Trigonometric and Oscillatory Examples in One Variable

By [L1] and [L2], $\mathrm{\mid }{\sin }^{\mathrm{\prime }}x\mathrm{\mid }=\mathrm{\mid }\cos x\mathrm{\mid }\le 1$ and $\mathrm{\mid }{\cos }^{\mathrm{\prime }}x\mathrm{\mid }=\mathrm{\mid }\sin x\mathrm{\mid }\le 1$ for every real $x$; both functions are continuous on $\mathbb{R}$ by [L3].

Apply [L4] to sine on the order-convex set $\mathbb{R}$ with $M=1$: $\mathrm{\mid }\sin u-\sin v\mathrm{\mid }\le \mathrm{\mid }u-v\mathrm{\mid }$.

Apply [L4] to cosine on the same set with $M=1$: $\mathrm{\mid }\cos u-\cos v\mathrm{\mid }\le \mathrm{\mid }u-v\mathrm{\mid }$.

For every $n\in \mathbb{N}$ and $x\in \mathbb{R}$, $\mathrm{\mid }{f}_n(x)\mathrm{\mid }=a^n\mathrm{\mid }\cos (b^n\pi x)\mathrm{\mid }\le a^n$.

Since $0<a<1$, the majorant series ${\sum }_{n=0}^{\mathrm{\infty }}{a}^n$ converges, including its first term $a^0=1$.

Cosine is continuous by [L4] and [L5]; the map $x\mapsto b^n\pi x$ is a polynomial and hence continuous by [L9], so each $f_n$ is continuous by [L7], and every finite partial sum $S_N$ is continuous by [L6].

Applying [L3] to steps 1.1 and 1.2 proves absolute convergence at every real point and uniform convergence of the partial sums to $W_{a,b}$ on $\mathbb{R}$.

The functions $S_N$ are continuous by step 1.3 and converge uniformly by step 2.1, so [L8] makes their sum $W_{a,b}$ continuous.

Apply [L2] to $b^m{x}_0+1\mathrm{/}2$. The resulting integer $k_m$ satisfies $k_m\le b^m{x}_0+1\mathrm{/}2<k_m+1$, hence $-1\mathrm{/}2\le z_m=b^m{x}_0-k_m<1\mathrm{/}2$.

Since $x_m-x_0=(1-z_m)\mathrm{/}{b}^m$, step 1.1 and $b^m>0$ give $0<x_m-x_0\le 3\mathrm{/}(2b^m)$.

Let $\epsilon >0$. By [L3], for all sufficiently large $m$ one has $b^m>3\mathrm{/}(2\epsilon )$, hence step 2.1 gives $0<x_m-x_0\le 3\mathrm{/}(2b^m)<\epsilon$. Thus $x_m\to x_0$.

For $n\ge m$, the integer $b^{n-m}$ is odd. The identities $b^n{x}_m=b^{n-m}(k_m+1)$ and $b^n{x}_0=b^{n-m}(k_m+z_m)$, followed by repeated use of [L4] to shift through integer multiples of $\pi$, give the two asserted cosine values; oddness preserves the parity of $k_m$ and reverses the parity of $k_m+1$.

The displayed finite sum defines $A_m$, and [L3] gives $A_0=0$.

Multiplying the finite sum ${\sum }_{n<m}(ab{)}^n$ by $ab-1$ and telescoping gives $(ab-1){\sum }_{n<m}(ab{)}^n=(ab{)}^m-1$, including at $m=0$; since $ab-1>0$, ${\sum }_{n<m}(ab{)}^n=\frac{(ab{)}^m-1}{ab-1}\mathrm{.}$

Repeated use of [L5], followed by [L1] on each summand and [L4], gives $\mathrm{\mid }{A}_m\mathrm{\mid }\le {\sum }_{n<m}{a}^n{b}^n\pi \mathrm{\mid }{x}_m-x_0\mathrm{\mid }=\pi (x_m-x_0){\sum }_{n<m}(ab{)}^n,$ where [L2] supplies $x_m-x_0>0$ and [L6] supplies $\pi >0$.

Substitute step 1.2 into step 2.1. For $m\ge 1$, one has $(ab{)}^m-1<(ab{)}^m$ and the other factors are positive, so the strict displayed bound follows; at $m=0$, the non-strict formula already gives $A_0=0$.

Absolute convergence in [L1] licenses subtraction of the two convergent series and defines the displayed tail $B_m$.

Since $\mathrm{\mid }{z}_m\mathrm{\mid }\le 1\mathrm{/}2$ and $\pi >0$ by [L7], parity and monotonicity in [L4], together with [L5], give $\cos (z_m\pi )=\cos (\mathrm{\mid }{z}_m\mathrm{\mid }\pi )\ge \cos (\pi \mathrm{/}2)=0$.

By [L2], every summand of $B_m$ is $-(-1{)}^{k_m}{a}^n(1+\cos (b^{n-m}{z}_m\pi ))\mathrm{.}$ The parenthesized factor is nonnegative by the range clause of [L4], so the partial sums share one weak sign. Their absolute values therefore converge to $\mathrm{\mid }{B}_m\mathrm{\mid }$ and dominate the absolute value of the $n=m$ term by [L6]; step 1.2 makes that term at least $a^m$. Hence $\mathrm{\mid }{B}_m\mathrm{\mid }\ge a^m$.

The upper bound in [L3] gives $1\ge (2b^m\mathrm{/}3)(x_m-x_0)$. Multiplying step 2.1 by this nonnegative bound yields $\mathrm{\mid }{B}_m\mathrm{\mid }\ge (2\mathrm{/}3)(ab{)}^m(x_m-x_0)$.

Fix $x_0$ and use [L6] for its probe sequence $(x_m)$. Continuity at $x_0$ already follows from [L1].

The hypothesis $ab>1+3\pi \mathrm{/}2$ and positivity of $\pi$ give $ab>1$ and $c:=\frac{2}{3}-\frac{\pi }{ab-1}>0.$

For $m\ge 1$, splitting the series increment at frequency $m$ gives $W_{a,b}(x_m)-W_{a,b}(x_0)=A_m+B_m$. The reverse triangle inequality and [L2] to [L3] yield $\mathrm{\mid }{W}_{a,b}(x_m)-W_{a,b}(x_0)\mathrm{\mid }\ge (ab{)}^m(\frac{2}{3}-\frac{\pi }{ab-1})(x_m-x_0)\mathrm{.}$

Since $x_m-x_0>0$, divide step 2.1 by it. Step 1.2 and [L4] show that the absolute values of the selected difference quotients are at least $c(ab{)}^m$ and tend to $+\mathrm{\infty }$.

Although $x_m\to x_0$, the difference quotients along this sequence have no finite limit by step 3.1, so [L5] rules out differentiability at $x_0$. The point $x_0$ was arbitrary, while [L1] gives continuity everywhere. Then $W_{a,b}$ is continuous at every real point and differentiable at no real point.

Choose $a=1\mathrm{/}2$ and $b=15$. The integer $15$ is odd, and [L7] gives $\pi <4$, so $1+3\pi \mathrm{/}2<7<15\mathrm{/}2=ab$; all hypotheses of [L3] hold.

Repeated application of [L4] and [L5] shows that every derivative of every finite partial sum $S_N$ is a finite linear combination of sine and cosine functions. Those derivatives are continuous by [L6], so each $S_N$ is smooth in the sense of [L1].

By [L2], the smooth functions from step 1.2 converge uniformly on $\mathbb{R}$ to $W_{1\mathrm{/}2,15}$.

The parameter check in step 1.1 lets [L3] identify this uniform limit as continuous everywhere and differentiable nowhere.

Thus the sequence in step 2.1 consists of smooth functions and converges uniformly to the nowhere-differentiable function in step 2.2.

By [L1], choose a polynomial $p$ with $\mathrm{\mid }f(t)-p(t)\mathrm{\mid }<\epsilon \mathrm{/}(4(b-a))$ for every $t\in [a,b]$.

Put $C_p:=\mathrm{\mid }p(a)\mathrm{\mid }+\mathrm{\mid }p(b)\mathrm{\mid }+{\int }_a^b\mathrm{\mid }{p}^{\mathrm{\prime }}(t)\mathrm{\mid }dt$. Using $v(t)=-\cos (nt)\mathrm{/}n$ in [L3], and then [L6], [L7], [L9], and [L11], gives $\mid {\int }_a^{b}p(t)\sin (nt)dt\mid \le C_p\mathrm{/}n\mathrm{.}$ Using $v(t)=\sin (nt)\mathrm{/}n$ gives the identical bound for the cosine integral.

Apply [L8] to $\epsilon \mathrm{/}(2(C_p+1))$ and choose a positive integer $N$ with $(C_p+1)\mathrm{/}N<\epsilon \mathrm{/}2$. Then $C_p\mathrm{/}n<\epsilon \mathrm{/}2$ whenever $n\ge N$.

The functions $f(t)\sin (nt)$ and $p(t)\sin (nt)$ are integrable, and [L2] with [L6] and [L9] gives $\mid {\int }_a^b(f(t)-p(t))\sin (nt)dt\mid \le \epsilon \mathrm{/}4<\epsilon \mathrm{/}2$ for every positive integer $n$; the same estimate holds with cosine.

For every $n\ge N$, linearity [L7] splits each integral into its $(f-p)$ part and its $p$ part. Steps 2.1, 1.2, and 1.3 make the absolute value of each integral less than $\epsilon$, for sine and for cosine.

Since $\epsilon >0$ was arbitrary, step 3.1 is exactly convergence of both sequences of integrals to zero by [L12]; [L10] permits the substitution $a=0$, $b=\pi$, and $g$ for $f$ in the stated special case.

For the nonperiodic case, assume $x\mathrm{\notin }2\pi \mathbb{Z}$ and define the auxiliary complex sums recursively by $S_0=0$ and $S_{j+1}=S_j+e^{i(j+1)x}$. Multiplication by $1-e^{ix}$ and the exponential addition law [L7] telescope directly to $(1-e^{ix})S_N=e^{ix}-e^{i(N+1)x}$. The half-angle identity $1-e^{ix}=-2i{e}^{ix\mathrm{/}2}\sin (x\mathrm{/}2)$ and [L3] show that the multiplier is nonzero.

For the periodic case, assume $x\in 2\pi \mathbb{Z}$. Then every $nx$ is a multiple of $2\pi$, so $\sin (nx)=0$ by [L3] and the finite sine sum is zero.

For the nonperiodic case, divide the identity in step 1.1 by its nonzero multiplier and use [L1] and [L2] to obtain $S_N=e^{i(N+1)x\mathrm{/}2}\frac{\sin (Nx\mathrm{/}2)}{\sin (x\mathrm{/}2)}\mathrm{.}$

For the nonperiodic case, take imaginary parts in step 2.1 and apply the product-to-sum consequence of [L2] to get ${\sum }_{n=1}^N\sin (nx)=\frac{\cos (x\mathrm{/}2)-\cos ((N+1\mathrm{/}2)x)}{2\sin (x\mathrm{/}2)}\mathrm{.}$

For the nonperiodic case, [L6] bounds the numerator in step 3.1 by $2$, so the absolute value of the sum is at most $1\mathrm{/}\mathrm{\mid }\sin (x\mathrm{/}2)\mathrm{\mid }$.

The nonperiodic branch gives the displayed formula and bound by steps 3.1 and 4.1, while the periodic branch gives the separate zero value by step 1.2; the two cases exhaust all real $x$.