Limit of field mode to null Cauchy surface Consider a two dimensional space-time with a mirror located at $x=0$. To the right of the mirror there is a massless scalar field, obeying $\square \phi=0$. The field modes are thus given by
$$\phi_w \sim  e^{-i  \omega x^+} - e^{-i  \omega x^-}.$$
As $\mathcal{I}^-$ is a Cauchy surface for this space-time, it suffices to specify the field on $\mathcal{I}^-$ to predict all future evolution. I would therefore expect an equation something like
$$\lim_{x^- \to -\infty} \phi_w \stackrel{?}{=}  e^{-i  \omega x^+}$$
I wonder how this is done rigorously? Is the limit simply defined as above?

 A: You should make your notation explicit. I'll assume that $x^\pm=x\pm t$, an your "mirror" is the boundary condition
$$
\phi(x=0,t)=0
$$
(which is not the only reasonable condition, another alternative would be $\partial_x\phi(x=0,t)=0$)
From your diagram $\mathcal I^-$ is the past null infinity at $x^-\to+\infty$ (I'll write this $\mathcal I^-_+$) when you're restricting to the half line (for the full line, you'd need to add $x^+\to -\infty$ as well, named $\mathcal I^-_-$, and $i^-$ the past time infinity at $t\to-\infty$).
In the general case of the line, this is done by noticing that the field equation can be rewritten as:
$$
\partial_{x^+}\partial_{x^-}\phi = 0
$$
which can be solved to:
$$
\phi = f_+(x_+)+f_-(x_-)
$$
with $f_\pm$ determined by the Cauchy condition at $\mathcal I^-_\pm,i^-$.
Your boundary condition can simply be implemented by imposing spatially odd parity of $\phi$, restricting your solutions to:
$$
\phi = f(x^+)-f(-x^-)
$$
with $f$ determined by the Cauchy condition at $\mathcal I^-_+$ by:
$$
f(z) = \phi(x^+=z,x^-\to+\infty)
$$
Getting back on your question, you can do a Fourier transform on $f$ to get your modes. In particular, the limits $x^-\to+\infty$ and $x^+\to+\infty$ are to be treated in the weak sense.
Hope this helps.
