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Write the partial sums as $S_N(x)={\sum }_{n=1}^N{f}_n(x)$. The series has uniform convergence to $s:E\to \mathbb{R}$ ifThe same threshold $N_0$ must control every point of $E$. Equivalently, the supremum error ${\sup }_{x\in E}|S_N(x)-s(x)|$ tends to zero, for nonempty $E$.

Suppose $f,g:X\to \mathbb{R}$ are bounded functions on the same nonempty set, and the supremum norm of their difference is at most $\delta$. Then their optimal values differ by at most the same amount:Indeed, $f(x)\le g(x)+\delta \le {\sup }_{X}g+\delta$ for every $x$, so ${\sup }_{X}f\le {\sup }_{X}g+\delta$; exchange $f$ and $g$ for the other direction. Adding a constant shifts a supremum by that constant, a basic compatibility between upper bounds and addition.^[ref]

For a chosen example on $[-1,1]$, take $f(x)=-x^2$ and $g(x)=-x^2+0.2x$. Their supremum error is $0.2$, since $|g(x)-f(x)|=0.2|x|$. Completing the square givesThus $f$ has maximum zero at $x=0$, while $g$ has maximum $0.01$ at $x=0.1$. The change in optimal value, $0.01$, is smaller than the certified bound $0.2$. For the vertical translation $h(x)=f(x)+0.2$, the bound is exact: its supremum error and optimal-value change are both $0.2$.

A stable optimal value does not guarantee a stable maximizing input. On $[-1,1]$, the two linear functions $f_{\epsilon }(x)=\epsilon x$ and $g_{\epsilon }(x)=-\epsilon x$ satisfyFor every positive $\epsilon$, their unique maximizing inputs remain two units apart, at $1$ and $-1$. Both functions approach the flat zero function, which has many maximizing inputs. Consequently an accurate objective function value is not by itself a certificate that the optimizing decision is accurate.

The estimate also gives a convenient consequence of uniform convergence. If $f_n$ converges uniformly to a bounded function $f$ on a nonempty $X$, then ${\sup }_X{f}_n\to {\sup }_{X}f$ for the bounded tail of the sequence. Uniform control handles the changing input that defines the best value. Pointwise convergence alone supplies no such control; the existing examples under uniform convergence illustrate that distinction.

For functions taking values in real numbers on a nonempty domain of a function $X$, the supremum error is the supremum of the absolute value of the approximation error over all inputs. It may be infinite and need not be attained. When finite, it is the supremum norm of $f_n-f$; uniform convergence is exactly $E_n\to 0$. Bounding this one quantity gives an accuracy guarantee for every input, including inputs that change with $n$.

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.

Let $f_n$ be continuously differentiable functions on an interval $I$. If $f_n^{\prime }\to g$ by uniform convergence on $I$ and $f_n(c)\to A$ at one fixed $c\in I$, then $f_n$ converge locally uniformly to a continuously differentiable function $f$ with $f^{\prime }=g$. If $I$ is bounded, the convergence is uniform on all of $I$.^[ref]

The anchor fixes the additive constant that a derivative cannot see. By the fundamental theorem of calculus, defineThe uniform limit theorem makes $g$ a continuous function, so the fundamental theorem of calculus gives $f^{\prime }=g$. The displayed estimate proves uniform convergence wherever $|x-c|$ has a common finite bound, and therefore proves locally uniform convergence on any interval.

For example, on $[0,2]$ take $f_n(x)=3+1/n+x^2/2+x/n$. Its derivative is $x+1/n$, with supremum error $1/n$ from $g(x)=x$, and $f_n(0)\to 3$. The reconstructed uniform limit is $f(x)=3+x^2/2$; the exact supremum error is $3/n$, equal to the bound above. Without an anchor, the constant functions $f_n(x)=n$ have zero derivatives but no finite pointwise limit. On the real line, $f_n(x)=x/n$ has convergent anchored values and uniformly convergent derivatives, but its supremum error from zero is infinite. The bounded-distance condition has a real role.

A function $f:E\to \mathbb{R}$ is the uniform limit of a sequence of functions $(f_n)$ whenEquivalently the supremum error ${\sup }_{x\in E}|f_n(x)-f(x)|$ tends to zero. The same index works for all inputs, including inputs chosen differently for each $n$. A uniform limit of continuous functions is continuous, by approximating both endpoint values with one common approximant.

For $f_n(x)=x^n$ on $[0,a]$, $0<a<1$, the zero function is a uniform limit because the supremum error is $a^n$. On $[0,1]$, the pointwise limit instead equals zero below one and one at one. Its supremum error remains one for every $n$, so there is no uniform limit there. The open supremum below the endpoint need not be attained to obstruct uniform convergence.