Proof of The Dirichlet and Fejer Kernel Identities

Each result cited is universally quantified over the data in its own statement. Write $3=2+1$. Fix $t\in\mathbb{R}$; since $t$ is arbitrary, proving each identity for every $N\in\mathbb{N}$ at this $t$ proves the claims. Abbreviate, for $N\in\mathbb{N}$,

Both claims are proved by applying Principle of Induction for the Natural Numbers to the set of those $N\in\mathbb{N}$ for which the identity in question holds at this $t$.

Claim 1. Let $T$ be the set of $N\in\mathbb{N}$ satisfying $D_{N}\sin(\pi t)=\sin((2N+1)\pi t)$.

The base point. By claim 1 of Properties of Finite Sums a sum with one summand equals that summand, so $\sum_{m=1}^{1}\cos(2\pi mt)=\cos(2\pi t)$ and $D_{1}=1+2\cos(2\pi t)$. Product-to-Sum Formulas for Sine and Cosine §sine-cosine, taken with $u=\pi t$ and $v=2\pi t$, gives

since $\pi t+2\pi t=3\pi t$ and $\pi t-2\pi t=-\pi t$. By Values at Zero, Parity, Addition and Double-Angle Identities for Sine and Cosine §parity the second term on the left is $-\sin(\pi t)$, so $2\cos(2\pi t)\sin(\pi t)=\sin(3\pi t)-\sin(\pi t)$ and therefore

Since $2\cdot 1+1=3$, this says $1\in T$.

The inductive step. Let $N\in T$. By the recursion in claim 1 of Properties of Finite Sums,

so $D_{N+1}=D_{N}+2\cos(2\pi(N+1)t)$, and hence, by the inductive hypothesis,

Product-to-Sum Formulas for Sine and Cosine §sine-cosine, taken with $u=\pi t$ and $v=2\pi(N+1)t$, gives

since $\pi t+2\pi(N+1)t=(2N+3)\pi t$ and $\pi t-2\pi(N+1)t=-(2N+1)\pi t$; and the second term on the left equals $-\sin((2N+1)\pi t)$ by Values at Zero, Parity, Addition and Double-Angle Identities for Sine and Cosine §parity. Substituting,

Since $2(N+1)+1=2N+3$, this says $N+1\in T$. By Principle of Induction for the Natural Numbers, applied to the inductive set $T$, one has $T=\mathbb{N}$, which is claim 1.

Claim 2. Let $T'$ be the set of $N\in\mathbb{N}$ satisfying $G_{N}(\sin(\pi t))^{2}=(\sin(N\pi t))^{2}$.

The base point. By claim 1 of Properties of Finite Sums a sum with one summand equals that summand, so $\sum_{m=1}^{1}(1-m)\cos(2\pi mt)=(1-1)\cos(2\pi t)=0$, the last step by claim 1 of Zero Products and Elementary Identities in a Field. Hence $G_{1}=1$ and $G_{1}(\sin(\pi t))^{2}=(\sin(\pi t))^{2}$, so $1\in T'$.

Consecutive sums. Let $N\in\mathbb{N}$. The summand of $G_{N+1}$ at $m=N+1$ carries the factor $(N+1)-(N+1)=0$ and so vanishes, by claim 1 of Zero Products and Elementary Identities in a Field; so the recursion in claim 1 of Properties of Finite Sums gives

For every $m\in[N]$ one has $((N+1)-m)-(N-m)=1$, so

by the distributivity of multiplication over addition in $\mathbb{R}$. Summing over $m\in[N]$ and using claims 2 and 3 of Properties of Finite Sums, the first for additivity and the second for the scalar $-1$,

Since $G_{N+1}-G_{N}$ equals $1$ plus twice the left-hand side, by the two displays and the definitions of $G_{N+1}$ and $G_{N}$, it follows that

The inductive step. Let $N\in T'$. By the relation just proved, by the inductive hypothesis, and by claim 1 of the present lemma in the form $D_{N}(\sin(\pi t))^{2}=\bigl(D_{N}\sin(\pi t)\bigr)\sin(\pi t)$,

Values at Zero, Parity, Addition and Double-Angle Identities for Sine and Cosine §difference-of-squares, taken with $x=(N+1)\pi t$ and $y=N\pi t$, for which $x+y=(2N+1)\pi t$ and $x-y=\pi t$, gives

Substituting this into the previous display leaves $G_{N+1}(\sin(\pi t))^{2}=(\sin((N+1)\pi t))^{2}$, so $N+1\in T'$. By Principle of Induction for the Natural Numbers, applied to the inductive set $T'$, one has $T'=\mathbb{N}$, which is claim 2.