Proof of A Nondecreasing Differentiable Majorant with Vanishing Derivative at Zero for a Little-o Modulus

Each result cited is universally quantified over the data in its own statement. For a real $t$ write $t^{+}=\max(t,0)$. Series of real numbers, their convergence and their sums are those of Series of Real Numbers §convergent; linearity of sums is Elementary Properties of Series of Real Numbers §linearity and comparison of sums is Elementary Properties of Series of Real Numbers §order; the comparison test is Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §comparison, the geometric series with ratio $\tfrac12$ (sum $1$, and $n$-th tail $(\tfrac12)^{n}$, with $(\tfrac12)^{n}\to0$) is Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric, and the tail bound for a dominated series of nonnegative terms is Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §tail-bound. Differentiability at a point of the interval $I=\mathbb{R}$, every point of which is an interior point, is that of Single-Variable Calculus on an Interval §derivative.

Step 1: the ramp. Define $P,Q,\sigma:\mathbb{R}\to\mathbb{R}$ by

$$P(t)=\tfrac12\,(t^{+})^{2},\qquad Q(t)=P(t)-P(t-1),\qquad \sigma(t)=t^{+}-(t-1)^{+}.$$

Sorting $t$ into the three ranges $t\le0$, $0\le t\le1$ and $t\ge1$ gives $Q(t)=0$ and $\sigma(t)=0$ for $t\le0$; $Q(t)=t^{2}/2$ and $\sigma(t)=t$ for $0\le t\le1$; and $Q(t)=t-\tfrac12$ and $\sigma(t)=1$ for $t\ge1$. In particular $0\le\sigma(t)\le1$ for every $t$.

We claim that $|P(t+u)-P(t)-t^{+}u|\le u^{2}/2$ for all real $t,u$. If $t\ge0$ and $t+u\ge0$ the left side is $|(t+u)^{2}/2-t^{2}/2-tu|=u^{2}/2$. If $t\ge0$ and $t+u<0$, then $0\le t<-u$ and the quantity inside the absolute value is $-t^{2}/2-tu=t(-u-t/2)$, which is nonnegative, and $u^{2}/2-t(-u-t/2)=(u+t)^{2}/2\ge0$. If $t<0$ and $t+u\ge0$, the quantity is $(t+u)^{2}/2$ and $0\le t+u<u$, so it lies between $0$ and $u^{2}/2$. If $t<0$ and $t+u<0$ it is $0$. Applying the claim at $t$ and at $t-1$ (note $(t-1)^{+}$ is the coefficient there) and using the triangle inequality gives

$$|Q(t+u)-Q(t)-\sigma(t)u|\le u^{2}\qquad\text{for all real }t,u. \tag{1}$$

Consequently $Q$ is differentiable at every $t\in\mathbb{R}$ with $Q'(t)=\sigma(t)$, in the sense of Single-Variable Calculus on an Interval §derivative: given $\varepsilon>0$ take $\delta=\varepsilon$; for $0<|u|<\delta$, dividing (1) by $|u|$ gives $|(Q(t+u)-Q(t))/u-\sigma(t)|\le|u|<\varepsilon$. By (1) and $|\sigma(t)|\le1$ we also have $|Q(t+u)-Q(t)|\le|u|+u^{2}$, so $Q$ is continuous at every point (given $\varepsilon>0$, $\delta=\min(1,\varepsilon/2)$ works: for $|u|<\delta$ we have $u^{2}\le|u|$, so $|u|+u^{2}\le2|u|<\varepsilon$), and hence continuous on every closed interval $[a,b]$ in the sense of The Real Line: Standing Notation and Background for Calculus §continuity. Let $a<b$. The restriction of $Q$ to $[a,b]$ is continuous on $[a,b]$, and it is differentiable at every point $x$ of $(a,b)$ with derivative $\sigma(x)$, because passing to a subinterval changes neither differentiability nor the derivative, as recorded in Single-Variable Calculus on an Interval §derivative. The mean value theorem Mean Value Theorem on a Closed Real Interval gives $c\in(a,b)$ with $Q(b)-Q(a)=\sigma(c)(b-a)$, and since $0\le\sigma(c)\le1$,

$$0\le Q(b)-Q(a)\le b-a\qquad\text{whenever }a\le b, \tag{2}$$

the case $a=b$ being trivial.

Step 2: rescaled ramps. For a real $\rho>0$ define $q_{\rho},\sigma_{\rho}:\mathbb{R}\to\mathbb{R}$ by $q_{\rho}(s)=\rho\,Q(s/\rho-1)$ and $\sigma_{\rho}(s)=\sigma(s/\rho-1)$. From Step 1 we read off the following, for all real $s,t,u$.

(a) If $s\le\rho$ then $s/\rho-1\le0$, so $q_{\rho}(s)=0$ and $\sigma_{\rho}(s)=0$; and always $0\le\sigma_{\rho}(s)\le1$.

(b) If $s\le t$ then $0\le q_{\rho}(t)-q_{\rho}(s)\le t-s$, by (2) applied to $s/\rho-1\le t/\rho-1$ and multiplied by $\rho$.

(c) $0\le q_{\rho}(s)\le s^{+}$. Indeed, if $s\le\rho$ this is (a); if $s>\rho$ then (b) and $q_{\rho}(\rho)=0$ give $0\le q_{\rho}(s)\le s-\rho\le s$.

(d) $q_{\rho}(s)\ge s-\tfrac32\rho$. Indeed, if $s\ge2\rho$ then $s/\rho-1\ge1$ and $q_{\rho}(s)=\rho(s/\rho-1-\tfrac12)=s-\tfrac32\rho$; if $s<2\rho$ then (b) gives $q_{\rho}(2\rho)-q_{\rho}(s)\le2\rho-s$, and $q_{\rho}(2\rho)=\rho/2$, so $q_{\rho}(s)\ge s-\tfrac32\rho$.

(e) Writing $e_{\rho}(s,u)=q_{\rho}(s+u)-q_{\rho}(s)-\sigma_{\rho}(s)u$, we have $|e_{\rho}(s,u)|\le u^{2}/\rho$ and $|e_{\rho}(s,u)|\le2|u|$. The first bound is (1) at the point $s/\rho-1$ with increment $u/\rho$, multiplied by $\rho$; the second follows from $|q_{\rho}(s+u)-q_{\rho}(s)|\le|u|$, which is (b), and $|\sigma_{\rho}(s)u|\le|u|$, which is (a).

Step 3: choice of constants. The constants are chosen in the following order. First, for each $k\in\mathbb{N}$ the hypothesis on $m$, applied with $\eta=2^{-k-3}$, gives a real $r'_{k}>0$ with $m(s)\le2^{-k-3}s$ for every $s\in[0,r]$ with $s<r'_{k}$; fix such a choice for every $k$. Such a sequence $(r'_{k})_{k\in\mathbb{N}}$ exists by Axiom of Dependent Choice, applied to the set $S$ of pairs $(k,\rho)$ with $k\in\mathbb{N}$ and $\rho>0$ real such that $m(s)\le2^{-k-3}s$ for every $s\in[0,r]$ with $s<\rho$, to the relation consisting of the pairs $\bigl((k,\rho),(k+1,\rho')\bigr)$ of elements of $S$, and to a starting element $(1,\rho_{1})\in S$: the hypothesis on $m$ (with $\eta=2^{-4}$) shows that $S$ has such an element, and (with $\eta=2^{-k-4}$) that every $(k,\rho)\in S$ is related to some element of $S$; the sequence $(a_{k})_{k\in\mathbb{N}}$ given by the axiom has first coordinate $k$ at every index $k$, by induction on $k$, and we let $r'_{k}$ be the second coordinate of $a_{k}$. Second, define $r_{1}=\min(r,r'_{1})$ and recursively $r_{k+1}=\min(r_{k}/2,r'_{k+1})$ for $k\in\mathbb{N}$. Then for every $k$: $r_{k}>0$, $r_{k+1}\le r_{k}/2$, $r_{k}\le r$, $r_{k}\le r'_{k}$, the sequence $(r_{k})$ is nonincreasing, and by induction $r_{k}\le(\tfrac12)^{k-1}r_{1}$. Hence

$$m(s)\le2^{-k-3}s\qquad\text{for every }k\in\mathbb{N}\text{ and every }s\in[0,r]\text{ with }s<r_{k}. \tag{3}$$

Third, since $B\ge m(0)=0$, the number $A=4B/r_{1}$ satisfies $A\ge0$.

We record a location fact. Let $0<s<r_{1}$. Since $(\tfrac12)^{n}\to0$ by the geometric series clause and $s/r_{1}>0$, there is $n'\in\mathbb{N}$ with $(\tfrac12)^{n'}<s/r_{1}$; with $n=n'+1$ we get $(\tfrac12)^{n-1}<s/r_{1}$, that is $(\tfrac12)^{n-1}r_{1}<s$, hence $r_{n}\le(\tfrac12)^{n-1}r_{1}<s$; so the set of $n\in\mathbb{N}$ with $r_{n}\le s$ is nonempty, and its least element $n_{0}$ satisfies $n_{0}\ge2$ because $r_{1}>s$. With $k=n_{0}-1\in\mathbb{N}$ we get

$$r_{k+1}\le s<r_{k}. \tag{4}$$

Step 4: definition of $h$. Write $q_{j}=q_{r_{j}}$, $\sigma_{j}=\sigma_{r_{j}}$ and $e_{j}=e_{r_{j}}$ for $j\in\mathbb{N}$. For $s\in\mathbb{R}$ we have $0\le2^{-j}q_{j}(s)\le s^{+}2^{-j}$ by (c), so by the comparison test the series $\sum_{j=1}^{\infty}2^{-j}q_{j}(s)$ converges, the dominating series $\sum_{j}s^{+}2^{-j}$ converging (with sum $s^{+}$) by the geometric series and linearity of sums, with sum in $[0,s^{+}]$ by comparison of sums. Likewise $0\le2^{-j}\sigma_{j}(s)\le2^{-j}$ by (a), so $\sum_{j=1}^{\infty}2^{-j}\sigma_{j}(s)$ converges with sum in $[0,1]$. Define $H,D:\mathbb{R}\to\mathbb{R}$ by

$$H(s)=A\,q_{2}(s)+\sum_{j=1}^{\infty}2^{-j}q_{j}(s),\qquad D(s)=A\,\sigma_{2}(s)+\sum_{j=1}^{\infty}2^{-j}\sigma_{j}(s).$$

Then $H(s)\ge0$ and $0\le D(s)\le A+1$ for every $s$. If $s<0$ then $s<r_{j}$ for every $j$, so every $q_{j}(s)$ vanishes by (a) and $H(s)=0$. Let $h:[0,\infty)\to[0,\infty)$ be the restriction of $H$ to $[0,\infty)$. Then the function $\bar h$ of the statement coincides with $H$ on all of $\mathbb{R}$.

Clause (Monotone). Let $0\le s\le t$. By (b), $A q_{2}(s)\le Aq_{2}(t)$ (as $A\ge0$) and $2^{-j}q_{j}(s)\le2^{-j}q_{j}(t)$ for every $j$, so comparison of sums gives $h(s)\le h(t)$; thus $h$ is nondecreasing on $[0,\infty)$ in the sense of Monotone Real Function §nondecreasing. Since $0\le r_{j}$ for every $j$, (a) gives $q_{j}(0)=0$ for every $j$, so $h(0)=0$.

Clause (Differentiable). Fix $s\in\mathbb{R}$ and $\varepsilon>0$. Choose, in this order, $J\in\mathbb{N}$ with $(\tfrac12)^{J}<\varepsilon/8$, which exists since $(\tfrac12)^{n}\to0$; then the positive number $C_{J}=A/r_{2}+\sum_{j=1}^{J}2^{-j}/r_{j}+1$; then $\delta=\varepsilon/(2C_{J})$. Let $u$ be real with $0<|u|<\delta$. The three series defining $H(s+u)$, $H(s)$ and $D(s)$ converge, so by linearity of sums

$$H(s+u)-H(s)-D(s)u=A\,e_{2}(s,u)+\sum_{j=1}^{\infty}2^{-j}e_{j}(s,u).$$

By (e), $0\le2^{-j}|e_{j}(s,u)|\le2|u|\,2^{-j}$, so by the comparison test the series $\sum_{j}2^{-j}|e_{j}(s,u)|$ converges, the dominating series $\sum_{j}2|u|\,2^{-j}$ converging by the geometric series and linearity of sums, and comparison of sums applied to $-2^{-j}|e_{j}(s,u)|\le2^{-j}e_{j}(s,u)\le2^{-j}|e_{j}(s,u)|$ (the left series converging by linearity) gives $\bigl|\sum_{j}2^{-j}e_{j}(s,u)\bigr|\le\sum_{j}2^{-j}|e_{j}(s,u)|$. The tail bound, applied with $\mu_{j}=2^{-j}$, $w_{j}=|e_{j}(s,u)|$, $M=2|u|$ and $n=J$, together with the geometric series tail $(\tfrac12)^{J}$, gives

$$\sum_{j=1}^{\infty}2^{-j}|e_{j}(s,u)|\le\sum_{j=1}^{J}2^{-j}|e_{j}(s,u)|+2|u|\,(\tfrac12)^{J}\le u^{2}\sum_{j=1}^{J}\frac{2^{-j}}{r_{j}}+2|u|\,(\tfrac12)^{J},$$

where the second inequality is the first bound of (e). Using the first bound of (e) also for $e_{2}$,

$$|H(s+u)-H(s)-D(s)u|\le u^{2}C_{J}+2|u|\,(\tfrac12)^{J}<|u|\,\frac{\varepsilon}{2}+|u|\,\frac{\varepsilon}{4}<\varepsilon|u|,$$

since $|u|C_{J}<\delta C_{J}=\varepsilon/2$. Dividing by $|u|$, $|(H(s+u)-H(s))/u-D(s)|<\varepsilon$. As $s$ is an interior point of $I=\mathbb{R}$ and $\varepsilon$ was arbitrary, $\bar h=H$ is differentiable at $s$ in the sense of Single-Variable Calculus on an Interval §derivative, with $\bar h'(s)=D(s)$, the derivative being unique as recorded there. Hence $h'(s)=D(s)$ for every $s\ge0$.

Clause (Derivative). Since $0<r_{j}$ for every $j$, (a) gives $\sigma_{j}(0)=0$ for every $j$, hence $h'(0)=D(0)=0$. With $K=A+1$ we have $|h'(s)|=D(s)\le K$ for every $s\ge0$, by Step 4. Let $\eta>0$. Choose $k\in\mathbb{N}$ with $k\ge2$ and $(\tfrac12)^{k}<\eta$, and put $r''=r_{k}>0$. Let $0\le s<r''$. For $j\le k$ we have $s<r_{k}\le r_{j}$, so $\sigma_{j}(s)=0$ by (a); in particular $\sigma_{2}(s)=0$ as $k\ge2$. The tail bound with $\mu_{j}=2^{-j}$, $w_{j}=\sigma_{j}(s)\in[0,1]$, $M=1$ and $n=k$, whose partial sum up to $k$ vanishes, then gives $0\le h'(s)=\sum_{j=1}^{\infty}2^{-j}\sigma_{j}(s)\le(\tfrac12)^{k}<\eta$.

Clause (Majorant). Let $s\in[0,r]$. If $s=0$ then $h(0)=0=m(0)$. If $s\ge r_{1}$, then, as the series part of $H(s)$ is nonnegative, (d) and $r_{2}\le r_{1}/2$ give

$$h(s)\ge A\,q_{2}(s)\ge A\bigl(s-\tfrac32r_{2}\bigr)\ge A\bigl(r_{1}-\tfrac34r_{1}\bigr)=\frac{Ar_{1}}{4}=B\ge m(s).$$

If $0<s<r_{1}$, let $k$ be as in (4). For $j\ge k+3$ we have $r_{j}\le r_{k+3}\le r_{k+1}/4\le s/4$, so (d) gives $q_{j}(s)\ge s-\tfrac38s\ge s/2$. Define $b_{j}=0$ for $j\le k+2$ and $b_{j}=2^{-j}s/2$ for $j\ge k+3$. Then $0\le b_{j}\le2^{-j}q_{j}(s)$ for every $j$, by (c) for $j\le k+2$. The sequence $(b_{j})$ differs from $(\tfrac{s}{2}2^{-j})$ by the finitely supported sequence equal to $\tfrac{s}{2}2^{-j}$ for $j\le k+2$ and $0$ afterwards, so by linearity of sums and the geometric series tail,

$$\sum_{j=1}^{\infty}b_{j}=\frac{s}{2}\Bigl(\sum_{j=1}^{\infty}2^{-j}-\sum_{j=1}^{k+2}2^{-j}\Bigr)=\frac{s}{2}\Bigl(\frac12\Bigr)^{k+2}=2^{-k-3}s.$$

Comparison of sums and $Aq_{2}(s)\ge0$ then give $h(s)\ge\sum_{j}2^{-j}q_{j}(s)\ge2^{-k-3}s$. Since $s\in[0,r]$ and $s<r_{k}$, estimate (3) gives $m(s)\le2^{-k-3}s\le h(s)$. This proves $h(s)\ge m(s)$ for every $s\in[0,r]$ and completes the proof.