How does the halting theorem interact with the recursion theorem? Recursion theorem (paraphrasing Sipser, Introduction to the Theory of Computation):

Let T be a program that computes a function t: N x N --> N. There is a program R that computes a function r: N --> N, where for every w,

        t(&lt;R&gt;, w) = r(w)


The proof is by construction. The program R thus constructed computes the same thing as program T with one input fixed at &lt;R&gt;. But what if the right side does not compute anything? In that event we can reason as follows:

T(&lt;R&gt;, w) halts  &lt;--&gt;  R(w) halts


and by modus tollens

T(&lt;R&gt;, w)  does not halt  &lt;--&gt;  R(w) does not halt


In particular we have 

R(w) halts  --&gt; T(&lt;R&gt;, w) halts
T(&lt;R&gt;, w)  does not halt  --&gt; R(w) does not halt


The transposition is based on the table below

T(&lt;R&gt;, w) halts__________________R(w) halts

   T                         T
   F                         F


But what if the value in the right column is unknown and unknowable? 

T(&lt;R&gt;, w) halts__________________R(w) halts

   T                         T
   ?                     unknowable 
   F                         F


We know that there is no universal halting decider, so the value in the right column may be unknowable to a given decider. So what if the decider concludes that T(&lt;R&gt;,w) does not halt? Is there anything wrong with the table below?

T(&lt;R&gt;, w) halts__________________R(w) halts

   T                         T
   F                     unknowable
   F                         F


In particular is there any compelling reason that if the value in the left column is F that the value in right column must also be F, i.e. that it cannot remain unknowable? It seems to me that if T(&lt;R&gt;, w) does not halt then it does not imply anything. 

Is it possible for a decider to say [for some T] that T(&lt;R&gt;, w) does NOT halt yet remain agnostic about R(w) halting??
The recursion theorem states that for every two-place partially computable function $t$ there is a program $R$ which computes a partial function $r$ satisfying
$$ r(w) = t(\langle R \rangle, w) $$
for all $w$. The function $t$ is given by a program that doesn't necessarily halt, and so we can think of it as a function from $\mathbb{N}^2$ to $\mathbb{N} \cup \{ \bot \}$, where $\bot$ means that the program doesn't halt. We call such a function a partial function, since a different way of specifying such a function is as a function from some subset $S \subseteq \mathbb{N}^2$ to $\mathbb{N}$; the function doesn't halt on inputs outside $S$.

The equality $r(w) = t(\langle R \rangle, w)$ is an equality of partial functions: the functions have the same domain (or the same set of inputs on which the output is a natural number rather than $\bot$), and they agree on their domain.