6
$\begingroup$

Bochner's theorem (for the real line version) asserts an infinite tower of inequalities, as a positivity condition. Taking each one, what do they mean, in an elementary fashion (at least at the start)?

For instance, the $1 \times 1$ matrix says that $Q(0)$ is positive. The $2 \times 2$ says that $|Q(x)| \le |Q(0)|$ (These two are commonly written down for necessary conditions of characteristic functions). What about 3 and 4?

$\endgroup$
4
  • 6
    $\begingroup$ Bochner (God rest his soul) proved more than one theorem. Perhaps you could state the result you are alluding to? $\endgroup$
    – Igor Rivin
    Commented Apr 25, 2012 at 20:08
  • 1
    $\begingroup$ I've added a link... $\endgroup$
    – genneth
    Commented Apr 26, 2012 at 9:32
  • 1
    $\begingroup$ What is $f$? The Wikipedia article you linked to starts with a finite positive Borel measure $\mu$ on $\mathbb R$, takes the Fourier transform to form $Q$, complex-valued a function on $\mathbb R$, and then forms a kernel $K$-- the theorem is that this is positive definite, so for any $n$, given $x_1,\cdots,x_n\in\mathbb R$, the matrix $(K(x_i,x_j)) = (Q(x_j-x_i))$ is positive (semi-)definite. You're asking, I think-- how to I interpret this for a fixed n? So I think your $f$ is Wikipedia's $Q$? $\endgroup$ Commented Apr 26, 2012 at 13:40
  • $\begingroup$ @MatthewDaws: that is correct. I will change the notation I have used to match the Wiki. $\endgroup$
    – genneth
    Commented Apr 26, 2012 at 16:16

1 Answer 1

3
$\begingroup$

I believe the $3\times3$ inequalities were first exploited by Kreĭn and Weil, who wrote $$ \begin{align} \bigl|\,Q(g) - Q(h)\,\bigr| &\leqslant \sqrt{2\mathrm{Re}\bigl(1 - Q(g^{-1} h)\bigr)}, \tag K\\[1ex] \bigl|\,Q(gh) - Q(g)Q(h)\,\bigr| &\leqslant \sqrt{1 - |Q(g)|{}^2}\sqrt{1 - |Q(h)|{}^2}. \tag W \end{align} $$ in (1940), resp. (1940, p. 57). I stated them for any positive-definite function $Q$ on any group (written multiplicatively, as none of this requires Bochner’s commutative setting), but normalized $Q(e)=1$; else rescale by $Q(e)$, which (as you noted) is nonnegative by the $1\times1$ case.

Kreĭn’s $(\mathrm K)$ says that $Q$ is uniformly continuous everywhere as soon as it’s continuous at $e$, and I like to think of Weil’s $(\mathrm W)$ as a group version of Heisenberg’s uncertainty inequalities. Deducing them from positive-definiteness is not so obvious: $(\mathrm W)$ is a rearrangement of the determinant criterion $$ \begin{vmatrix} 1&Q(a^{-1}b)&Q(a^{-1}c)\\ \overline{Q(a^{-1}b)}&1&Q(b^{-1}c)\\ \overline{Q(a^{-1}c)}&\overline{Q(b^{-1}c)}&1 \end{vmatrix} \geqslant0, \tag{$*$} $$ written with $g=a^{-1}b$ and $h=b^{-1}c$; and proving $(\mathrm K)$ from $(\mathrm W)$ alone is possible but not very enlightening. I prefer to observe that positive-definiteness of $Q$ means positive-definiteness of the sesquilinear form $(c,d)\mapsto Q(c^*\cdot d)$ on the group *-algebra $\mathbf C[G]=$ {functions $c:G\to\mathbf C$ with finite support} where product, *-operation, and linear form $Q$ are defined on the basis of (Kronecker) $\delta^g$’s by $$ \delta^g\cdot\delta^h=\delta^{gh}, \qquad (\delta^g)^*=\delta^{g^{-1}}, \qquad Q(\delta^g) = Q(g). $$ Then $(\mathrm K)$ and $(\mathrm W)$ are just the Cauchy-Schwarz inequality $$ |Q(c^*\cdot d)|^2\leqslant Q(c^*\cdot c)Q(d^*\cdot d), $$ written out for $(c^*,d) = (\delta^e,\delta^g-\delta^h)$, resp. $(\delta^g-Q(g)\delta^e,\delta^h-Q(h)\delta^e)$.

(Which triples give $(\mathrm K)$ and $(\mathrm W)$ can then be found by remembering that Cauchy-Schwarz amounts to writing $Q(f^*\cdot f)\geqslant 0$ where $f=Q(c^*\cdot c)d-Q(c^*\cdot d)c$. Hewitt-Ross (1970, pp. 255, 289) have another proof, and the attribution to Kreĭn & Weil. It might be interesting to know if the $4\times4$ version of $(*)$ has some nice consequences...?)

$\endgroup$

You must log in to answer this question.

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