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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.

For a locally compact Hausdorff space $X$, $C_c(X)$ consists of continuous functions whose support is compact. A positive linear functional on this space, bounded on each fixed compact support, is represented by a Radon measure. For a locally compact group with Haar measure, these functions are dense in $L^p$ for $1\le p<\infty$.

For a bounded operator on a complex Banach space, the spectrum is a nonempty compact subset of the complex plane contained in the closed disk of radius equal to the operator norm.

For whole-line Cauchy data with support contained in $[-a,a]$, the displayed inclusion holds at $t\ge 0$. A short check uses D'Alembert's formula: if $x>a+ct$, both sampled endpoints and the whole interval of integration lie to the right of $a$, so all contributions vanish; the argument for $x<-a-ct$ is identical. The inclusion can be strict, and a nonzero response at the first permitted arrival is not guaranteed.