Consider an operator (T_a) acting on elements of $L^2(\mathbb{R})$, defined to symmetrically translate a function by plus and minus $a$. This operator is mathematically expressed as: $$ (T_a f)(x) = \frac{1}{2}[f(x-a) + f(x+a)] $$ which translates a function symmetrically by plus and minus, and its action in the frequency domain is represented by multiplication with $\cos(a\xi)$, where $\xi$ is the frequency variable: $$ \mathcal{F}\{T_a f\}(\xi) = \mathcal{F}\{f\}(\xi) \cdot \cos(a\xi) $$
Given this setup, a question arises about the invertibility of this operation (in general, but also) in the frequency domain, specifically, whether it can be inverted by multiplying by $1/\cos(a\xi)$, despite the zeros of $\cos(a\xi)$ introducing singularities.
Question 1: Can distribution theory be applied to achieve the inversion of the operation in the Fourier domain, considering the singularities introduced by the zeros of $\cos(a\xi)$? (I think maybe because the zeros of cosine are on a set of measure zero.)
Question 2: Is the operator $T_a$ injective? Specifically, can it be shown that if $T_a f = T_a g$ for two functions $f, g \in L^2(\mathbb{R})$, then $f = g$?
I'll try to answer your question in case I understood you right.
What about injectivity? If $f \in L^2(\mathbb{R})$ and $T_af = 0$ then for any $x \in \mathbb{R}$ we have $f(x-a)=-f(x+a)$ which immediately yields that $f$ is periodic with period $4a$. Consequently, $f = 0$ because there are no periodic functions in $L^2(\mathbb{R})$ besides $0$ and so $T_a$ is injective.
However, a potential inverse of $T_a$ on $L^2(\mathbb{R})$ could never be bounded. This follows from the fact that $T_a$ maps $4a$-periodic functions to $0$: Let's consider a cut-off of a $4a$-periodic function:
$$ f_n(x) = \sin\left(\frac{\pi}{2a}x\right) \mathbf{1}_{[-2an,2an]}(x). $$
Then $\Vert f_n \Vert_2^2 = 2an$ but $T_a(f_n)$ is $0$ everywhere (due to periodicity) except around the boundaries of the cutoff, yielding $\Vert T_af_n \Vert_2^2 = 2a$. Thus, $T_a$ is not bounded from below. This also implies that $T_a$ is not surjective by the bounded inverse theorem. A similar counterexample can be constructed on the Fourier domain using a function that spikes exactly where the zeros of $cos$ are.
This shows that extending the operator to a larger domain won't solve the problem as long as you are interested in $L^2(\mathbb{R})$. If you meant to completely dispense with $L^2(\mathbb{R})$ then it is a different question but in that case I suggest you tell us a little bit more about what your goals are.