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Put ${\epsilon }_n={\sup }_{[-1,1]}|F_n-F|$. Uniform convergence means ${\epsilon }_n\to 0$, and continuity ensures that the functions have Riemann integrals. The integral triangle inequality givesThis proves that uniform convergence preserves integrals on a compact interval.

For the derivative assertion, take $F_n(x)=\sin (n^{2}x)/n$ and $F(x)=0$. These functions are differentiable, with $\sup |F_n|\le 1/n$, so their convergence is uniform. However, $F_n^{\prime }(0)=n$ whereas $F^{\prime }(0)=0$. The derivatives therefore need not converge even at one interior point.

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.

A sequence of functions $f_n:X\to Y$ converges uniformly to $f:X\to Y$, where $Y$ is a metric space with metric $d$, precisely whenFor real numbers, use the absolute value $d(u,v)=|u-v|$. Compare the universal quantification and existential quantification:The decisive change is whether $N$ can depend on $x$. Uniform convergence implies pointwise convergence.^[ref]

For functions taking values in real numbers on a nonempty domain of a function, the supremum error is $E_n={\sup }_{x\in X}|f_n(x)-f(x)|$, allowing $+\infty$. Then uniform convergence is equivalent to $E_n\to 0$. The supremum need not be a maximum: pointwise strict bounds below $\epsilon$ only give $E_n\le \epsilon$, so use tolerance $\epsilon /2$ when proving this equivalence. This single error bound controls every possible input, including one chosen after seeing the index.

For $f_n(x)=x^n$ and its pointwise limit on $[0,1]$, the error is zero at one but approaches one as $x\to 1$ from below. Consequentlyfor every positive integer $n$. This supremum is never attained: below one the error is $x^n<1$, and at one it is zero. Deleting the endpoint leaves the same supremum error on $[0,1)$. To disprove uniform convergence directly, choose $\epsilon =1/2$: for any proposed $N$, take $n=N$ and $x=2^{-1/N}<1$. The error equals $1/2$, violating the strict tolerance. Thus a moving input defeats every proposed common threshold.

On the smaller interval $[0,a]$, for a fixed $0\le a<1$, the pointwise limit is zero andFor $0<a<1$ and $0<\epsilon <1$, the exact first successful common threshold isThe logarithm and floor function turn an infinite approximation statement into a finite stopping rule. At $a=0.8$, $0.8^{20}=0.01152921504606846976$ fails tolerance $0.01$, while $0.8^{21}=0.009223372036854775808$ succeeds. At $a=1/2$ and $\epsilon =1/8$, the ratio is exactly three: $N=4$ is necessary because equality is insufficient. For $a=0$, every error is zero. The domain of a function determines whether one threshold can certify every input.

For another explicit stopping rule, let $h_n(x)=x/(n+1)$ on $[0,1]$. Its uniform limit is zero and its supremum error is $1/(n+1)$. For every $\epsilon >0$, $N=\max \{1,\lceil 1/\epsilon \rceil \}$ works, using the ceiling function. Unlike $x^n$ on $[0,1]$, it controls even the endpoint.

The practical beauty of uniform convergence is that one quantitative bound can survive further operations: the uniform limit theorem preserves continuity, and uniform convergence preserves integrals on a compact interval controls accumulated error. Differentiation needs additional hypotheses, such as uniform derivative convergence with an anchored value. These results distinguish an approximation that looks accurate at selected inputs from one that justifies replacing a whole function in a calculation.