- Let \(\textstyle F ( s ) = \sum _ { n = 1 } ^ { \infty } a _ { n } / n ^ { s }\) , where \(| a _ { n } | \leq M\) for all n.
- Then
\[ \lim _ {T \to \infty} \frac {1}{2 T} \int_ {- T} ^ {T} | F (\sigma + i t) | ^ {2} d t = \sum_ {n = 1} ^ {\infty} \frac {| a _ {n} | ^ {2}}{n ^ {2 \sigma}} \quad \text { if } \sigma > 1. \]
How is this reminiscent of the Parseval-Plancherel theorem? See e.g. Chapter 3 in Book I.
- Show as a consequence the uniqueness of Dirichlet series: If \(\scriptstyle F ( s ) = \sum _ { n = 1 } ^ { \infty } a _ { n } n ^ { - s }\) where the coeficients are assumed to satisfy \(| a _ { n } | \leq c n ^ { k }\) for some k, and \(F ( s ) \equiv 0\) , then \(a _ { n } = 0\) for all n.
Hint: For part (a) use the fact that
\[ \frac {1}{2 T} \int_ {- T} ^ {T} (n m) ^ {- \sigma} n ^ {- i t} m ^ {i t} d t \to \left\{ \begin{array}{l l} n ^ {- 2 \sigma} & \text {if} n = m, \\ 0 & \text {if} n \neq m. \end{array} \right. \]
Solution
For Problem 1(a), assume first \(\sigma>1\). Since \(|a_n|\le M\), the series for \(F(\sigma+it)\) converges absolutely and uniformly in \(t\). Hence \[ |F(\sigma+it)|^2 =\sum_{n,m\ge1}a_n\overline{a_m}(nm)^{-\sigma}e^{-it\log(n/m)}, \] and the double series is absolutely summable because \[ \sum_{n,m\ge1}|a_na_m|(nm)^{-\sigma} \le M^2\zeta(\sigma)^2<\infty. \] Therefore we may integrate termwise: \[ \frac1{2T}\int_{-T}^T|F(\sigma+it)|^2\,dt =\sum_{n,m\ge1}a_n\overline{a_m}(nm)^{-\sigma}K_T(n,m), \] where \[ K_T(n,m)=\frac1{2T}\int_{-T}^Te^{-it\log(n/m)}\,dt. \] If \(n=m\), then \(K_T(n,n)=1\). If \(n\ne m\), then \[ K_T(n,m)=\frac{\sin(T\log(n/m))}{T\log(n/m)}\longrightarrow0. \] Also \(|K_T(n,m)|\le1\). Dominated convergence for the absolutely summable double series therefore gives \[ \lim_{T\to\infty}\frac1{2T}\int_{-T}^T|F(\sigma+it)|^2\,dt =\sum_{n=1}^\infty\frac{|a_n|^2}{n^{2\sigma}}. \] This is the Dirichlet-series analogue of Parseval–Plancherel: averaging in the vertical variable kills the cross terms and leaves the square sum of the coefficients.
For Problem 1(b), suppose \[ F(s)=\sum_{n\ge1}a_nn^{-s}\equiv0, \qquad |a_n|\le cn^k. \] Set \(b_n=a_nn^{-k}\) and \[ G(s)=F(s+k)=\sum_{n\ge1}b_nn^{-s}. \] Then \(|b_n|\le c\) and \(G\equiv0\). Applying part (a), for every \(\sigma>1\), \[ 0=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T|G(\sigma+it)|^2\,dt =\sum_{n=1}^\infty\frac{|b_n|^2}{n^{2\sigma}}. \] Every term on the right is nonnegative, so every \(b_n=0\), hence every \(a_n=0\). This proves uniqueness of Dirichlet series under the stated polynomial growth condition.