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 $z\in K\_1+K\_2$?
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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 $z\in K\_1+K\_2$?
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> 
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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 $C\ge 2$. 



Continuous selections from sums of compact sets
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 $f\_1(x)+f\_2(x)=x$.
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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.


