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?

  • 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 Apr 25 '12 at 20:08
  • 1
    $\begingroup$ I've added a link... $\endgroup$ – genneth Apr 26 '12 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 $Q$? $\endgroup$ – Matthew Daws Apr 26 '12 at 13:40
  • $\begingroup$ @MatthewDaws: that is correct. I will change the notation I have used to match the Wiki. $\endgroup$ – genneth Apr 26 '12 at 16:16

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...?)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.