An invertible system with memory

Question

Suppose $\mathcal{L}$ be invertible system with memory. Does $\mathcal{L}^{-1}$ have memory necessarily?

Intuitively I think the answer is "yes". There are many examples showing that. For instance $\mathcal{L}(x(t)) = x(t-2)$ and $\mathcal{L}(x(t)) = x(\frac t 3)$. Another example which seems problematic to me is $$\mathcal{L}(x(t)) = \int_{-\infty}^{t}x(\lambda)d\lambda$$The inverse is $$\mathcal{L}^{-1}(x(t)) = \frac{dx(t)}{dt}$$Does differentiator have memory? Of course the main question here is about memory of an invertible system which has memory. Note that here $\mathcal{L}$ can be nonlinear as well.

For clarity, I add some related definitions from Oppenheim's book:

Invertible system: A system is said to be invertible if distinct inputs lead to distinct outputs.

Causal system: A system is causal if the output at any time depends only on values of the input at the present time and in the past.

Memoryless system: A system is said to be memoryless if its output for each value of the independent variable at a given time is dependent only on the input at that same time.

Answer 1

For time-based systems, I understand that it is difficult to imagine a memory of the future. But for general systems, $-t$ and $t$ are just left and right (think of a spatial system). Other discussions are in LTI system $y(t)=x(t−T)$ with or without memory, What is a memory less system?, or A question about the concept of the time.

By definition of invertibility, $\mathcal{L}^{-1}$ is such that $\mathcal{L}^{-1}( \mathcal{L}(x))=x$. But also that $\mathcal{L}( \mathcal{L}^{-1}(x))=x$ (by the way, derivatives and integrals are not inverses). Let us suppose the converse: $ \mathcal{L}^{-1}$ has no memory. Hence $\mathcal{L}^{-1}(x[n]))$ can only use the present state, and $\mathcal{L}$ as well to yield $(x[n])$.

So, if $\mathcal{L}^{-1}$ is memoryless, $\mathcal{L}$ is memoryless as well. By contraposition, the converve is true