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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.

Statement (ii) is true. LetEach $K_n$ is compact and has connected complement. By the polynomial Runge theorem, choose a polynomial $P_n$ withEvery fixed $z\in E$ belongs to $K_n$ for all sufficiently large $n$, so $P_n(z)\to f(z)$. Thus the convergence is pointwise on $E$.

A function $f:(X,d_X)\to (Y,d_Y)$ is uniformly continuous if for every $\epsilon >0$ there is $\delta >0$ such thatfor all $x,x^{\prime }\in X$.

Suppose $f_n\to f$ uniformly and every $f_n$ is uniformly continuous. Given $\epsilon >0$, choose $N$ such thatfor every $x$. Uniform continuity of $f_N$ supplies $\delta >0$ such that $d_X(x,x^{\prime })<\delta$ impliesThe triangle inequality then gives $d_Y(f(x),f(x^{\prime }))<\epsilon$. Thus the uniform limit theorem for uniformly continuous functions proves that $f$ is uniformly continuous.

Pointwise convergence is insufficient. On $[0,1]$, the uniformly continuous functions $f_n(x)=x^n$ converge pointwise towhich is discontinuous and therefore not uniformly continuous.

Continuity of $\phi$ at zero and $\phi (0)=1$ allow $u>0$ so small thatPointwise convergence and dominated convergence make the corresponding expression for ${\phi }_n$ less than $\epsilon$ for all sufficiently large $n$. Part (ii) then boundsThere are only finitely many remaining $n$, so enlarge $M\ge 2/u$ until each of their tails is also at most $\epsilon$. This proves uniform tightness.

Lévy's continuity theorem states that if characteristic functions ${\varphi }_n$ converge pointwise to a function $\varphi$ continuous at zero, then $\varphi$ is a characteristic function and the corresponding laws converge weakly to its law. Conversely, convergence in distribution implies pointwise convergence of characteristic functions.

Every continuous $f:\mathbb{R}\to \mathbb{R}$ can be approximated with pointwise convergence by uniformly continuous functions. Set $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one and $f{|}_{[-n,n]}$ is uniformly continuous by the Heine-Cantor theorem, so $f_n$ is uniformly continuous on the whole line. On every fixed bounded interval, $f_n=f$ eventually.