Decidability of the TM's computing a non-empty subset of total functions I have this HW problem:
Let $F$  be the set of computable total functions, and let $\emptyset\subsetneq S\subseteq F$. Denote 
$$L_S=\{ \langle M \rangle | M \text{ is a TM that computes a function and } f_M\in S \}$$

where $f_M$ is the function $M$ computes. 

Prove that for every such none-trivial $S$, $L_S \notin \mathcal{R}$.

I tried to  construct


$L_{f}=\left\{ x\#y\in{\Sigma^*}|y=f\left(x\right)\right\} $
$\tilde{S}=\left\{ L_{f}|f\in S\right\} $ 
$L_{\tilde{S}}=\left\{ \left\langle M\right\rangle |L\left\langle M\right\rangle \in\tilde{S}\right\} $


and then show with Rice that $L_{\tilde{S}} \notin \mathcal{R}$, when the idea behind it was to eventually show that $L_{\tilde{S}} \leq_m L_S $.
But the problem here is that I couldn't show a mapping reduction from $L_{\tilde{S}}$ to $L_S$ without assuming $L_{\tilde{S}} \in \mathcal{RE}$ (which I'm quite sure is not true).
So any other directions will be warmly welcomed!
For any function $f: \Sigma^* \rightarrow \Sigma^*$, let 

$$L_f = \{ x \# y \mid y = f(x) \}$$

For any Turing machine $M$, we can construct the description of a Turing machine $M'$ such that $M'$ recognizes $L_f$ if and only if $M$ computes $f$. $M'$ will receive a string, check that it is of the form $x \# y$ and then simulate $M$ with input $x$. If the simulation returns $w$, $M'$ accepts if $w = y$. Therefore, the language

$$ A_f = \{ \langle M \rangle \mid M \text{ is a TM that computes the function $f$} \}$$

reduces to the language

$$B_f = \{\langle M \rangle \mid L(M) = L_f \}$$

The class

$$L_S = \{ L_f \mid f \in S \}$$

is a nontrivial class of Turing-recognizable languages and therefore the decision problem $B_f$ is undedicable by Rice's theorem.