Pointwise convergence ($f_n(x)\to f(x)$)

For a sequence of functions $f_n:X\to \mathbb{R}$, pointwise convergence to a function $f:X\to \mathbb{R}$ meansIt is the ordinary limit of a sequence at each fixed input. The threshold $N=N(x,\epsilon )$ may depend on the input. Uniform convergence instead requires one threshold for the entire domain of a function.^[ref]

Consider the polynomial functions $f_n(x)=x^n$, for positive integers $n$, on the interval $[0,1]$. For $0\le x<1$ the geometric sequence $(x^n)$ tends to zero, while $f_n(1)=1$. ThusEach $f_n$ is a continuous function, but its pointwise limit is discontinuous at one. Holding an input still hides what happens near that endpoint. The analogous even-power example on $[-1,1]$ appears in the source.^[ref]

For $0<x<1$ and $0<\epsilon <1$, taking the logarithm of $x^n<\epsilon$ and dividing by the negative real number $\log x$ reverses the inequality:The floor function followed by adding one handles the strict inequality even when the ratio is an integer. For tolerance $\epsilon =0.01$, the first successful indices at $x=0.5$, $0.8$, and $0.99$ are respectively $7$, $21$, and $459$. At $x=0.5$, $2^{-6}=0.015625$ fails and $2^{-7}=0.0078125$ succeeds. At $x=0$ and $x=1$, the error from this pointwise limit is already zero, so $N=1$ works. Nevertheless, $N(x,\epsilon )$ grows without bound as $x\to 1$ from below. A finite list of accurate test inputs cannot establish uniform convergence.

An even sharper warning concerns integration. On $[0,1]$, define the continuous functionsTheir supports are $[0,1/n]$, with height $n$ at $1/(2n)$ and value zero at both ends. Every fixed $x>0$ eventually lies beyond the support, and $g_n(0)=0$ always, so pointwise convergence gives the zero function. Nevertheless, each integral is the area of a triangle: ${\int }_0^1{g}_n(x)dx=\frac{1}{2}(1/n)n=\frac{1}{2}$. Shrinking width compensates increasing height. Hence the limit of a sequence of these integrals differs from the integral of the pointwise limit; interchanging them requires additional conditions, such as uniform convergence preserves integrals on a compact interval.

On $[0,1]$, for $n\ge 2$, these continuous functions have support $[0,2/n]$, height $n$, and integral one. Their pointwise limit is zero: at zero every value vanishes, while every fixed positive argument eventually leaves the support. Thus pointwise convergence of continuous functions on a compact interval does not suffice for convergence of their integrals. The mass becomes more concentrated as its height grows.

For fixed $m$, the inner limit is one precisely when $m!x$ is an integer, and zero otherwise, because $|\cos (\pi y)|=1$ exactly at integers and powers of a number with modulus below one tend to zero. If $x=p/q$ is a rational number, $q$ divides $m!$ for every $m\ge q$, making the inner limit eventually one. If $x$ is an irrational number, multiplication by a nonzero integer cannot make it integral, making every inner limit zero. The order of the two limits is essential to this argument.

For $n\ge 2$, these continuous functions on $[0,1]$ have maximum one and integral $1/n^2$. Their supports shrink towards zero while their value at zero remains zero. They converge pointwise to zero and their integrals converge to zero, but their convergence is not uniform convergence. Thus convergence of integrals and continuity of the pointwise limit do not imply uniform convergence.

A pointwise limit assigns to each input the limit of a sequence of function values there, on the specified domain of a function. It is unique wherever it exists, by uniqueness of the numerical limit of a sequence. Its existence supplies no common approximation rate over the domain of a function. For $f_n(x)=x^n$ on $[0,1]$, it equals zero below one and one at one; restricting to $[0,a]$ with $a<1$ gives the zero function.