is insufficient. Section 2.7 introduces uniform convergence, which is exactly the upgrade needed: it preserves continuity, allows interchange with integration, and (with a derivative condition) allows interchange with differentiation.

Mark section complete →

References

• Garrity, T. (2002). All the Mathematics You Missed. Cambridge University Press, ch. 2.
• Rudin, W. (1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill, ch. 7.
• Abbott, S. (2015). Understanding Analysis (2nd ed.). Springer, ch. 6.
• Apostol, T. M. (1974). Mathematical Analysis (2nd ed.). Addison-Wesley, ch. 9.
• Tao, T. (2016). Analysis I (3rd ed.). Hindustan Book Agency, ch. 14.