Uniform convergence strengthens pointwise convergence by making the index $n_0$ independent of the point of the domain. The focus keyword uniform convergence is central because this condition preserves continuity under limits. Supremum estimates and uniform Cauchy criteria are the main technical tools for working with uniform limits. :::definition[Uniform Convergence] Let $f_n:X\to Y$ and $f:X\to Y$, where $(Y,d_Y)$ is a metric space. The sequence $\{f_n\}$ converges uniformly to $f$ if for every $\epsilon>0$, there exists $n_0$ such that $d_Y(f_n(x),f(x))<\epsilon$ for all $n\geq n_0$ and all $x\in X$. ::: :::theorem[Supremum Criterion for Uniform Convergence] For real-valued or metric-valued functions, $f_n\to f$ uniformly on $X$ if and only if $\sup_{x\in X}d_Y(f_n(x),f(x))\to0$. ::: :::proof Given the sequence of functions. To prove the equivalence. Uniform convergence says that for each $\epsilon>0$, the inequality $d_Y(f_n(x),f(x))<\epsilon$ holds for every $x$ and all large $n$. This is exactly the assertion that the supremum of these distances is eventually less than $\epsilon$. Hence, the supremum tends to $0$, and the converse is the same statement read backwards. $\square$ ::: :::example[Arctangent Sequence] Let $f_n(x)=\tan^{-1}(nx)$ for $x\geq0$. The pointwise limit is $0$ at $0$ and $\pi/2$ for $x>0$. The convergence is uniform on $[a,\infty)$ for every $a>0$, but not uniform on $[0,\infty)$ because the limit is discontinuous at $0$. ::: :::theorem[Uniform Limit of Continuous Functions] Let $f_n:X\to Y$ be continuous functions between metric spaces. If $f_n\to f$ uniformly on $X$, then $f$ is continuous on $X$. ::: :::proof Given that each $f_n$ is continuous and $f_n\to f$ uniformly. To prove that $f$ is continuous. Let $a\in X$ and $\epsilon>0$. Choose $N$ such that $d_Y(f_N(x),f(x))<\epsilon/3$ for all $x$. Choose $\delta>0$ so that $d_Y(f_N(x),f_N(a))<\epsilon/3$ when $d_X(x,a)<\delta$. Then the triangle inequality gives $d_Y(f(x),f(a))<\epsilon$. Hence, $f$ is continuous. $\square$ ::: :::theorem[Cauchy Criterion for Uniform Convergence] Let $\{f_n\}$ be functions from $X$ into a complete metric space $Y$. Then $\{f_n\}$ converges uniformly on $X$ if and only if for every $\epsilon>0$, there exists $n_0$ such that $d_Y(f_m(x),f_n(x))<\epsilon$ for all $m,n\geq n_0$ and all $x\in X$. ::: :::proof Given that $Y$ is complete. To prove the criterion. Uniform convergence implies the uniform Cauchy condition by the triangle inequality with the limit function. Conversely, the uniform Cauchy condition makes each pointwise sequence $\{f_n(x)\}$ Cauchy in $Y$, so it converges. Define $f(x)=\lim f_n(x)$. Passing to the limit in the uniform Cauchy inequality gives uniform convergence to $f$. $\square$ ::: :::solved-problem Let $f_n$ be continuous on $X$, $f_n\to f$ uniformly, and $x_n\to x$. Prove that $f_n(x_n)\to f(x)$. ::: :::solution By the uniform limit theorem, $f$ is continuous. Given $\epsilon>0$, for large $n$ we have $|f_n(y)-f(y)|<\epsilon/2$ for all $y$, and also $|f(x_n)-f(x)|<\epsilon/2$. Hence $|f_n(x_n)-f(x)|\leq|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<\epsilon$. ::: :::exercise 1. Define uniform convergence without supremum notation. 2. Why is completeness of $Y$ needed in the Cauchy criterion? 3. Give pointwise convergence with discontinuous limit. ::: :::answer 1. One $n_0$ must work for all $x$. 2. Pointwise Cauchy sequences must converge in $Y$. 3. $x^n$ on $[0,1]$. :::