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A sequence of functions $f_n:A\to \mathbb{R}$ converges uniformly to $f$ when, for every $\epsilon >0$, there is an $N$ such thatfor every $x\in A$. Equivalently, $\Vert f_n-f{\Vert }_{\infty }\to 0$ in the supremum norm.

Yes. Fix $x_0\in A$ and $\epsilon >0$. Uniform convergence gives $N$ such that $|f_N(x)-f(x)|<\epsilon /3$ for every $x\in A$. Since the continuous function $f_N$ is continuous at $x_0$, there is a $\delta >0$ such that $|x-x_0|<\delta$ implies $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality then givesThis proves the uniform limit theorem: $f$ is continuous.

The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.

Apply Arzela-Ascoli theorem on $[-m,m]$. Pointwise boundedness and equicontinuity give a subsequence converging uniformly there. Successively extract subsequences for $m=1,2,\dots$, and take the diagonal subsequence. It converges uniformly on every closed bounded interval. Its limits on nested intervals agree, so they define $f$ on the real numbers; the locally uniform limit of continuous functions is continuous.

This is always true. Given $\epsilon >0$, choose $N_A,N_B$ such thaton $A$ for $n\ge N_A$ and on $B$ for $n\ge N_B$. For $n\ge \max (N_A,N_B)$ the same inequality holds at every point of $X=A\cup B$. Hence $f_n\to f$ uniformly on $X$.

A sequence $f_n:S\to \mathbb{R}$ converges uniformly to $f$ if, for every $\epsilon >0$, there is $N$ such thatfor every $x\in S$ and every $n\ge N$. A map $h:M\to N$ is uniformly continuous if, for every $\epsilon >0$, there is $\delta >0$ such thatfor all $x,y\in M$.

If $f\in C_0({\mathbb{R}}^d)$, choose a ball outside which $|f|<1$. On the closed ball, continuity gives boundedness, so $f$ is bounded everywhere.

Now let $(f_n)$ be Cauchy in the uniform metric. For each $x$, $(f_n(x))$ is Cauchy in $\mathbb{R}$; let its limit be $f(x)$. Passing to the pointwise limit in the uniform Cauchy estimate shows that $f_n\to f$ uniformly. Hence $f$ is continuous. Given $\epsilon >0$, choose $n$ with $\Vert f-f_n{\Vert }_{\infty }<\epsilon /2$, and then choose $K$ so that $|f_n(x)|<\epsilon /2$ for $\Vert x\Vert >K$. It follows that $f$ also vanishes at infinity. Thus $C_0({\mathbb{R}}^d)$ is complete, as in completeness of continuous functions vanishing at infinity.

Every $f\in C_0({\mathbb{R}}^d)$ is uniformly continuous. Given $\epsilon >0$, choose $R$ so that $|f(x)|<\epsilon /2$ outside the ball of radius $R$. On the compact ball of radius $R+1$, $f$ is uniformly continuous; choose the corresponding $\delta \le 1$. If two points at distance below $\delta$ are not both in that ball, then both lie outside the ball of radius $R$, and their function values differ by less than $\epsilon$.

For the final sequence, continuity of $\epsilon$ at zero gives, for each fixed $x$,Thus pointwise convergence is compulsory. Uniform convergence need not hold: with $\epsilon (t)=t^2$,which is unbounded as a function of $x$ for every fixed $n$.

Under the additional bound $\epsilon (t)\le M|t|$, however,for $x\ne 0$, and the difference is zero at $x=0$. Hence convergence is uniform, by the uniform square-root perturbation under linear growth estimate. The pointwise answer remains yes.

Fix $x\in X$. By assumption there is a neighbourhood $U$ of $x$ on which $f_n\to f$ with uniform convergence. Each restriction $f_n{|}_U$ is continuous, and a uniform limit of continuous functions is continuous. Thus $f{|}_U$ is continuous, in particular at $x$. Since $x$ was arbitrary, $f$ is continuous on $X$.

Now let $K\subseteq X$ be compact and let $\epsilon >0$. For each $x\in K$, choose a neighbourhood $U_x$ on which convergence is uniform. The sets $U_x$ cover $K$, so compactness gives a finite subcoverFor each $j$, choose $N_j$ such thatTaking $N={\max }_j{N}_j$, every $y\in K$ belongs to one of these finitely many neighbourhoods, and thereforeThus $f_n\to f$ uniformly on every compact subset. This proves the local uniform convergence on compact subsets principle.

The polynomial form of Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ such that

Statement (i) is false. Take the entire function $f(z)=e^{-z}$. It is bounded on the closed first quadrant $E$. If polynomials $P_n$ converged uniformly to $f$ on $E$, then each sufficiently late $P_n$ would be bounded on the positive real axis. A polynomial bounded on that ray must be constant. No sequence of constants converges uniformly to $e^{-x}$ on $[0,\infty )$, since its values at zero and near infinity differ by one.

Statement (iv) is true. Apply the Weierstrass approximation theorem separately to the real and imaginary parts of the continuous function $f$ on the compact interval $[0,1]$. Combining the two real polynomial sequences gives complex polynomials converging uniformly to $f$ on $I$.

The inequalities implyThus the two metrics induce the same open sets and are equivalent.

The converse fails. On $\mathbb{R}$, letThe metrics are equivalent because $\arctan :\mathbb{R}\to (-\pi /2,\pi /2)$ is a homeomorphism, butso no positive lower comparison constant exists. This is an equivalent metrics need not be bi-Lipschitz equivalent example.

Equivalent metrics on the codomain also need not give the same uniform convergence. Take $X=\mathbb{N}$, $Y=\mathbb{R}$, and use the two metrics above. DefineThenso $f_n\to f$ uniformly for $d_2$, whereasso convergence is not uniform for $d_1$. This is the equivalent codomain metrics need not preserve uniform convergence phenomenon.

Apply the residue theorem to $f(z)/(z-w{)}^{n+1}$; its residue at $w$ is $f^{(n)}(w)/n!$, which proves Cauchy's derivative formula.

For a compact subdisc choose a slightly larger contour still inside $D(a,R)$. Cauchy's derivative formula bounds the supremum of $g_k^{(n)}-g^{(n)}$ on the smaller disc by a fixed constant times the supremum of $g_k-g$ on the contour. Local uniform convergence therefore passes to every derivative.

On $|z-a|=\epsilon$, $g$ has a positive minimum modulus. For large $k$, $|g_k-g|<|g|$ there. Rouché's theorem then says $g_k$ and $g$ have the same number of zeros inside, namely $m$, counting multiplicity.

Uniform convergence means: for every $\epsilon >0$ there is $N$ such that $n\ge N$ implies $|f_n(x)-f(x)|<\epsilon$ for every $x\in E$.

If $|f_n(x)|\le M_n$ and $\sum M_n<\infty$, then every tail is bounded uniformly by the corresponding numerical tail. The uniform Cauchy criterion therefore proves uniform convergence.

For each $a>1$, $n^{-x}\le n^{-a}$ on $[a,\infty )$ and the p-series converges. The M-test gives locally uniform convergence there, so the sum $\zeta$ is continuous on $(1,\infty )$.

Runge theorem says that if $K\subset \mathbb{C}$ is compact and $\mathbb{C}\setminus K$ is connected, every function holomorphic near $K$ is uniformly approximable on $K$ by polynomials.

Choose a compact exhaustion $K_j$ of the slit disc $\Delta \setminus L^{+}$, rounding the two sides of the slit so that $0\notin K_j$ and $\mathbb{C}\setminus K_j$ is connected. Runge applied to $1/z$ gives a polynomial $p_j$ with ${\sup }_{K_j}|p_j-1/z|<1/j$. Every compact subset of the slit disc lies in all sufficiently large $K_j$, proving locally uniform convergence.

Replace the two $G$ quantiles by conditional bootstrap quantiles ${\hat{q}}_u$ in the interval from (a). Uniform convergence implies ${\hat{q}}_u\to q_u$ in probability at continuity points, so Slutsky and continuity of $G$ give coverage tending to $1-\alpha$.

For $2\pi$-periodic continuous functions, convergence of positive linear operators on continuous functions to the identity on $1,\sin x,\cos x$ implies uniform convergence on every function. Indeed $1-\cos (t-x)$ is nonnegative and has just one zero on the circle, so the compact Korovkin test space criterion applies.

For a Schwartz function on the real numbers, its periodization is a smooth periodic function of period one. Rapid decay gives uniform convergence on $[0,1]$ of the series and each differentiated series. Under the Fourier transform convention $\hat{f}(\xi )=\int f(x)e^{-2\pi ix\xi }dx$, its $m$th Fourier coefficient is $\hat{f}(m)$: integrate termwise over one period and combine the translated intervals. The rapidly decreasing coefficients give a Fourier series with uniform convergence. Evaluation at zero proves the Poisson summation formula.

A pointwise limit assigns to each point the limit of the sequence of function values there. The convergence rate may depend on that point. Uniform convergence additionally bounds the error simultaneously over the whole domain.

The space of test functions has the locally convex inductive-limit topology of the spaces of smooth functions supported in a fixed compact set $K$. Each ${\mathcal{D}}_K$ uses seminorms $p_m(\phi )={\max }_{|\alpha |\le m}\Vert {\partial }^{\alpha }\phi {\Vert }_{\infty }$. Sequential convergence means a common compact support and uniform convergence of every derivative. A linear functional on this space is a distribution exactly when its restriction to each ${\mathcal{D}}_K$ satisfies a finite order of a distribution estimate; the order and constant may depend on $K$.