Let $U\_\infty$ be a compact space, and let $U\_r$ be an increasing family of compact subspaces whose closure is all of $U\_\infty$. That is, $U\_r \subseteq U\_{r'}$ if $r \le r'$ and $U\_\infty = \overline{\bigcup U\_r}$. 


For $r \in [1,\infty]$, let $Y\_r = C(U\_r,\mathbb R)$ be the Banach space of real-valued continuous functions over $U\_r$ with the supremum norm. For $r \le r'$, let $\phi\_{r,r'} : Y\_{r'} \to Y\_r$ be the restriction maps, so that $Y\_\infty$ is the inverse limit of the spaces $Y\_r$. Write $\phi\_r : Y\_\infty \to Y\_r$ for the restriction map $\phi\_{r,\infty}$.


Suppose there exists a family of continuous linear operators $m\_r : Y\_r \to Y\_\infty$ such that $\|m\_r\| \le M$ for all $r$, and $\phi\_r \circ m\_r$ is the identity map on $Y\_r$.


**Question:** Suppose $\Gamma \subseteq Y\_\infty$ is compact. Does $m\_r \circ \phi\_r$ converge strongly to the identity operator on $\Gamma$? That is, for all $\epsilon > 0$, does there exist $R > 0$ such that if $r \ge R$, then $$\sup\_{y \in \Gamma} \left\| (m\_r \circ \phi\_r)(y) - y \right\|\_{Y\_\infty} < \epsilon?$$



Convergence of operators to the identity on Banach spaces
Contrary to my original muddled guess, the answer is no: the problem is that your `extension operators' don't give enough control over what happens in the gap between $U\_\infty$ and $U\_r$.


For a concrete example, take $U\_r$ to be the closed interval $[r^{-1},2]$ (for $1\leq r\leq\infty$), which clearly satisfies the conditions of the question. Now define the extension operator $m\_r$ as follows: given $f$ continuous and real-valued on $U\_r$, extend it to all of $[0,2]$ by putting $m\_r(f)(0)=0$ and interpolating linearly, i.e.


$$ m\_r(f)(t) = r^{-1}t f(1/r) \hbox{ if } 0\leq t\leq r \hbox{ and } m\_r(f)(t)=f(t)
 \hbox{ if } 1/r\leq t \leq 2.$$
Clearly each $m\_r$ is a linear extension operator with norm $1$.


Now let ${\bf 1}$ be the function on $[0,2]$ with constant function $1$. Then


$$ \Vert m\_r\circ\phi\_r({\bf 1}) - {\bf 1} \Vert \geq
 \vert m\_r\circ\phi\_r({\bf 1})(0) - {\bf 1}(0) \vert = 1 $$
for all $r<\infty$.


