## Hardy and the zeta function, Part 1

Today we’ll begin with an interesting and important theorem about the Riemann zeta function. Recall that this is defined when ${\Re(s)>1}$ by the absolutely convergent series $\displaystyle \zeta(s)=\sum_{n=1}^\infty n^{-s}.$

This can be analytically continued to a function meromorphic on the entire complex plane, holomorphic everywhere except for a simple pole at ${s=1}$. Also, we know that it obeys the following functional equation whenever ${s\neq1}$: $\displaystyle \zeta(s)=\zeta(1-s)2^s\pi^{s-1}\Gamma(1-s)\sin(\pi s/2).$

It isn’t hard to show that ${\zeta(s)\neq0}$ whenever ${s>1}$, and hence the functional equation shows that apart from possible zeros in the critical strip ${0\leq s\leq 1}$ the only zeros of ${\zeta}$ are the so-called ‘trivial zeroes’ at ${s=-2,-4,\hdots}$.

The functional equation also shows that zeros are symmetrically distributed about the line ${\Re(s)=1/2}$. The infamous Riemann hypothesis conjectures that this symmetry collapses so that all non-trivial zeros of ${\zeta}$ occur on the line ${\Re(s)=1/2}$.

This is a whirlwind summary, and we’ll introduce or elaborate on other properties of ${\zeta}$ as they are needed. It is not trivial that ${\zeta}$ should have any zeros in the critical strip at all, though it actually does. We now know the first ${10^{13}}$ or so zeros, and sure enough, they all lie on the critical line ${\Re(s)=1/2}$.

It had been shown soon after its introduction that ${\zeta}$ had infinitely many non-trivial zeros. Very little was known about the distribution of these zeros in regards to the critical line was known until 1914 when Hardy proved the following.

Theorem 1 There are infinitely many zeros of ${\zeta}$ on the critical line ${\Re(s)=1/2}$.

This is a result seemingly more often quoted than proved, and the proof is rather technical and difficult. I have always found it astounding, however, and intrigued by what methods could accomplish this feat. Hence I shall blog the proof of this theorem in a ‘top down’ fashion: first giving the outline of the proof, and then filling in the technical lemmas as needed. I will be following closely the proof given in Montgomery and Vaughan’s “Multiplicative Number Theory” in chapter 14.

Firstly, we note that the function we are really bothered about for this theorem is the function of the real variable ${t}$ defined by $\displaystyle \zeta(1/2+it).$

After all, we’re not going to even attempt to discover anything about ${\zeta}$ away from this critical line. If we can show that this function of ${t}$ has infinitely many zeros, then we’re done. This should be easier for a start, since now we’re only dealing with a function of a real variable rather than a complex variable.

But, unfortunately, this function is not always real-valued itself. Looking back at the functional equation, we can see why — ${\zeta}$ has a fundamentally ‘asymmetric’ nature. To fix this, we recast the functional equation by first defining $\displaystyle \xi(s)=\frac{s(s-1)}{2}\zeta(s)\Gamma(s/2)\pi^{-s/2},$

and then noting that after some analytic manipulation the functional equation can be rephrased as $\displaystyle \xi(s)=\xi(1-s)$

for all ${s\in\mathbb{C}}$. In particular, since we also have the reflection property ${\xi(\overline{s})=\overline{\xi(s)}}$, we see that this functional equation implies that ${\xi(1/2+it)}$ is real for all real ${t}$. Furthermore, ${\xi(1/2+it)}$ has a zero at ${t\in\mathbb{R}}$ if and only if ${\zeta(1/2+it)}$ does.

It suffices, then, to show that the function ${\xi(1/2+it)}$ has infinitely many zeros ${t\in\mathbb{R}}$. This is already much easier to handle, since it is a function from ${\mathbb{R}}$ to ${\mathbb{R}}$. This is nearly the actual function we will consider. However, in some ways ${\zeta}$ remains easier to analyse than ${\xi}$, and so we’ll normalise so that the absolute value of the function always agrees with that of ${\zeta(1/2+it)}$.

Putting all this together, we can finally define the Hardy ${Z}$-function as $\displaystyle Z(t)=f(t)\xi(1/2+it)$

where the function ${f:\mathbb{R}\rightarrow\mathbb{R}}$ is chosen so that ${\lvert Z(t)\rvert=\lvert\zeta(1/2+it)\rvert}$ for all ${t\in\mathbb{R}}$. We can be explicit, looking at the definition of ${\xi}$, and define $\displaystyle Z(t)=\zeta(1/2+it)\frac{\Gamma(1/4+it/2)\pi^{-1/4-it/2}}{\lvert\Gamma(1/4+it/2)\pi^{-1/4-it/2}\rvert}.$

The function ${Z(t)}$ will change sign ${\gamma}$ if and only if ${\zeta(1/2+i\gamma)}$ has a zero at ${1/2+i\gamma}$ of odd multiplicity. There is an easy way to detect when a function has a sign change in the interval ${[T,2T]}$ — if and only if the inequality $\displaystyle \left\lvert\int_T^{2T}Z(t)\,\mbox{d}t\right\rvert<\int_T^{2T}\lvert Z(t)\rvert\,\mbox{d}t$

is sharp. If we could show that was true for arbitrarily large ${T}$, we’d be done. This is possible, but hard.

To make things easier, we will first multiply by some suitably chosen kernel ${W(t)}$. We’ll make sure that ${W(t)>0}$ for all ${t\in\mathbb{R}}$, so it won’t mess things up. It will hopefully make the analysis simpler.

We’ll choose ${W(t)}$ later. For now, remember that we’ve boiled down the proof to deriving a contradiction from the assumption that for all sufficiently large ${T}$, $\displaystyle \left\lvert\int_T^{2T}W(t)Z(t)\,\mbox{d}t\right\rvert=\int_T^{2T}W(t)\lvert Z(t)\rvert\,\mbox{d}t.$

This entry was posted in Uncategorized. Bookmark the permalink.

### 3 Responses to Hardy and the zeta function, Part 1

1. Paul says:

Hi, I’m trying to learn about the zeta function and this seems really helpful, thanks! Are you ever going to get around to writing pt 2?

2. gowers says:

I second that. This seems like a great blog: I really like the explanations I’ve read so far and would be delighted to see more of them.