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Originally asked on MSE.

Let $X$ be a normal (or even metrizable) topological space and let $Y$ be a closed subset of $X$. Let $C(X)$ be the linear space of all continuous scalar functions on $X$ endowed with the compact-open topology. Consider a map $R:C(X)\to C(Y)$ defined by $Rf=f|_Y$. This map is obviously linear, and it is easy to see that it is continuous. By Tietze theorem it is also surjective. It seems that this is in fact a quotient map. However, what interests me is the following question:

Let $B$ be a closed convex balanced subset of $C(X)$. Is $RB$ closed in $C(Y)$?

Under additional assumption that $B$ is weakly compact, this is true, since then $B$ is compact in the pointwise topology, and so $RB$ is pointwise compact, and so pointwise closed, and so closed in $C(Y)$. I can also show that this is wrong if we don't require $Y$ to be closed in $X$.

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  • $\begingroup$ many years ago I read a theorem in a paper that there is linear isometric embedding from $C(Y)$ to $C(X)$ which act by extension. that is: there is an isometric section for the map you consider. I really do not not remember that paper because I read it more than 10 years ago. but this obviously implies that the restriction map is closed. $\endgroup$
    – Ali Taghavi
    Commented Jul 1, 2018 at 8:44
  • $\begingroup$ may be Michael selection lemma is helpful to construct an appropriate right inverse(section)/ $\endgroup$
    – Ali Taghavi
    Commented Jul 1, 2018 at 8:57
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    $\begingroup$ @AliTaghavi do I understand you correctly that there is an isometry $S$ from $C(Y)$ into $C(X)$ such that $RS=Id$? I guess that $X$ is assumed to be compact in this statement. I don't see immediately why this implies that $R$ is a closed map.. $\endgroup$
    – erz
    Commented Jul 1, 2018 at 8:59
  • $\begingroup$ I am sorry I was mistaken. my comment was inappropriate. $\endgroup$
    – Ali Taghavi
    Commented Jul 1, 2018 at 9:38
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    $\begingroup$ @AliTaghavi: selection theorems do give a right inverse to any bounded linear surjective operator between Banach spaces, which is however in general not linear (thus also nowhere differentiable). In the present case, the extension is a bounded linear operator by construction. $\endgroup$
    – Pietro Majer
    Commented Jul 1, 2018 at 13:10

1 Answer 1

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The answer to the main question is negative:

Consider the compact subset $X=[0,1]\cup \{2\}$ of the real line and let $Y=\{2\}$ be a singleton in $X$. In the function space $C(X)$ consider the closed convex balanced subset $$B:=\{f\in C(X):\sup_{x\in [0,1]}|f(x)|\le 1,\;f(0)=0,\;f(2)=\int_0^1 f(t)dt\}.$$ It is easy to see that for the restriction operator $R:C(X)\to C(Y)$, the image $R(B)=\{f\in C(Y):|f(2)|<1\}$ is not closed in $C(Y)$.

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