Let $f\in L^2(\mathbb R)$ and suppose that $u(t,x)=f(x+t)$. Consider the equation $$\partial_t u=\partial_x u\iff\left(\partial_t-\partial_x\right)u=0.$$ I see that since we have assumed $f\in L^2(\mathbb R)$ only, we cannot make sense of either $t$- or $x$- differentiation. In other words, we cannot show that $u\in\text{dom}(\partial_t)\cup\text{dom}(\partial_x)$. On the other hand, I have read that one can show that $u\in\text{dom}(\overline{\partial_t-\partial_x})$, where the overline denotes the operator closure.
My problems: how do you actually compute the closure of the operator $\partial_t-\partial_x$ and show this?
My thoughts: Since the claim is that $u\in\text{dom}(\overline{\partial_t-\partial_x})$ for arbitrary $f\in L^2(\mathbb R)$, it makes me suspect that $\overline{\partial_t-\partial_x}=L^2(\mathbb R)\times L^2(\mathbb R)$. And from this case, all one has is that $\overline{\partial_t-\partial_x}u\in L^2(\mathbb R)$.