3
$\begingroup$

Does any one know an example of a bilinear functional $B:C(X)\times C(Y)\to {\bf R}$ ($X$ and $Y$ are open subsets of Euclid spaces) which cannot be extended continuously to a measure $\mu:C(X\times Y)\to {\bf R}$ (in the sense that $<\mu, f \times g >=B(f,g)$ where $f\in C(X)$, $g\in C(Y)$).

Many thanks!

$\endgroup$
2
  • 1
    $\begingroup$ To my eyes, your question would seem more natural if $X$ and $Y$ were compact (measures correspond to functionals on $C_0$ or $C_c$). In any case, I think the word to look for online or in MathSciNet is "bimeasure" $\endgroup$
    – Yemon Choi
    Jul 1, 2014 at 15:14
  • $\begingroup$ We probably also want to assume that $B$ is bounded? $\endgroup$ Jul 2, 2014 at 0:05

1 Answer 1

8
$\begingroup$

Take $x_1,\ldots,x_n \in X$ and $y_1,\ldots,y_n\in Y$ and consider the bilinear functional $$B_n(f,g)=\sum_{i\neq j}(i-j)^{-1}f(i)g(j).$$ By Hilbert's inequality, $|B_n(f,g)|\le\pi\|f\|_2\|g\|_2\le\pi n \|f\|_\infty\|g\|_\infty$. On the other hand, $$\sum_{i \neq j}|i-j|^{-1}=2\sum_{k=1}^{n-1}\frac{n-k}{k}\sim2n\log n.$$ So the measure $\mu_n$ representing $B_n$ has norm of order $n\log n$. Hence, for $\lambda_n>0$ such that $\sum_n \lambda_n n <\infty$ and $\sum_n \lambda_n n\log n=\infty$, one obtains a bounded bilinear form $\sum_n \lambda_n B_n$ which does not extends to a bounded measure. This construction is somewhat canonical. If you want to study it more, look up Grothendieck's inequality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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