Friday, June 26, 2015

On Kronecker's lemma

While reading some parts of the nice lecture notes on Analytic Number Theory by Hildebrand, I encountered Kronecker's lemma on page 58.
If $f : \mathbb N \to \mathbb C$ is a function, and $s \in \mathbb C$ with $\Re(s) > 0$ is a complex number such that the Dirichlet series $L(f,s) = \displaystyle\sum_{n=1}^{\infty} \frac{f(n)}{n^s}$ converges, then $\displaystyle\sum_{n \le x } f(n) = o(x^s) \quad\text{for}\quad x \to + \infty$

As an example, take $f(n) = 1$, then $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{s}}$ converges for $s > 1$ and indeed $\displaystyle\sum_{n \le x } 1 = \lfloor x \rfloor = o(x^s)$ for all $s > 1$.

As a second example, take $f(n) = \frac{1}{n}$, then $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1+s}}$ converges for $s > 0$ and indeed $\displaystyle\sum_{n \le x } \frac{1}{n} = \log x = o(x^s)$ for all $s > 0$.

What I like about Kronecker's lemma is that its statement and proof involve real analysis only; one does not need to know anything about the behaviour of the function $L(f,s)$ for complex numbers $s$. On the other hand, in analytic number theory, one often uses Perron's formula to estimate sums $\sum_{n \le x } f(n)$ for number theoretic functions $f$. One then needs to have information about the zeros of the Dirichlet series, and its behaviour for large imaginary values of $s$ to make Perron's formula work. All this information is thus not needed when using Kronecker's lemma.

Although a proof of Kronecker's lemma can be found in Hildebrand's lecture notes, I write here a proof for the case $s \in \mathbb R$. This proof is based on the French Wikipedia article about Kronecker's lemma and is very transparent.

The above version of Kronecker's lemma is a special case of
If $b_n$ is a sequence of real numbers such that $0 \le b_1 \le b_2 \le \cdots$ and $\displaystyle\lim_{n \to \infty} b_n = + \infty$, and $a_n$ is a sequence of complex numbers such that $\displaystyle\sum_{n=1}^{\infty} a_n$ converges, then $\displaystyle\sum_{ k \le n} b_k a_k = o(b_n) \quad\text{for}\quad n \to + \infty$

The special case above follows from the general lemma by taking $a_n = f(n) / n^s$ and $b_n = n^s$.

Proof of the lemma: write $A_n = \displaystyle\sum_{ k = n}^{\infty} a_k$, then partial summation gives (setting $b_0 = 0$ for convenience )
\begin{equation*}
\sum_{k=1}^n b_k a_k = - b_n A_{n+1} +\sum_{k=1}^n (b_k - b_{k-1}) A_k
\end{equation*}
thus
\begin{equation*}
\frac{1}{b_n} \sum_{k=1}^n b_k a_k = - A_{n+1} +\frac{1}{b_n} \sum_{k=1}^n (b_k - b_{k-1})
A_k
\end{equation*}
Taking the limit $n \to \infty$, the first term converges to zero because $\sum_{n=1}^{\infty} a_n$ converges and the second term converges to zero because of the (generalized) lemma of Cesàro. QED.

Postscript: For a while I thought that because $\displaystyle\sum_{n=1}^{\infty} \frac{\mu(n)}{n} = \frac{1}{\zeta(1) } = 0$, Kronecker's lemma implied that $M (x) = \sum_{n \le x } \mu(n) = o(x)$. Hence I thought that Kronecker's lemma implied the prime number theorem. This reasoning is incorrect however: one needs more work to make the calculation $\displaystyle\sum_{n=1}^{\infty} \frac{\mu(n)}{n} =0$ sound; although this calculation only involves swapping $\lim_{ n \to \infty}$ with $\lim_{s \to 1}$. Is the proof on the prime number theorem nothing more than proving one can change the order of these two limits?