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summation by parts, with $S(x)={\sum }_{n\le x}\chi (n)$, gives for an integer $N$The boundary term tends to zero when $\mathrm{Re}s>0$ because $S$ is bounded. In the original region of absolute convergence this provesFor a compact subset of $\mathrm{Re}s>0$, choose $\delta >0$ below all its real parts and bound $|s|$ there. The integral tail is bounded by a constant times ${\int }_N^{\infty }{x}^{-1-\delta }dx$, uniformly on that compact set. Its truncated integrals are holomorphic function in $s$, so locally uniform convergence proves a analytic continuation throughout $\mathrm{Re}s>0$. This is holomorphy of a Dirichlet series from bounded partial sums.

The Euler product, valid for $\mathrm{Re}s>1$, isExpand the logarithms of these absolutely convergent Euler products. Writing $\theta =mt\log p$, the logarithm of the product in the desired inequality isFor odd $p$, $\chi (p{)}^m$ is $1$ or $-1$. In the first case the numerator is the given nonnegative trigonometric polynomial. In the second replace $\theta$ by $\theta +\pi$, which changes the sign of $\cos \theta$ and leaves $\cos 2\theta$ unchanged. At $p=2$ the numerator is $3+\cos 2\theta \ge 2$. Every summand is nonnegative. Exponentiation provesThis is the three-four-one inequality for Euler products applied to the real character modulo four.

Suppose $t\ne 0$ and $L(1+it)=0$. Holomorphy gives $L(\sigma +it)=O(\sigma -1)$ as $\sigma \downarrow 1$. The only pole of the Riemann zeta function is at one, so $\zeta (\sigma +2it)$ remains bounded, while $\zeta (\sigma {)}^3=O((\sigma -1{)}^{-3})$. The displayed product is then $O(\sigma -1)$, contradicting its lower bound. Hence $L(1+it)\ne 0$ for $t\ne 0$.

At $t=0$ the last zeta factor also has a pole, so that contradiction no longer follows from the inequality. Instead evaluate the convergent alternating series:For rigor, integrate the finite geometric identity for $(1+x^2{)}^{-1}$; the remainder has absolute integral at most $1/(2N+3)$ and tends to zero. summation by parts above identifies this value with the analytic continuation at one.

For a positive integer $N$, a Dirichlet character $\psi$ modulo $N$ is a group homomorphism $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }\to {\mathbb{C}}^{\times }$, extended periodically to the integers and set equal to zero on integers not coprime to $N$. Its nonzero values are roots of unity. For $\mathrm{Re}s>1$, its Dirichlet L-function is the absolutely convergent Dirichlet seriesComplete multiplicativity of $\psi$ proves the Euler product by expanding each factor as a geometric series and applying unique integer factorization; absolute convergence justifies these rearrangements.

Suppose $\psi$ is a nonprincipal Dirichlet character. Choose a unit $u$ with $\psi (u)\ne 1$. Multiplication by $u$ permutes a complete residue system, so ${\sum }_{a=1}^N\psi (a)=\psi (u){\sum }_{a=1}^N\psi (a)$, and the sum is zero. Consequently $A(x)={\sum }_{n\le x}\psi (n)$ is bounded: complete periods contribute zero and the remaining incomplete period has at most $N$ terms. The Abel summation formula gives, initially for $\mathrm{Re}s>1$,For $\mathrm{Re}s>0$ the integral converges, locally uniformly together with its derivatives in $s$, because on a compact set with $\mathrm{Re}s\ge \delta >0$ the differentiated integrands are bounded by constants times $x^{-1-\delta }(1+\log x{)}^k$. Thus holomorphy of a Dirichlet series from bounded partial sums proves

For a nonreal $\psi$, the character ${\psi }^2$ is also nonprincipal. The supplied logarithmic Euler product yields, for real $\sigma >1$,At a prime dividing $N$ the numerator is $3$. At every other prime put $\psi (p{)}^k=e^{i\theta }$; then the numerator is $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$. This Euler product positivity for L-function nonvanishing proves that the product inside the logarithm is at least one. If $L_N(\psi ,1)=0$, its holomorphy would give $L_N(\psi ,\sigma )=O(\sigma -1)$. Meanwhile $L_N({\psi }^2,\sigma )$ is bounded near one and the Riemann zeta function has a simple pole there. The product would therefore be $O(\sigma -1)$ and tend to zero, a contradiction.

For a real nonprincipal $\psi$, suppose instead that $L_N(\psi ,1)=0$. The product $H(s)=\zeta (s)L_N(\psi ,s)$ would then be holomorphic on the whole half-plane $\mathrm{Re}s>0$, because the zero cancels the only zeta pole there. In $\mathrm{Re}s>1$ it has coefficientsThese are nonnegative zeta-times-real-L coefficients. They are multiplicative, and their prime-power values are $k+1$ if $\psi (p)=1$, one or zero according as $k$ is even or odd if $\psi (p)=-1$, and one if $\psi (p)=0$. In particular $b_n\ge 0$ and $b_{m^2}\ge 1$ for every $m$.

The Taylor series of the assumed holomorphic $H$ about $2$ converges in the disk $|s-2|<2$. Termwise differentiation of its absolutely convergent Dirichlet series at $2$ gives $(-1{)}^k{H}^{(k)}(2)={\sum }_n{b}_n(\log n{)}^k/n^2$. Evaluate the Taylor series at $s=1/2$ and apply the Tonelli theorem to the resulting nonnegative terms:This contradicts holomorphy at $1/2$. The argument is the positive-coefficient mechanism of the Landau theorem for a Dirichlet series with nonnegative coefficients, with the needed step proved explicitly. Combining the real and nonreal cases gives the nonvanishing of a nonprincipal Dirichlet L-function at one:

Now let $a$ be coprime to $N$. The Dirichlet theorem on primes in arithmetic progressions asserts that infinitely many prime numbers satisfy $p\equiv a(\mathrm{mod}N)$. To prove it, separate the first-power terms in the given logarithmic Euler product:For $\sigma \ge 1$, the remainder is bounded uniformly byFor a nonprincipal character, holomorphy and nonvanishing near one give a local holomorphic logarithm. Along a short real interval to the right of one, this logarithm and the supplied continuous logarithm differ by a fixed integer multiple of $2\pi i$. Hence the prime-character sum near one is bounded:For the principal Dirichlet character ${\psi }_0$,The latter follows from the simple zeta pole and the bounded remainder. Finally Orthogonality of Dirichlet characters givesHere $\phi$ is the Euler totient function. The right-hand side tends to infinity as $\sigma \downarrow 1$, whereas a finite set of primes would give a bounded left-hand side. Therefore