Function sequences require tracking both the input variable and the sequence index. The central distinction is whether convergence is pointwise or uniform.
Pointwise and uniform convergence
Pointwise convergence fixes the input first; uniform convergence controls all inputs at once.
Definition. For maps \(f_n:X\to(Y,d)\), \[ \begin{aligned} f_n\to f\text{ pointwise} &\quad\Longleftrightarrow\quad \forall x\ \forall\varepsilon>0\ \exists N=N(x,\varepsilon)\ \forall n\ge N:\ d(f_n(x),f(x))<\varepsilon,\\ f_n\to f\text{ uniformly} &\quad\Longleftrightarrow\quad \forall\varepsilon>0\ \exists N=N(\varepsilon)\ \forall n\ge N\ \forall x:\ d(f_n(x),f(x))<\varepsilon. \end{aligned} \] When the supremum is defined, uniform convergence is equivalent to \(\sup_{x\in X}d(f_n(x),f(x))\to0\).
Proposition. Uniform convergence implies pointwise convergence, but the converse need not hold.
Proof. Uniform convergence immediately bounds each fixed \(x\). For the converse, \(f_n(x)=x^n\) on \([0,1]\) converges pointwise to \(0\) on \([0,1)\) and \(1\) at \(1\), so the limit is not continuous despite all \(f_n\) being continuous; uniform convergence of continuous functions would preserve continuity.
Uniform Cauchy criterion
Uniform convergence can be tested without knowing the limit when the target is complete.
Theorem. If \(Y\) is complete, then \(f_n:X\to Y\) converges uniformly if and only if for every \(\varepsilon>0\) there is \(N\) such that \(m,n\ge N\Rightarrow \sup_x d(f_m(x),f_n(x))\lt \varepsilon\).
Proof. Uniform convergence implies the uniform Cauchy estimate by the triangle inequality. Conversely, for each \(x\), \(f_n(x)\) is Cauchy in \(Y\), so define \(f(x)=\lim_n f_n(x)\). Letting \(m\to\infty\) in the uniform Cauchy inequality gives \(\sup_x d(f_n(x),f(x))\le\varepsilon\) for large \(n\).
Corollary. If \(Y\) is Banach, then the space \(C_b(X;Y)\) of bounded continuous maps, equipped with \[ \|f\|_\infty=\sup_{x\in X}\|f(x)\|_Y, \] is Banach.
Proof. A sup-norm Cauchy sequence is uniformly Cauchy, hence converges uniformly to a \(Y\)-valued function by completeness of \(Y\). The uniform limit theorem makes the limit continuous, and a uniform limit of bounded maps is bounded.
Uniform convergence of series of functions
Theorem (Weierstrass \(M\)-test). If \(|f_n(x)|\le M_n\) for every \(x\in X\) and \(\sum_nM_n<\infty\), then \(\sum_nf_n\) converges uniformly and absolutely.
Proof. For \(m>n\), \[ \sup_{x\in X}\left|\sum_{k=n+1}^{m}f_k(x)\right| \le\sum_{k=n+1}^{m}M_k. \] The scalar series has Cauchy tails, so the function-series partial sums are uniformly Cauchy. Completeness of \(\mathbb R\) or \(\mathbb C\) and the preceding criterion give uniform convergence. The same estimate applied to \(|f_n|\) gives absolute convergence at every \(x\).
Continuity under uniform limits
Uniform convergence permits epsilon estimates to be split into three controlled pieces.
Theorem. If \(f_n:X\to Y\) are continuous and \(f_n\to f\) uniformly, then \(f\) is continuous.
Proof. Fix \(x\) and \(\varepsilon>0\). Choose \(n\) with \(d(f_n(z),f(z))\lt \varepsilon/3\) for all \(z\). Then choose \(\delta\) from continuity of \(f_n\) at \(x\). The triangle inequality gives \(d(f(y),f(x))\lt \varepsilon\) for \(d(x,y)\lt \delta\).
Interchanging limits and integrals
Uniform convergence is strong enough for Riemann integration on compact intervals.
Proposition. If \(f_n\) are Riemann integrable on \([a,b]\) and \(f_n\to f\) uniformly, then \(f\) is Riemann integrable and \(\int_a^b f_n\,\mathrm{d}x\to\int_a^b f\,\mathrm{d}x\).
Proof. Uniform limits of Riemann integrable functions are Riemann integrable because upper and lower sums of \(f\) are controlled by those of \(f_n\) plus \(2(b-a)\|f-f_n\|_\infty\). The integral difference satisfies \(|\int(f_n-f)\,\mathrm{d}x|\le(b-a)\|f_n-f\|_\infty\).
Differentiation requires stronger hypotheses
Uniform convergence of derivatives, not just functions, controls derivatives of limits.
Theorem. If \(f_n\in C^1([a,b])\), \(f_n(x_0)\) converges at one point, and \(f_n'\to g\) uniformly, then \(f_n\) converges uniformly to a differentiable \(f\) with \(f'=g\).
Proof. By the fundamental theorem, \(f_n(x)=f_n(x_0)+\int_{x_0}^x f_n'(t)\,\mathrm{d}t\). The right side is uniformly Cauchy because the constants converge and the derivatives converge uniformly. Passing to the limit gives \(f(x)=c+\int_{x_0}^x g(t)\,\mathrm{d}t\), hence \(f'=g\).
Example (why derivative control is needed). The smooth functions \[ f_n(x)=\sqrt{x^2+\frac1n},\qquad x\in[-1,1], \] converge uniformly to \(|x|\), because \(0\le f_n(x)-|x|\le n^{-1/2}\). Their limit is not differentiable at \(0\). Correspondingly, \[ f_n'(x)=\frac{x}{\sqrt{x^2+1/n}} \] does not converge uniformly on any interval containing \(0\).