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Let $X$ and $Y$ be compact metric spaces and $f: X \times Y \to \mathbb{R}$ be a continuous function.

We know that, for every $n \in \mathbb{N}$, by the Stone-Weierstrass theorem, there exist $k_n \in \mathbb{N}$ and continuous functions $f^{(1,n)}_i$, $i \in \{1, \ldots, k_n\}$, on $X$ and continuous functions $f^{(2,n)}_i$, $i \in \{1, \ldots, k_n\}$, on $Y$, such that

$$ \sup_{x \in X, y \in Y } \bigg| f(x,y) - \sum_{i=1}^{k_n} f^{(1,n)}_i(x) f^{(2,n)}_i(y) \bigg| < \frac{1}{n} .$$

Is it possible to choose these functions such that

$$ \sup_{n \in \mathbb{N}} k_n < + \infty, $$ $$ \sup_{n \in \mathbb{N}} \sup_{i \in \{1, \ldots, k_n\}}\sup_{x \in X} \Big| f^{(1,n)}_i(x) \Big| < +\infty, \quad \quad \sup_{n \in \mathbb{N}} \sup_{i \in \{1, \ldots, k_n\}}\sup_{y \in Y} \Big| f^{(2,n)}_i(y) \Big| < +\infty \quad ? $$

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  • $\begingroup$ Doesn't one use partions of unity multiplied by values of $f$ to get those univariate functions, and doesn't that show that the answer is yes? $\endgroup$ – Dirk Werner Jan 10 at 19:41
  • $\begingroup$ Sorry - partitions of unity... $\endgroup$ – Dirk Werner Jan 10 at 21:04
  • $\begingroup$ @Dirk Sorry can you give more details? One possible choice of partition of unity of $X \times Y$ consists of products of univariate functions of $X$ and $Y$. But after multiplying each function by $f$ ( not necessarily a product of univariate functions), the product itself is still not necessarily a product of univariate functions... $\endgroup$ – Richard Jan 11 at 0:11
  • $\begingroup$ @DirkWerner I just realised that, in my proof, I also need the sequence $k_n$ to be bounded in $n$. Do you think that partition of unity still works? Thanks. $\endgroup$ – Richard Jan 11 at 7:41
  • $\begingroup$ @Richard Do you want $k_n\leq n$ or some other bound will suffice? $\endgroup$ – Piotr Hajlasz Jan 11 at 17:19
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Instead of the function $f$ consider the associated function $F: X \to C(Y)$, which is continuous. Let $\varepsilon>0$, and for each $x_0$ take a neighbourhood $U(x_0)$ where $\|F(x)-F(x_0)\|<\varepsilon$, from this cover of $X$ take a finite subcover $U(x_1),\dots,U(x_n)$ and a partition of unity $f_1,\dots,f_n$ subordinate to this subcover. Write $g_k=F(x_k)$ and $g(x,y)= \sum_{k=1}^n f_k(x) g_k(y)$. Then $\|f-g\|\le3\varepsilon$, all $f_k$ satisfy $0\le f_k\le 1$ and of course $\sum f_k = 1$. Also $\|g_k\|\le \|f\|$. So the bounds on $f_k$ and $g_k$ and the sum condition on the $f_k$ are ok.

However, in general one cannot have the sum condition $\sup_y \sum |g_k(y)| < \infty$ as well. If so, one would have an estimate for the ``weak $\ell_2$-norm'' $\sup_y (\sum |g_k(y)|^2)^{1/2}$ as well and likewise for $\sup_x [\dots]$. In terms of tensor norms this means that the injective tensor norm (corresponding to the sup-norm of $f$) is equivalent to Grothendieck's Hilbertian norm $w_2$, defined on the tensor product of two Banach spaces as follows: If $\|(x_j)\|_2^w := \sup_{x^*\in B_{E^*}} \|(x^*(x_j))\|_{\ell_2}$, put $\|u\|_{w_2}=\inf \|(x_j)\|_2^w\|(y_j)\|_2^w$, where the infimum is taken over all representations of the tensor $u\in E\otimes F$ as $u=\sum x_k \otimes y_k$. The punchline here is that for spaces of continuous functions, the Hilbertian norm is equivalent to the projective norm by Grothendieck's inequality. Hence the injective and the projective tensor norms would be equivalent on spaces of continuous functions, which they aren't.

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