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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 $$\sup_{y \in \Gamma} \left\| (m_r \circ \phi_r)(y) - y \right\|_{Y_\infty} < \epsilon?$$

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$Y_{\infty}$ is Banach? – santker heboln Apr 3 '10 at 3:24
@santker heboln: by definition $Y_\infty$ is the space of real-valued continuous functions on the compact space $U_\infty$, equipped with the sup norm. – Yemon Choi Apr 3 '10 at 3:34
Oh, I missed the definition of $U_{\infty}$. Nvm – santker heboln Apr 3 '10 at 3:49
(deleted daft comment claiming answer to question was yes. in fact answer appears to be no; see below.) – Yemon Choi Apr 3 '10 at 4:13
up vote 3 down vote accepted

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 $r<\infty$.

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Ee glaiky o'mesel. Cheers, marra. – Tom LaGatta Apr 3 '10 at 23:14

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