Given an arbitrary TM, can you decide whether it's a LBA?
An LBA is limited to working only on the space defined by the input. That means it won't ever move right on reading a blank space (if your model doesn't allow writing blank, only a fake blank, that is). That is easy to check by looking at the transitions of the TM.
To check if the language is accepted by an LBA is undecidable by Rice's theorem.
No, assume towards contradiction you would have been able to solve it, with some Turing machine $M$. Notice, that it's easy to check given an LBA whether it halts or not (there are a finite number of total possible states) and let's call a TM that solves this by $M'$.
Now, let us construct the new Turing machine $\hat M$ that will solve the halting problem: Given $\langle T,x \rangle$ we want to decide if $T$ halts on $x$.
- Run $M$ with input $\langle T_x \rangle$, where $T_x$ is a Turing machine that ignores the input and computes $T(x)$.
- If the answer is false, then also return false.
- If the answer is true, then we know that $T_x$ is an LBA, and use $M'$ to decide whether it halts or not.
I will leave verifying this proof and completing it to you.