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 Ces├áro. 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?

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