Three function classes recur throughout analysis, and each turns out to control the others: monotone functions always have one-sided limits, functions of bounded variation are exactly differences of monotone functions, and convex functions inherit continuity from the same monotonicity argument. This chapter builds all three on top of the functional-limit and continuity machinery already established.
Monotone Functions
For \(f:I\to\mathbb{R}\) on an interval \(I\subseteq\mathbb{R}\), \(f\) is
- monotone increasing if \(x_1<x_2\implies f(x_1)\leq f(x_2)\) for all \(x_1,x_2\in I\);
- monotone decreasing if \(x_1<x_2\implies f(x_1)\geq f(x_2)\) for all \(x_1,x_2\in I\);
- strictly monotone increasing (decreasing) if the inequality on \(f(x_1),f(x_2)\) above is strict as well, \(f(x_1)<f(x_2)\) (\(f(x_1)>f(x_2)\)).
\(f(x)=x^3\) is strictly monotone increasing on all of \(\mathbb{R}\). \(f(x)=x^2\) is not monotone on \(\mathbb{R}\) at all — it decreases then increases — but it is strictly monotone decreasing on \((-\infty,0]\) and strictly monotone increasing on \([0,\infty)\); restricting the domain is exactly what turns a non-monotone function into two monotone pieces.
A strictly monotone function is automatically injective: \(x_1\neq x_2\) forces \(x_1<x_2\) or \(x_2<x_1\) (trichotomy on \(\mathbb{R}\)), and either way strict monotonicity gives \(f(x_1)\neq f(x_2)\). This holds with no continuity assumption at all; the converse — that a continuous injective function on an interval must already be strictly monotone — is the harder direction, proved in the chapter on continuity.
One-Sided Limits
For \(f:(\alpha,\beta)\to\mathbb{R}\) and \(x_0\in(\alpha,\beta)\), the left-hand and right-hand limits
\[ f(x_0^-):=\lim_{x\to x_0,\,x<x_0}f(x), \qquad f(x_0^+):=\lim_{x\to x_0,\,x>x_0}f(x) \]
are the ordinary functional limit restricted to sequences approaching \(x_0\) from below, respectively above — already mentioned in passing when one-sided limits were first defined. The two-sided limit \(\lim_{x\to x_0}f(x)\) exists exactly when both one-sided limits exist and agree. A point \(x_0\) is a jump discontinuity if both \(f(x_0^-)\) and \(f(x_0^+)\) exist but \(f(x_0^-)\neq f(x_0^+)\).
Limits of Monotone Functions
Let \(f\) be monotone increasing on an interval \(I\) with endpoints \(a<b\) (either may be infinite). Then at every interior point \(y\in I\), both one-sided limits exist, with
\[ f(y^-)=\sup_{x<y}f(x)\leq f(y)\leq\inf_{x>y}f(x)=f(y^+), \]
and at the endpoints, \(f(a^+)=\inf_{x>a}f(x)\) and \(f(b^-)=\sup_{x<b}f(x)\) (possibly \(\pm\infty\) if \(f\) is unbounded). The mirror statement, with \(\sup\)/\(\inf\) swapped, holds for monotone decreasing \(f\).
Proof. Let \(S:=\sup_{x<y}f(x)\), which exists in \(\mathbb{R}\cup\{\infty\}\) since \(f\) restricted to \(\{x\in I:x<y\}\) is bounded above by \(f(y)\) whenever such \(x\) exist. Given \(\epsilon>0\) (if \(S\) is finite), there is \(x_\epsilon<y\) with \(f(x_\epsilon)>S-\epsilon\); by monotonicity, \(S-\epsilon<f(x_\epsilon)\leq f(x)\leq S\) for every \(x\in(x_\epsilon,y)\), so \(f(x)\to S\) as \(x\uparrow y\), i.e. \(f(y^-)=S\leq f(y)\). If \(S=\infty\), the same argument with \(f(x_\epsilon)>\) any bound shows \(f(x)\to\infty\). The right-hand limit and the endpoint cases follow identically, working with \(\inf_{x>y}f(x)\) instead.
A monotone function has at most countably many discontinuities, and every one is a jump. Since \(f(y^-)\leq f(y)\leq f(y^+)\) always exists for monotone \(f\), \(f\) is continuous at \(y\) exactly when \(f(y^-)=f(y^+)\); every discontinuity is therefore a jump, with a nonempty open interval \(\bigl(f(y^-),f(y^+)\bigr)\) attached to it. For \(y_1<y_2\) both discontinuities, monotonicity forces \(f(y_1^+)\leq f(y_2^-)\), so the intervals \(\bigl(f(y_1^-),f(y_1^+)\bigr)\) and \(\bigl(f(y_2^-),f(y_2^+)\bigr)\) are disjoint. Choosing one rational number inside each such interval gives an injection from the set of discontinuities into \(\mathbb{Q}\), so that set is countable.
A few standard limits illustrate the endpoint case: \(e^x\to\infty\) as \(x\to\infty\); \(\ln x\to-\infty\) as \(x\downarrow0\); \(1/x\to+\infty\) as \(x\downarrow0\) and \(1/x\to-\infty\) as \(x\uparrow0\); and \(\arctan x\to\pi/2\) as \(x\to\infty\), \(\arctan x\to-\pi/2\) as \(x\to-\infty\), matching the general theorem's \(\inf\)/\(\sup\) description of each endpoint.
Functions of Bounded Variation
For \(f:[a,b]\to\mathbb{R}\) and a partition \(a=t_0<t_1<\cdots<t_n=b\), the variation of \(f\) along that partition is \(\sum_{i=1}^n|f(t_i)-f(t_{i-1})|\). The total variation is the supremum over all partitions,
\[ V_a^b(f):=\sup_{\text{partitions}}\sum_{i=1}^n|f(t_i)-f(t_{i-1})|, \]
and \(f\) has bounded variation on \([a,b]\) (is BV) if \(V_a^b(f)<\infty\).
If \(f\) is monotone, every partition sum telescopes to exactly \(|f(b)-f(a)|\) (each term has the same sign), so \(f\) is BV with \(V_a^b(f)=|f(b)-f(a)|\). If \(f\) is Lipschitz with constant \(L\), every partition sum is bounded by \(L\sum(t_i-t_{i-1})=L(b-a)\), so \(f\) is BV with \(V_a^b(f)\leq L(b-a)\).
Continuity alone does not imply bounded variation. Take \(f(x):=x\cos(1/x)\) for \(x\in(0,1/\pi]\) and \(f(0):=0\); \(f\) is continuous everywhere on \([0,1/\pi]\) (at \(0\), \(|f(x)|\leq|x|\to0\) by the squeeze theorem). For \(k=1,\ldots,N\), let \(x_k:=1/(k\pi)\), so \(\cos(1/x_k)=\cos(k\pi)=(-1)^k\) and \(f(x_k)=(-1)^k/(k\pi)\). Using the partition \(0,x_N,x_{N-1},\ldots,x_1\) of \([0,1/\pi]\),
\[ \sum_{k=1}^{N-1}|f(x_k)-f(x_{k+1})| =\sum_{k=1}^{N-1}\left(\frac{1}{k\pi}+\frac{1}{(k+1)\pi}\right) \geq\frac{1}{\pi}\sum_{k=2}^{N}\frac{1}{k}, \]
since consecutive \(f(x_k)\) alternate in sign. The right side is a partial sum of the (divergent) harmonic series, so it grows without bound as \(N\to\infty\): \(V_0^{1/\pi}(f)=\infty\), even though \(f\) is continuous on a compact interval.
The total variation is additive and linear enough to make BV functions on \([a,b]\) a vector space: for \(f,g\) BV and \(\lambda,\mu\in\mathbb{R}\), the triangle inequality applied term-by-term to any partition gives \(V_a^b(\lambda f+\mu g)\leq|\lambda|V_a^b(f)+|\mu|V_a^b(g)<\infty\).
The Jordan Decomposition
\(f:[a,b]\to\mathbb{R}\) has bounded variation if and only if \(f=g-h\) for two monotone increasing functions \(g,h\) on \([a,b]\).
Proof. (\(\Leftarrow\)) Both \(g\) and \(h\) are BV (monotone functions are BV, shown above), and BV functions form a vector space, so \(f=g-h\) is BV. (\(\Rightarrow\)) Given \(f\) BV, define
\[ g(x):=V_a^x(f),\qquad h(x):=g(x)-f(x), \]
with \(V_a^a(f):=0\). \(g\) is monotone increasing: for \(x<y\), appending the single extra point \(y\) to any partition of \([a,x]\) shows \(V_a^y(f)\geq V_a^x(f)+|f(y)-f(x)|\geq V_a^x(f)\), i.e. \(g(y)-g(x)\geq|f(y)-f(x)|\geq0\); moreover this same inequality gives \(g(y)-g(x)\geq f(y)-f(x)\), i.e. \(h(y)-h(x)=\bigl(g(y)-f(y)\bigr)-\bigl(g(x)-f(x)\bigr)\geq0\), so \(h\) is monotone increasing too. Since \(f=g-h\) by construction, this is the required decomposition.
Combined with the countable-discontinuity theorem for monotone functions, this gives an immediate corollary: any BV function has at most countably many discontinuities, and every one is a jump (a discontinuity of \(f=g-h\) must be a discontinuity of \(g\) or \(h\), each of which has only countably many, all jumps).
Convex and Concave Functions
\(f:I\to\mathbb{R}\) on an interval \(I\) is convex if
\[ f\bigl(tx+(1-t)y\bigr)\leq tf(x)+(1-t)f(y)\quad\text{for all }x,y\in I,\ t\in[0,1], \]
strictly convex if the inequality is strict whenever \(x\neq y\) and \(t\in(0,1)\), and (strictly) concave if \(-f\) is (strictly) convex. Geometrically, the chord between any two points of the graph lies on or above the graph. \(e^x\) is strictly convex on \(\mathbb{R}\), \(\ln x\) is strictly concave on \((0,\infty)\), and \(x^n\) is strictly convex on \(\mathbb{R}\) for even \(n\) — these follow directly once a second-derivative convexity criterion is available, but the difference-quotient characterization below already pins down convexity without needing derivatives at all.
The Difference-Quotient Characterization
\(f\) is convex on \(I\) if and only if, for all \(x<y<z\) in \(I\),
\[ \frac{f(y)-f(x)}{y-x}\ \leq\ \frac{f(z)-f(x)}{z-x}\ \leq\ \frac{f(z)-f(y)}{z-y}, \]
i.e. the slope of every secant is squeezed between the slopes of the secants to its left and right; replacing \(\leq\) by \(<\) throughout characterizes strict convexity the same way.
Proof. Write \(y=tx+(1-t)z\) with \(t:=\frac{z-y}{z-x}\in(0,1)\), so \(1-t=\frac{y-x}{z-x}\). (\(\Rightarrow\)) Convexity gives \(f(y)\leq tf(x)+(1-t)f(z)\). Subtracting \(f(x)\) from both sides,
\[ f(y)-f(x)\leq(t-1)f(x)+(1-t)f(z)=(1-t)\bigl(f(z)-f(x)\bigr)=\frac{y-x}{z-x}\bigl(f(z)-f(x)\bigr), \]
and dividing by \(y-x>0\) gives the first inequality. Subtracting \(f(z)\) instead,
\[ f(y)-f(z)\leq t\bigl(f(x)-f(z)\bigr)=\frac{z-y}{z-x}\bigl(f(x)-f(z)\bigr), \]
so \(f(z)-f(y)\geq\frac{z-y}{z-x}\bigl(f(z)-f(x)\bigr)\), and dividing by \(z-y>0\) gives the second inequality. (\(\Leftarrow\)) Reversing these two algebraic steps recovers \(f(y)\leq tf(x)+(1-t)f(z)\) from the two difference-quotient inequalities, for every \(x<y<z\) in \(I\) — which is exactly convexity, since every \(y=tx+(1-t)z\) with \(t\in(0,1)\) and \(x\neq z\) arises this way.
Convex Functions Are Continuous
A convex function on an open interval \(I\) is continuous on \(I\).
Proof. Fix \(x_0\in I\) and consider the difference-quotient function \(\varphi(y):=\frac{f(y)-f(x_0)}{y-x_0}\) for \(y\in I\setminus\{x_0\}\). The difference-quotient characterization above says exactly that \(\varphi\) is monotone increasing in \(y\) (for \(y_1<y_2\) both different from \(x_0\), whichever of \(x_0,y_1,y_2\) is in the middle, the chain of inequalities gives \(\varphi(y_1)\leq\varphi(y_2)\)). By the monotone-limits theorem, \(\varphi(x_0^-)\) and \(\varphi(x_0^+)\) both exist (as elements of \(\mathbb{R}\cup\{\pm\infty\}\)), and since \(I\) is open, \(x_0\) is an interior point with \(\varphi\) defined immediately on both sides.
It remains to show these one-sided limits are finite (ruling out \(\pm\infty\)) and that \(\lim_{y\to x_0}f(y)=f(x_0)\). Pick any \(a<x_0<b\) in \(I\); by the chain of inequalities with \(x=a,y=x_0,z=b\) rearranged, \(\varphi(y)\) for \(y\) near \(x_0\) on either side is squeezed between the fixed finite numbers \(\varphi(a)\) and \(\varphi(b)\), so both one-sided limits of \(\varphi\) are finite. Writing \(f(y)-f(x_0)=\varphi(y)(y-x_0)\), the right side is a product of a function bounded near \(x_0\) and a factor \((y-x_0)\to0\), so \(f(y)-f(x_0)\to0\), i.e. \(f(y)\to f(x_0)\) as \(y\to x_0\). This is exactly continuity of \(f\) at \(x_0\).
On a closed sub-interval \([\alpha,\beta]\subset I\) strictly inside \(I\), the argument above bounds every difference quotient \(\varphi\) between the (finite) slopes to a fixed point just outside \([\alpha,\beta]\), so \(f\) is in fact Lipschitz, not merely continuous, on \([\alpha,\beta]\) — convexity can fail to be Lipschitz only as the interval's own endpoints are approached (\(\sqrt{-\ln x}\)-type blow-ups of the slope near an open endpoint are the standard obstruction).