A condition for $\emptyset \neq S\subset RE$ under which $L_S \notin RE$

I read some computation theory lecture notes and after citing and proving the proposition: $ \emptyset \in S \Rightarrow L_S = \{\langle M \rangle : L(M)\in S\} \notin RE$ it says that $ \emptyset\in S$ is not a sufficient condition, i.e $ L_S \notin RE$ doesn’t yield $ \emptyset \in S$ , by giving the counter example $ L_{\Sigma ^*}\notin RE$ . However it says that there is a necessary and sufficient condition under which $ L_S\notin RE$ . I searched in Sipser’s book for this condition but couldn’t find it. I would really appreciate a reference for this condition.


Edit : given the answer by @dkaeae , I wish to know what is the stronger property one can deduce about a non-trivial $ S\subset RE$ in case $ L_S \notin RE$ .

If $\lambda \notin \sigma(A)$ then $\lambda \in \sigma (B)$ iff $\lambda \in \sigma_p(B)$ or $\overline{\lambda} \in \sigma_p(B^*)$

Let $ A$ , $ B$ and $ C$ the differential operators defined by $ $ D(A)=H^{2n}(\mathbb{R}), \ D(B)=H^{2n}(0, \infty)\cap H^{n}_0(0, \infty), D(C)=H^{2n}(-\infty,0)\cap H^{n}_0(-\infty,0),$ $ where $ H^{k}(I)$ is the Sobolev space $ W^{k,2}(I)$ , and $ $ Af:=Tf, Bg:=Tg, Ch:=Th \mbox{ for } \ f \in D(A), \ g \in D(B), \ h \in D(C),$ $ where $ $ T:=\sum_{r=0}^{2n}a_r\frac{d^r}{dx^r}$ $ with $ a_r$ complex constants.

I know that $ \sigma(A)=\sigma_{ess}(A)$ , i.e., $ \lambda \in \sigma(A)$ iff $ A-\lambda I$ is not a Fredholm operator (an operator is Fredholm if its range is closed and both its kernel and its cokernel are finite-dimensional).

I need to show that if $ \lambda \notin \sigma(A)$ then $ \lambda \in \sigma (B)$ iff $ \lambda$ is an eigenvalue of $ B$ or $ \overline{\lambda}$ is an eigenvalue of $ B^{*}$ . I’m reading a proof of that assertion, but I don’t understand the argument. The proof is the following:

Since $ B\oplus C$ differs from $ A$ only by imposing homogeneous Dirichlet boundary conditions at $ 0$ (I think that I can say that $ A$ is a finite-dimensional extension of $ B\oplus C$ ), the difference of the resolvents is of finite rank (If $ A$ is a finite-dimensional extension of $ B\oplus C$ , the difference of the resolvents of that operators is of finite rank (Lemma 6.32 on page 188 of Kato’s book)). So if $ \lambda \notin \sigma(A)$ then it follows that $ B-\lambda I$ and $ C-\lambda I$ are Fredholm operators (why?). Thus $ \lambda \in \sigma (B)$ iff $ \lambda$ is an eigenvalue of $ B$ or $ \overline{\lambda}$ is an eigenvalue of $ B^{*}$ (why?).

Thank you for reading my question. Can you help me to understand the proof?

How to proof problem L $\notin \texttt{DSPACE}(f)$

I want to proof a language $ L$ is not in $ \texttt{DSPACE}(f(n))$ the languages, that a deterministic Turing machine can decide with fixed tape length of $ f(n)$ (wiki). Show:

$ $ \begin{align*} L \notin \texttt{DSPACE}(f(n)) \end{align*}$ $

How would I do this?

My first attempt was to reduce a language $ L’$ from $ \texttt{DSPACE}$ to the language $ L$ , but I don’t know of any language in $ \texttt{DSPACE}$ and I don’t know if I can “simply” reduce it. Does anyone have a reduction that I see and try to understand?

edit:

as david-richerby mentioned: $ f(n) \in \Omega(\log{n})$

The exact exercise is: $ L = \{ w\mid w \in T(M_w),M_w \text{ needs space |w|}\}$

Is $ L = \{ a^n\ |\ a^n \not\in L_n \} $ Turing recognizable (recursively enumerable)?


Say $ \Sigma = \{a\} $ , $ M_1, M_2, … $ is an enumeration of all TMs that recognize languages over $ \Sigma$ and $ L_1, L_2, … $ are respectively the languages that are recognized by those TMs. We construct the language $ L = \{ a^n\ |\ a^n \not\in L_n \} $ . Is $ L$ Turing recognizable (recursively enumerable)?

My thought process is this: Say $ L$ is recognizable. Then given a string $ a^k$ , we find $ L_k$ based on the enumeration of $ L_i$ ‘s, and then we run $ M_k$ on input $ a^k$ to find if indeed $ a^k \not\in L_k$ . We must accept if $ a^k \not\in L_k$ , but $ M_k$ is not guaranteed to halt in that case. So $ L$ is not recognizable.

Now I have 2 questions: 1) Is my thought correct?, and 2) How can I make this more formal?