Proof of The One-Dimensional Slice Bound for a Continuously Differentiable Periodic Function

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named below. We use silently that the order of $\mathbb{R}$ is reflexive, transitive and antisymmetric, that equal real numbers satisfy $\le$ in both directions, and that $x<y$ entails $x\le y$.

Let $x\in\mathbb{R}^{n}$ be fixed until the end of the proof and abbreviate $x[s]=x[i{:}s]$ for $s\in\mathbb{R}$. Write

$$\theta_{x}(s)=\Theta_{i}(x[s])\qquad(s\in\mathbb{R}),$$

let $\tilde{\theta}_{x}=\widetilde{(\Theta_{i})}_{i,x}$ be the zero extension of $\theta_{x}$ taking the value $\theta_{x}(s)$ at $s\in[0,1]$ and the value $0$ at every $s\in\mathbb{R}$ with $s\notin[0,1]$, and put

$$F=(P_{i}\Theta_{i})(x)=\int_{\mathbb{R}}\tilde{\theta}_{x}\,d\lambda .$$

By The Slice Average of a Continuous Periodic Function (i), applied to $\Theta_{i}\in C_{\mathrm{per}}$, the map $\tilde{\theta}_{x}$ is measurable with respect to $\mathcal{B}(\mathbb{R})$ and integrable with respect to $\lambda$, and $0\le F$, the map $\Theta_{i}$ taking nonnegative values. The assertion to be proved is $|u(x)|\le F$.

Since $u\in C^{1}_{\mathrm{per}}$, the map $u$ is of class $C^{1}$ on $\mathbb{R}^{n}$ and $\mathbb{Z}^{n}$-periodic, and $u\in C_{\mathrm{per}}$ and $\partial_{i}u\in C_{\mathrm{per}}$: a map of class $C^{1}$ is continuous by claim 3 of Euclidean Space Is Open in Itself, periodicity is the same condition for the classes concerned by Lattice-Periodic Functions and the Periodic Function Classes (iii), and $\partial_{i}u\in C_{\mathrm{per}}$ by Elementary Properties of Lattice-Periodic Functions (ii). By The Slice Average of a Continuous Periodic Function (v) the maps $|u|$ and $|\partial_{i}u|$ also lie in $C_{\mathrm{per}}$. Let $p,q:\mathbb{R}\to\mathbb{R}$ denote the maps

$$p(s)=u(x[s]),\qquad q(s)=\partial_{i}u(x[s])\qquad(s\in\mathbb{R}),$$

so that $\theta_{x}(s)=|p(s)|+|q(s)|$. By The Slice Average of a Continuous Periodic Function (i), applied in turn to the members $u$, $\partial_{i}u$, $|u|$ and $|\partial_{i}u|$ of $C_{\mathrm{per}}$, the maps $p$, $q$, $|p|$ and $|q|$ are continuous from $(\mathbb{R},d_{\mathbb{R}})$ to $(\mathbb{R},d_{\mathbb{R}})$, and the zero extension off $[0,1]$ of each of them is measurable with respect to $\mathcal{B}(\mathbb{R})$ and integrable with respect to $\lambda$; the zero extensions of $|p|$ and $|q|$ take nonnegative values, so their integrals are nonnegative by the monotonicity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, the map with constant value $0$ being integrable with integral $0$ by that same claim applied with both coefficients $0$.

(P1) The slice is differentiable with derivative $q$. Let $a,b\in\mathbb{R}$ with $a<b$ and let $t\in\mathbb{R}$ with $a<t<b$. Then the restriction of $p$ to $[a,b]$ is differentiable at $t$ with derivative $q(t)$.

Indeed, apply Slice Function and the Partial Derivative with the open set $\mathbb{R}^{n}$, which is open in itself by claim 1 of Euclidean Space Is Open in Itself, the map $u$, the point $a^{\ast}=x[t]$ whose $i$th coordinate is $t$, and the index $i$. Claim 1 of that lemma provides a real $\rho>0$ with $a^{\ast}[s]\in\mathbb{R}^{n}$ for $|s-t|<\rho$, and claim 2 provides the interval $I=\{s\in\mathbb{R}:t-\rho<s\text{ and }s<t+\rho\}$, with $t$ an interior point of $I$, and the slice function $g:I\to\mathbb{R}$, $g(s)=u(a^{\ast}[s])$. Since $a^{\ast}=x[t]$ has the same $k$th coordinate as $x$ for every $k\ne i$, we have $a^{\ast}[s]=x[s]$ and hence $g(s)=p(s)$ for every $s\in I$. The partial derivative of $u$ with respect to the $i$th variable exists at $a^{\ast}$, since $u$ is of class $C^{1}$ on $\mathbb{R}^{n}$ and that definition requires exactly the existence of the first-order partial derivatives at every point, so claim 2 of that lemma gives that $g$ is differentiable at $t$ with $g'(t)=\partial_{i}u(a^{\ast})=q(t)$.

It remains to transfer this to the restriction $P$ of $p$ to $[a,b]$, for which $t$ is an interior point since $a<t<b$. Let $\varepsilon$ be a real number with $0<\varepsilon$. By Derivative of a Real Function on an Interval applied to $g$ there is a real $\delta>0$ such that every $h\in\mathbb{R}$ with $0<|h|<\delta$ and $t+h\in I$ satisfies

$$\Bigl|\frac{g(t+h)-g(t)}{h}-q(t)\Bigr|<\varepsilon .$$

Put $\delta'=\min\{\delta,\rho\}$, a positive real by claim 9 of Elementary Order Arithmetic in an Ordered Field. Let $h\in\mathbb{R}$ with $0<|h|<\delta'$ and $t+h\in[a,b]$. Then $|h|<\rho$, so $t+h\in I$, and $|h|<\delta$; moreover $P(t+h)=p(t+h)=g(t+h)$ and $P(t)=g(t)$. Hence the displayed inequality holds with $P$ in place of $g$. As $\varepsilon$ was arbitrary, $P$ is differentiable at $t$ with derivative $q(t)$, by Derivative of a Real Function on an Interval.

(P2) The increment bound. Put

$$M=\int_{\mathbb{R}}\tilde{\kappa}\,d\lambda ,$$

where $\tilde{\kappa}$ is the map taking the value $|q(s)|$ at $s\in[0,1]$ and $0$ elsewhere; it is the zero extension off $[0,1]$ of $|q|$, so it is measurable and integrable, and $0\le M$, by the opening paragraphs of this proof. Then

$$|p(t)-p(s)|\le M\qquad\text{for all }s,t\in[0,1].$$

To see this, let $s,t\in[0,1]$. If $s=t$ the left-hand side is $0$ and $0\le M$, as just noted. Otherwise let $a$ be the smaller and $b$ the larger of $s$ and $t$, so $a<b$ and $[a,b]\subseteq[0,1]$. The restriction of $q$ to $[a,b]$ is continuous, since for a point of $[a,b]$ and a real $\varepsilon>0$ any $\delta$ witnessing the continuity of $q$ at that point also witnesses that of the restriction, the requirement being imposed at fewer points; hence Riemann integrable on $[a,b]$ by Continuous Functions on Compact Intervals are Riemann Integrable. The restriction $P$ of $p$ to $[a,b]$ is continuous by the same argument, and by (P1) it is differentiable at every $t'$ with $a<t'<b$, with derivative $q(t')$. So Fundamental Theorem of Calculus, Part II: Evaluating an Integral by an Antiderivative applies and gives

$$\int_{a}^{b}q(r)\,dr=P(b)-P(a)=p(b)-p(a).$$

By claim 3 of Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval, applied to the restriction of $q$ to $[a,b]$, the Riemann integral on the left equals $\int_{\mathbb{R}}\tilde{q}_{[a,b]}\,d\lambda$, where $\tilde{q}_{[a,b]}$ takes the value $q(r)$ on $[a,b]$ and $0$ elsewhere. Hence, using claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion,

$$|p(b)-p(a)|=\Bigl|\int_{\mathbb{R}}\tilde{q}_{[a,b]}\,d\lambda\Bigr|\le\int_{\mathbb{R}}\bigl|\tilde{q}_{[a,b]}\bigr|\,d\lambda .$$

Now $\bigl|\tilde{q}_{[a,b]}(r)\bigr|\le\tilde{\kappa}(r)$ for every $r\in\mathbb{R}$: for $r\in[a,b]$ both sides equal $|q(r)|$, since $[a,b]\subseteq[0,1]$; for $r\notin[a,b]$ the left-hand side is $|0|=0$ and the right-hand side is nonnegative. The map $\bigl|\tilde{q}_{[a,b]}\bigr|$ is measurable by claim 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps and integrable, being dominated by the integrable $\tilde{\kappa}$: this follows from claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion together with the criterion in Integrable Functions and Their Lebesgue Integral. So the monotonicity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives $\int_{\mathbb{R}}\bigl|\tilde{q}_{[a,b]}\bigr|\,d\lambda\le M$. Since $|p(t)-p(s)|=|p(b)-p(a)|$ by claim 2 of Properties of the Absolute Value in an Ordered Field, the bound follows.

(P3) Averaging over one period. Let $t\in[0,1]$. Then $|p(t)|\le F$.

Let $s\in[0,1]$. By claim 5 of Properties of the Absolute Value in an Ordered Field and (P2),

$$|p(t)|=|p(s)+(p(t)-p(s))|\le|p(s)|+|p(t)-p(s)|\le|p(s)|+M .$$

Let $\tilde{\pi}$ be the map taking the value $|p(s)|$ at $s\in[0,1]$ and $0$ elsewhere, measurable and integrable as above, and let $c$ denote the real number $|p(t)|$. Let $\mathbf{1}_{[0,1]}$ be the indicator of $[0,1]$, which lies in $\mathcal{B}(\mathbb{R})$ and satisfies $\lambda([0,1])=1$ by claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness. It is measurable by claim 1 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, and $\int_{\mathbb{R}}\mathbf{1}_{[0,1]}\,d\lambda=\lambda([0,1])=1$ by The Integral of an Indicator Function is the Measure of the Set. Since $\mathbf{1}_{[0,1]}$ takes only the values $0$ and $1$, which are nonnegative, $|\mathbf{1}_{[0,1]}|=\mathbf{1}_{[0,1]}$ by claim 1 of Properties of the Absolute Value in an Ordered Field together with the sign of those values, so $\int_{\mathbb{R}}|\mathbf{1}_{[0,1]}|\,d\lambda=1$ is finite and $\mathbf{1}_{[0,1]}$ is integrable by the criterion recorded in Integrable Functions and Their Lebesgue Integral.

The displayed inequality says that $c\,\mathbf{1}_{[0,1]}(s)\le\tilde{\pi}(s)+M\,\mathbf{1}_{[0,1]}(s)$ for every $s\in\mathbb{R}$: for $s\in[0,1]$ this is the inequality just proved, and for $s\notin[0,1]$ both sides are $0$, the right-hand side because $\tilde\pi(s)=0$. Both sides are integrable by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, so the monotonicity and linearity in that claim give

$$c=c\cdot 1=\int_{\mathbb{R}}c\,\mathbf{1}_{[0,1]}\,d\lambda\le\int_{\mathbb{R}}\tilde{\pi}\,d\lambda+M\int_{\mathbb{R}}\mathbf{1}_{[0,1]}\,d\lambda=\int_{\mathbb{R}}\tilde{\pi}\,d\lambda+M .$$

Finally $\tilde{\pi}(s)+\tilde{\kappa}(s)=\tilde{\theta}_{x}(s)$ for every $s\in\mathbb{R}$, both sides being $|p(s)|+|q(s)|$ on $[0,1]$ and $0$ elsewhere, so claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives $\int_{\mathbb{R}}\tilde{\pi}\,d\lambda+M=F$. Hence $|p(t)|\le F$.

Proof of claim 1. By Existence and Uniqueness of the Integer Part of a Real Number there is an integer $m$ with $m\le x_{i}$ and $x_{i}<m+1$; put $t=x_{i}-m$, so that $0\le t$ and $t<1$, whence $t\in[0,1]$. Let $e$ be the point of $\mathbb{R}^{n}$ whose $i$th coordinate is $m$ and whose other coordinates are $0$; then $e\in\mathbb{Z}^{n}$, the integer lattice, and $x[t]+e=x[t+m]=x[x_{i}]=x$. Since $u$ is $\mathbb{Z}^{n}$-periodic,

$$u(x)=u(x[t]+e)=u(x[t])=p(t).$$

By (P3), $|u(x)|=|p(t)|\le F$. Since $x\in\mathbb{R}^{n}$ was arbitrary, claim 1 is proved.