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This question is somehow related to the last open problem from Grothendieck's thesis about completeness of locally convex inductive limit. However, a particular case of the problem boils down to a very concrete question in Banach spaces: For two compact and absolutely convex sets $K_1, K_2$ the (Minkowski-) sum $K_1+K_2$ just consists of all sums $x_1+x_2$ with $x_j \in K_j$. One might ask whether one can choose the $x_j$ depending continuously on the sum, that is

Are there continuous decomposition maps $f_j: K_1+K_2 \to K_j$ with $f_1(z)+f_2(z)=z$ for all $z\in K_1+K_2$?

The answer to this question is negative, as I learned from Petr Holicky: Let $e_n$ be the unit vectors of $\ell_2(\mathbb N_0)$, $K_1=\overline{\Gamma(e_0+\frac{1}{n} e_n: n\in\mathbb N)}$ (where $\Gamma$ denotes the absolutly convex hull) and $K_2 = \overline{ \Gamma(-e_0+\frac{1}{n} e_n: n\in\mathbb N)}$.

In this example one has $K_1 \subseteq 3 K_2$ and there is a trivial continuous selection $f_j: K_1+K_2 \to C K_j$ with $f_1(z)+f_2(z)=z$ for $C=3$ (namely, $f_1(z)=z$, $f_2(z)=0$).

Now the question becomes:

Is there a universal constant $C>0$ such that for all absolutely convex compact sets $K_1,K_2$ in a Banach space there are continuous decomposition maps $f_j: K_1+K_2 \to C K_j$ with $f_1(z)+f_2(z)=z$ for all $z\in K_1+K_2$?

A close look at Holicky's example in an article "(LB)-spaces of vector-valued continuous functions", Bull. Lond. Math. Soc. 40 (2008), no. 3, 505–515 showed that necessarily $C\ge 2$.

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I think you meant to say "compact absolutely convex sets" above, otherwise finding a counterexample is rather easy. –  George Lowther Mar 3 '12 at 12:34
    
Of course. I have corrected the question. –  Jochen Wengenroth Mar 5 '12 at 15:05
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1 Answer

up vote 4 down vote accepted

No, there does not exist any such universal constant C.

I'll build up a counterexample inductively. First, suppose that we have the following.

(i) Let $K_1,K_2$ be compact and absolutely convex subsets of Hilbert space $H$, and $C\ge0$ be a constant such that there do not exist continuous functions $f_j\colon K_1+K_2\to CK_j$ such that $f_1(x)+f_2(x)=x$.

For example, as long as $K_1+K_2$ contains at least one nonzero element then they satisfy (i) for any constant $C < 1$. I'll show that we can construct new compact and absolutely convex sets $\tilde K_1,\tilde K_2$ for which (i) holds with $C$ replaced by $C+2$. Applying this process inductively shows that there exist sets $K_1,K_2$ satisfying (i) for any fixed $C \ge 0$.

Let $K_1,K_2,C$ be as in (i). By embedding $H$ in a larger Hilbert space if necessary, we can assume that we have unit vectors $u_{mn},v_{1mnk},v_{2mnk}$ ($m,n,k=1,2,\ldots$), all of which are mutually orthogonal and orthogonal to $K_1+K_2$. Let $(x_{1,1},x_{2,1}),(x_{1,2},x_{2,2}),(x_{1,3},x_{2,3}),\ldots$ be a sequence dense in $K_1\times K_2$. Set $w_{jmnk}=x_{jm}+(-1)^j\frac1{mn}u_{mn}+\frac1{mnk}v_{jmnk}$. Define the compact sets $\tilde K_j$ by $$ \tilde K_j=\overline{\Gamma\left(w_{jmnk}\colon m,n,k=1,2,\ldots\right)}. $$ Suppose that $f_j\colon\tilde K_1+\tilde K_2\to(C+2)\tilde K_j$ are continuous functions satisfying $f_1(x)+f_2(x)=x$. I'll show that this contradicts (i).

For $x$ in $H$, define $\theta_j(x)$ to be the infimum of $c\in\mathbb{R}_+$ such that $x\in c\tilde K_j$. Set $w_{jmn}=x_{j,m}+(-1)^j\frac1{mn}u_{mn}=\lim_{k\to\infty}w_{jmnk}$. Any $x$ in the subspace generated by $\tilde K_j$ can be decomposed as $$ \begin{align} x = \lambda_j\tilde x+\sum_{m,n}\lambda_{jmn}w_{jmn}+\sum_{m,n,k}\lambda_{j,m,n,k}w_{jmnk}&&{\rm(1)} \end{align} $$ for some $\tilde x\in K_j$. The real coefficients $\lambda_{jmn},\lambda_{jmnk}$ are uniquely determined, and I will regard them as functions of $x$. Also, we can choose $\lambda_j$ to be nonnegative and as small as possible, in which case it is also uniquely determined. It can be seen that $$ \theta_j(x)=\lambda_j+\sum_{m,n}\vert\lambda_{jmn}\vert+\sum_{m,n,k}\vert\lambda_{jmnk}\vert. $$

Let $\pi\colon H\to H$ be the orthogonal projection onto the closed subspace generated by $K_1+K_2$. Then $K_j=\pi(\tilde K_j)$. We have $\pi f_1(x)+\pi f_2(x)=x$ for all $x\in K_1+K_2$ so, by (i) and continuity of $f_j$, there exists $\bar m\in\mathbb{N}$ and $j\in\lbrace 1,2\rbrace$ such that $\pi f_j(x_{1\bar m}+x_{2\bar m})\not\in CK_j$. Wlog, suppose that $j=1$. Then, $f_1(x_{1\bar m}+x_{2\bar m})\not\in C\tilde K_1$ or, equivalently, $\theta_j(f_1(x_{1\bar m}+x_{2\bar m})) > C$. Set $y_j=f_j(x_{1\bar m}+x_{2\bar m})$. Also, set $\epsilon=\theta_1(y_1)-C > 0$. Then fix $\bar n$ large enough that $\vert\lambda_{1\bar m\bar n}(y_1)\vert < \epsilon/2$.

Since $w_{1mnk}+w_{2mnk}\to w_{1mn}+w_{2mn}=x_{1m}+x_{2m}$ as $k\to\infty$, continuity of $f_j$ allows us to write $$ f_j(w_{1\bar m\bar nk}+w_{2\bar m\bar nk})=w_{j\bar m\bar nk}+y_j-w_{j\bar m\bar n}+(-1)^jr_k $$ where $r_k\to0$, and is in the subspace generated by $\tilde K_1\cap \tilde K_2$. Note that the coefficients $\lambda_1,\lambda_{1mn},\lambda_{1mnk}$ in expansion (1) are all the same for $y_1-w_{1\bar m\bar n}$ as for $y_1$, except for $\lambda_{1\bar m\bar n}$ which satisfies $$ \vert\lambda_{1\bar m\bar n}(y_1-w_{1\bar m\bar n})\vert\ge1-\vert\lambda_{1\bar m\bar n}(y_1)\vert > 1-\epsilon/2 > 1+\vert\lambda_{1\bar m\bar n}(y_1)\vert-\epsilon. $$ So, $$ \theta_1(y_1-w_{1\bar m\bar n}) > 1+\theta_1(y_1)-\epsilon = C+1. $$ As $r_k\to0$ we have $\theta_1(y_1-w_{1\bar m\bar n}-r_k) > C+1$ for large $k$. Similarly, the coefficients in expansion (1) are the same for $w_{1\bar m\bar nk}+y_1-w_{1\bar m\bar n}-r_k$ as for $y_1-w_{1\bar m\bar n}-r_k$, except $\lambda_{1\bar m\bar n k}$. Also, $r_k$ is in the subspace generated by $\tilde K_1\cap\tilde K_2$ and, hence, $\lambda_{1\bar m\bar n k}(r_k)=0$. Therefore, $$ \begin{align} &\vert\lambda_{1\bar m\bar n k}(w_{1\bar m\bar nk}+y_1-w_{1\bar m\bar n}-r_k)\vert-\vert\lambda_{1\bar m\bar n k}(y_1-w_{1\bar m\bar n}-r_k)\vert\cr &\qquad\ge 1-2\vert\lambda_{1\bar m\bar n k}(y_1-w_{1\bar m\bar n})\vert\to1 \end{align} $$ as $k\to\infty$. So, $$ \begin{align} \liminf_{k\to\infty}\theta_1(w_{1\bar m\bar nk}+y_1-w_{1\bar m\bar n}-r_k)&\ge\liminf_{k\to\infty}\theta_1(y_1-w_{1\bar m\bar n}-r_k)+1\cr & > C+2. \end{align} $$ So, $f_1(w_{1\bar m\bar n k}+w_{2\bar m\bar n k})\not\in(C+2)\tilde K_1$ for large $k$, giving the required contradiction.

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Thanks a lot for this impressing construction. If it leads to something (concerning Grothendieck's question) I will contact you. –  Jochen Wengenroth Mar 5 '12 at 15:03
    
You can simplify the construction slightly by removing dependence on the index $n$. It doesn't really make the proof any easier though. –  George Lowther Mar 6 '12 at 2:22
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