Let $U$ be an open bounded subset of $\mathbb{R}^n$. I'm trying to make sense of the equation $$\langle\Delta u,\psi\rangle=\langle u,\Delta\psi\rangle,\tag{$*$}$$ where $u\in W^{2,p}(U)$ and $\psi\in C_c^\infty(U)$, and I was told that the angle brackets represent the pairing between distributions and test functions.
If I take the exposition literally, $\Delta u$ would be a distribution, but what is it really? Up to now, I haven't found any relevant definitions in my books on distributions. What I've got so far is that $u$ can be seen as a distribution defined by $$\psi\in C_c^\infty(U)\mapsto\int_U u\psi.$$ That way, $\langle u,\Delta\psi\rangle$ would be understood as $$\int_U u\Delta\psi.$$ As for $\Delta u$ in $(*)$, I'm not sure if it is defined by differentiating distributions, which is defined to be the operation $D^\alpha$ yielding another distribution $$D^\alpha u:\psi\in C_c^\infty(U)\mapsto(-1)^{|\alpha|}u(D^\alpha\psi),\tag{$**$}$$ where $\alpha$ is a multi-index. That's because I saw from an online source that $$(\Delta u)(\psi)=\int_U u\Delta\psi.\tag{$***$}$$ Is the Laplacian in $(***)$ equivalent to that defined via $(**)$? Any help is appreciated. Thank you.
Yes, what you read is correct, distributions are (bounded) linear forms acting on $C^\infty_c$ functions (called test functions). If $f$ is a distribution, one often denotes by $$ \langle f,\varphi\rangle = f(\varphi) $$ its action on a rest function varphi. In the case of a classical locally integrable function (or a measure), then one identifies it with the distribution acting as $$ \langle f,\varphi\rangle = \int f\,\varphi. $$ But this is not always possible. In your case, for any distribution $u$, $$ f_u : \varphi \mapsto \langle u,\Delta\varphi\rangle $$ is indeed a linear form on $C^\infty_c$, hence defines a distribution. Now people chose to call it $\Delta u$, so that it extends the usual definition of the Laplacian, and with the notation above, one obtains your first formula $$ \langle f_u,\varphi\rangle = \langle \Delta u,\varphi\rangle = \langle u,\Delta\varphi\rangle. $$ Hence, here, the two first expressions are just notations to write the last one. Since your $u$ is a locally integrable function, then the last expression is indeed an integral so $(***)$ is indeed equivalent to $(*)$.
Indeed one can also define derivatives of distributions by $\langle \nabla f,\varphi\rangle = -\langle f ,\nabla\varphi\rangle$. In particular, since $\Delta\varphi = \nabla \cdot \nabla\varphi$, it follows from the definitions that $$ \langle \Delta u,\varphi\rangle = \langle\nabla \cdot \nabla u,\varphi\rangle $$ and so $(***)$ is equivalent to $(**)$.