Originally [asked](https://math.stackexchange.com/questions/2831043/is-restriction-a-closed-map) 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:



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



Is restriction a closed map?
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 $C(Y)$.


