3
$\begingroup$

Now, I am new to functional analysis. So please dont be harsh.

I was going through some papers by Balazard & Saias, Baez-Duarte, etc. that discussed and delved deep into details of approximating the characteristic function $\chi(0,1]$.

Now, what appears to me is, if we can construct a function, using functions of the form $f(x) = \sum c_k \rho(\frac{\theta_k}{x})$ where $\rho(x)$ is the fractional part of $x$ that stay constant in the interval (0,1] and satisfying $\sum c_k \theta_k = 0$, then we have a proof of RH.

But then I saw that they have also discussed the function $g(x) = - \sum \mu(n) \rho(1/nx) = \chi(t)$ which they say holds for every positive $t$ but in the sense of pointwise convergence, and then stated that in the words of Balazard and Saias's paper, "Unfortunately, this feeling is, at least to a certain extent, a mirage"

My question is:

Why is g(x) not the fulfilling the criteria of being that "function" which mathematicians are trying to find to serve as the approximation for $\chi(0,1]$? I would really appreciate any sort of clarification for this characteristic function that is being searched.

Thanks,

The papers that I am reading includes:

The Nyman-Beurling equivalent form for the Riemann hypothesis -- Michel Balazard and Eric Saias

Convergence and the Riemann hypothesis -- Jungseob Lee

A Note on Nyman-Beurling's equivalent formulation of the Riemann hypothesis -- Jean Francois Burnol

A Closure Problem related to the Riemann zeta funtion -- Arne Beurling

New versions of the Nyman-Beurling criterion for the Riemann hypothesis -- Luis Baez Duarte

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

The Nyman-Beurling Theorem is: $\zeta(s)$ has no zeroes for $\sigma>1/p$ if and only if the subspace of functions of the form $\sum_kc_k\rho(\alpha_k/x)$, for $0\le\alpha_k\le 1$ with $\sum_k c_k\alpha_k$=0, is dense in $L^p(0,1)$ (which is equivalent to the closure containing the characteristic function $\chi_{(0,1)}$).

The series $\sum_k\mu(k)\rho\big(1/(kt)\big)$ converges to $\chi_{(0,1)}(t)$ pointwise and in $L^1$, but not in $L^p$ for any $p>1$. Note that, in this formulation, $L^1$ convergence doesn't even imply nonvanishing on the edge of the critical strip. Similar sums have been tried. It's known that sums of this form converge very slowly (if at all, of course).

By the way, I gathered this from page 2 of this paper of Baez-Duarte (plus a little extra googling).

Improve this answer
$\endgroup$
8
  • $\begingroup$ Can you elaborate the part: The series $\sum_k \mu(k)\rho(1/kt)$ converges in $L^1$ and not in $L^P$? Why not in $L^p$? $\endgroup$
    – Roupam Ghosh
    Commented Jan 26, 2011 at 11:00
  • $\begingroup$ In my above comment, suppose p = 2... $\endgroup$
    – Roupam Ghosh
    Commented Jan 26, 2011 at 11:04
  • $\begingroup$ By not converging in $L^p$ does it mean that, $\sum_k \left( \mu(k)\rho(1/kt) \right)^2$ does not converge? $\endgroup$
    – Roupam Ghosh
    Commented Jan 26, 2011 at 11:37
  • 1
    $\begingroup$ Not converging in $L^p$ means $$\int_0^1|1-\sum_{k=1}^N \mu(k)\rho\big(1/(kt)\big)|^p$$ does not converge to 0 as $N\rightarrow\infty$. $\endgroup$
    – B R
    Commented Jan 26, 2011 at 18:08
  • 2
    $\begingroup$ Pointwise convergence does not imply $L^p$ convergence. For example, $(n+1)x^n$ converges to $0$ on $(0,1)$ pointwise but not in $L^1$, since $\int_0^1 |(n+1)x^n|\ dx=1$. And $L^1$ convergence does not imply $L^2$ convergence. For example, $\sqrt{n}\chi_{(0,1/n)}$ converges to $0$ in $L^1$, since $\int_0^1 |\sqrt{n}\chi_{(0,1/n)}|\ dx=\sqrt{n}/n$ (which goes to $0$ as $n$ goes to infinity), but not in $L^2$, since $\int_0^1 |\sqrt{n}\chi_{(0,1/n)}|^2\ dx=n/n=1$. $\endgroup$
    – B R
    Commented Jan 27, 2011 at 4:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .