For \(f\colon I\to\RR\) on an interval \(I\): \[ f \text{ differentiable with bounded } f' \implies f \text{ Lipschitz} \implies f \text{ absolutely continuous} \implies f \text{ uniformly continuous} \implies f \text{ continuous}. \]
If \(\abs{f'}\leq C\), the mean value theorem gives \(\abs{f(x)-f(y)}\leq C\abs{x-y}\). If \(f\) is \(C\)-Lipschitz and \(\sum_i\abs{b_i-a_i}<\varepsilon/C\) for disjoint intervals \((a_i,b_i)\), the triangle inequality gives \(\sum_i\abs{f(b_i)-f(a_i)}<\varepsilon\). Absolute continuity applied to a single interval is uniform continuity, and uniform continuity implies continuity by definition.
Discontinuity sets
Function discontinuous on the rationals (Tag L-75UZY)
There is a function \(f\colon\RR\to\RR\) whose set of points of discontinuity is exactly \(\QQ\).
Enumerate \(\QQ=\theset{q_1,q_2,\ldots}\) and put \(f(x)\coloneqq\sum_{n\,:\,q_n<x}2^{-n}\). The function \(f\) is nondecreasing and bounded by \(1\). Fix \(x\in\RR\) and put \(g(x)\coloneqq\sum_{n\,:\,q_n\le x}2^{-n}\). For \(\delta>0\), \[ f(x)-f(x-\delta)=\sum_{n\,:\,x-\delta\le q_n<x}2^{-n}, \qquad f(x+\delta)-g(x)=\sum_{n\,:\,x<q_n<x+\delta}2^{-n}. \] Given \(N\), for \(\delta\) small enough neither index set contains any \(n\le N\), so both sums are at most \(\sum_{n>N}2^{-n}=2^{-N}\). Hence \(\lim_{t\to x^-}f(t)=f(x)\) and \(\lim_{t\to x^+}f(t)=g(x)\). If \(x=q_n\), then \(g(x)-f(x)=2^{-n}>0\); if \(x\notin\QQ\), then \(g(x)=f(x)\). So \(f\) is discontinuous at every rational and continuous at every irrational.
Let \((r_n)_{n\geq1}\) enumerate \(\QQ\), and let \(f(r_n) \coloneqq \frac 1 n\) and \(f(x)\coloneqq0\) for \(x\notin\QQ\). For every \(x_0\) and \(N\), some neighborhood of \(x_0\) omits \(r_1,\ldots,r_N\) (other than possibly \(x_0\) itself), so \(\lim_{x\to x_0}f(x) = 0\). Hence \(f\) is continuous exactly where \(f(x_0)=0\), that is, on \(\RR\setminus\QQ\).
No function $\RR\to\RR$ is discontinuous exactly on the irrationals (Tag PR-F5V7D)
There is no function \(f\colon\RR\to\RR\) whose set of points of discontinuity is \(\RR\setminus\QQ\).
Properties of the set of discontinuities $D_f$ of a function (Tag FF-AVBFU)
Let \((X, d)\) be a metric space, let \(f\colon X\to\RR\), and let \(D_f\subseteq X\) be the set of points at which \(f\) is not continuous. For \(x\in X\), the oscillation of \(f\) at \(x\) is \[ \omega_f(x)\coloneqq\inf_{\delta>0}\,\sup\theset{\abs{f(y) - f(z)} \suchthat y, z\in B_\delta(x)}\in[0,\infty]. \] Then \(f\) is continuous at \(x\) if and only if \(\omega_f(x) = 0\), each set \(\theset{x\in X\suchthat \omega_f(x)\geq 1/k}\) is closed, and \[ D_f = \bigcup_{k\geq 1}\theset{x\in X\suchthat \omega_f(x)\geq 1/k}. \] In particular, \(D_f\) is an $F_\sigma$ set.
If \(f\) is continuous at \(x\) and \(\varepsilon>0\), choose \(\delta>0\) with \(\abs{f(y)-f(x)}<\varepsilon\) for \(y\in B_\delta(x)\); then \(\abs{f(y)-f(z)}<2\varepsilon\) for \(y,z\in B_\delta(x)\), so \(\omega_f(x)\leq 2\varepsilon\). Conversely, if \(\omega_f(x) = 0\) and \(\varepsilon>0\), some \(\delta>0\) has \(\abs{f(y)-f(x)}<\varepsilon\) for all \(y\in B_\delta(x)\). This proves the characterization of continuity, and hence the formula for \(D_f\).
For closedness, suppose \(\omega_f(x)<1/k\), and choose \(\delta>0\) with \(s\coloneqq\sup\theset{\abs{f(y)-f(z)}\suchthat y,z\in B_\delta(x)}<1/k\). For \(x'\in B_\delta(x)\), the ball \(B_{\delta - d(x,x')}(x')\) is contained in \(B_\delta(x)\), so \(\omega_f(x')\leq s<1/k\). Hence \(\theset{\omega_f<1/k}\) is open.
Let \(X\) be a metric space, and let \(f_n\colon X\to\RR\) be continuous functions converging pointwise to \(f\colon X\to\RR\). Then \(D_f\) is meager in \(X\).
(Lebesgue criterion)
Let \(a<b\) and let \(f\colon[a,b]\to\RR\) be bounded. Then \(f\) is Riemann integrable on \([a,b]\) if and only if \(D_f\) has Lebesgue measure \(0\).
Let \(I\subseteq\RR\) be an interval and let \(f\colon I\to\RR\) be monotone. Then \(D_f\) is countable, and \(f\) is differentiable at almost every point of \(I\).
We prove countability; differentiability almost everywhere is Lebesgue’s differentiation theorem for monotone functions. Assume \(f\) is nondecreasing. At each interior point \(x\) of \(I\) the one-sided limits \(f(x^-)\leq f(x)\leq f(x^+)\) exist, and \(f\) is discontinuous at \(x\) if and only if \(f(x^-)<f(x^+)\). For interior points \(x<x'\) of \(D_f\), \(f(x^+)\leq f(x'^-)\), so the open intervals \((f(x^-), f(x^+))\) for interior \(x\in D_f\) are pairwise disjoint. Choosing a rational number in each gives an injection of the interior points of \(D_f\) into \(\QQ\), and \(I\) has at most two endpoints.
Discontinuity sets of real-valued functions (Tag FF-5EBCJ)
Let \((X, d)\) be a metric space, let \(f\colon X\to\RR\), and let \(D_f\subseteq X\) be the set of points at which \(f\) is not continuous. For \(x\in X\), the oscillation of \(f\) at \(x\) is \[ \omega_f(x)\coloneqq\inf_{\delta>0}\,\sup\theset{\abs{f(y) - f(z)} \suchthat y, z\in B_\delta(x)}\in[0,\infty]. \] Then \(f\) is continuous at \(x\) if and only if \(\omega_f(x) = 0\), each set \(\theset{x\in X\suchthat \omega_f(x)\geq 1/k}\) is closed, and \[ D_f = \bigcup_{k\geq 1}\theset{x\in X\suchthat \omega_f(x)\geq 1/k}. \] In particular, \(D_f\) is an $F_\sigma$ set.
If \(f\) is continuous at \(x\) and \(\varepsilon>0\), choose \(\delta>0\) with \(\abs{f(y)-f(x)}<\varepsilon\) for \(y\in B_\delta(x)\); then \(\abs{f(y)-f(z)}<2\varepsilon\) for \(y,z\in B_\delta(x)\), so \(\omega_f(x)\leq 2\varepsilon\). Conversely, if \(\omega_f(x) = 0\) and \(\varepsilon>0\), some \(\delta>0\) has \(\abs{f(y)-f(x)}<\varepsilon\) for all \(y\in B_\delta(x)\). This proves the characterization of continuity, and hence the formula for \(D_f\).
For closedness, suppose \(\omega_f(x)<1/k\), and choose \(\delta>0\) with \(s\coloneqq\sup\theset{\abs{f(y)-f(z)}\suchthat y,z\in B_\delta(x)}<1/k\). For \(x'\in B_\delta(x)\), the ball \(B_{\delta - d(x,x')}(x')\) is contained in \(B_\delta(x)\), so \(\omega_f(x')\leq s<1/k\). Hence \(\theset{\omega_f<1/k}\) is open.
Let \(X\) be a metric space, and let \(f_n\colon X\to\RR\) be continuous functions converging pointwise to \(f\colon X\to\RR\). Then \(D_f\) is meager in \(X\).
(Lebesgue criterion)
Let \(a<b\) and let \(f\colon[a,b]\to\RR\) be bounded. Then \(f\) is Riemann integrable on \([a,b]\) if and only if \(D_f\) has Lebesgue measure \(0\).
Let \(I\subseteq\RR\) be an interval and let \(f\colon I\to\RR\) be monotone. Then \(D_f\) is countable, and \(f\) is differentiable at almost every point of \(I\).
We prove countability; differentiability almost everywhere is Lebesgue’s differentiation theorem for monotone functions. Assume \(f\) is nondecreasing. At each interior point \(x\) of \(I\) the one-sided limits \(f(x^-)\leq f(x)\leq f(x^+)\) exist, and \(f\) is discontinuous at \(x\) if and only if \(f(x^-)<f(x^+)\). For interior points \(x<x'\) of \(D_f\), \(f(x^+)\leq f(x'^-)\), so the open intervals \((f(x^-), f(x^+))\) for interior \(x\in D_f\) are pairwise disjoint. Choosing a rational number in each gives an injection of the interior points of \(D_f\) into \(\QQ\), and \(I\) has at most two endpoints.
The discontinuity set \(D_f = \bigcup_{k\geq1}\theset{\omega_f\geq 1/k}\) is an \(F_\sigma\) set, since \(f\) is continuous at \(x\) if and only if \(\omega_f(x) = 0\) and each \(\theset{\omega_f\geq1/k}\) is closed. If \(D_f = \RR\setminus\QQ\), then \(\RR\setminus\QQ = \bigcup_k F_k\) with \(F_k\) closed; each \(F_k\) contains no rational, so has empty interior, and \(\QQ = \bigcup_{q\in\QQ}\theset q\) together with the \(F_k\) writes \(\RR\) as a countable union of closed sets with empty interior, contradicting the Baire category theorem.
Uniform continuity
A function \(f\colon\RR\to\RR\) is uniformly continuous if and only if \(\norm{f(\wait+y)-f}_\infty \to 0\) as \(y \to 0\).
Lipschitz maps are uniformly continuous (Tag PR-SX6NO)
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces and let \(f\colon X\to Y\) be Lipschitz: there is \(C>0\) with \(d_Y(f(x),f(y))\leq C\,d_X(x,y)\) for all \(x,y\in X\). Then \(f\) is uniformly continuous.
Given \(\varepsilon>0\), take \(\delta\coloneqq\varepsilon/C\). If \(d_X(x,y)<\delta\), then \[ d_Y(f(x),f(y)) \leq C\,d_X(x,y) < C\delta = \varepsilon . \]
Heine--Cantor theorem (Tag T-O4UD3)
Let \(X\) and \(Y\) be metric spaces with \(X\) compact. Every continuous function \(f\colon X\to Y\) is uniformly continuous [@Mun00]; [@Rud76].
Let \(U\subseteq\RR^n\) and let \(f\colon U\to \RR\) be continuous. For every compact \(K \subseteq U\), the restriction \(\ro{f}{K}\) is uniformly continuous.
Let \(\varepsilon>0\). For each \(x\in X\) choose \(\delta_x>0\) with \(d_Y(f(x),f(y))<\varepsilon/2\) whenever \(d(x,y)<\delta_x\). The balls \(B_{\delta_x/2}(x)\) cover \(X\); choose a finite subcover by balls centered at \(x_1,\ldots,x_m\) and let \(\delta\coloneqq\min_i\delta_{x_i}/2\). If \(d(x,y)<\delta\), then \(d(x,x_i)<\delta_{x_i}/2\) for some \(i\), and \(d(y,x_i)<\delta+\delta_{x_i}/2\leq\delta_{x_i}\), so \(d_Y(f(x),f(y))\leq d_Y(f(x),f(x_i))+d_Y(f(x_i),f(y))<\varepsilon\).
Equicontinuity
Equicontinuity at a point (Tag D-2CCDB)
Let \(X\) be a topological space, let \(x\in X\), and let \(\mathcal F\subseteq C(X)\) be a family of continuous functions \(X\to\CC\). The family \(\mathcal F\) is equicontinuous at \(x\) if \[ \forall \varepsilon > 0 \quad \exists \text{ an open set } U \ni x \quad \forall y\in U \quad \forall f\in \mathcal{F}: \quad \abs{f(y) - f(x)} < \varepsilon. \]
Equicontinuous families (Tag FD-TGBYP)
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces, let \(x_0\in X\), and let \(\mathcal{F}\) be a family of functions \(X\to Y\).
The family \(\mathcal{F}\) is equicontinuous at \(x_0\) if for every \(\varepsilon > 0\) there exists \(\delta=\delta(\varepsilon, x_0)>0\) such that for all \(x\in X\) and all \(f\in\mathcal{F}\), \[ x\in B_\delta(x_0) \implies f(x) \in B_\varepsilon(f(x_0)). \]
The family \(\mathcal{F}\) is uniformly equicontinuous if for every \(\varepsilon>0\) there exists \(\delta=\delta(\varepsilon)>0\) such that for all \(x_1,x_2\in X\) and all \(f\in\mathcal{F}\), \[ x_1\in B_\delta(x_2) \implies f(x_1) \in B_\varepsilon(f(x_2)). \]
Equicontinuous and uniformly equicontinuous families (Tag FD-XVMEE)
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces, let \(\mcf\) be a family of functions \(X\to Y\), and let \(x_0\in X\).
The family \(\mcf\) is equicontinuous at \(x_0\) if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(d_Y(f(x), f(x_0)) < \varepsilon\) for all \(f\in\mcf\) and all \(x\in X\) with \(d_X(x, x_0) < \delta\).
The family \(\mcf\) is uniformly equicontinuous if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(d_Y(f(x_1), f(x_2)) < \varepsilon\) for all \(f\in\mcf\) and all \(x_1, x_2\in X\) with \(d_X(x_1, x_2) < \delta\).
In equicontinuity at \(x_0\) the number \(\delta\) depends on \(\varepsilon\) and \(x_0\) but not on \(f\in\mcf\); in uniform equicontinuity one \(\delta\) depending only on \(\varepsilon\) serves every \(f\in\mcf\) and every pair of points of \(X\).
Equicontinuity and uniform equicontinuity (Tag FF-63IWC)
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces, let \(\mcf\) be a family of functions \(X\to Y\), and let \(x_0\in X\).
\(\mcf\) is equicontinuous at $x_0$ if and only if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(d_Y(f(x), f(x_0)) < \varepsilon\) for all \(f\in\mcf\) and all \(x\in X\) with \(d_X(x, x_0) < \delta\).
\(\mcf\) is uniformly equicontinuous if and only if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(d_Y(f(x_1), f(x_2)) < \varepsilon\) for all \(f\in\mcf\) and all \(x_1, x_2\in X\) with \(d_X(x_1, x_2) < \delta\).
Arzelà--Ascoli theorem (Tag FF-XZGIY)
Let \(K\) be a compact metric space, let \(C(K)\) be the space of continuous functions \(K\to\CC\) with the norm \(\norm{f}_\infty\coloneqq\sup_{x\in K}\abs{f(x)}\), and let \(\mcf\subseteq C(K)\). Then the closure of \(\mcf\) in \(C(K)\) is compact if and only if \(\mcf\) is uniformly bounded, that is, \(\sup_{f\in\mcf}\norm{f}_\infty<\infty\), and equicontinuous at every point of \(K\).
In particular, every sequence \((f_n)\) in \(C(K)\) that is uniformly bounded and equicontinuous at every point of \(K\) has a subsequence that converges uniformly on \(K\).
Since \(K\) is compact, a family \(\mcf\subseteq C(K)\) that is equicontinuous at every point of \(K\) is uniformly equicontinuous.