Show that $\int_U (\Delta^2 f) g \ d\lambda^2 = \int_U (\Delta f) (\Delta g) d\lambda^2.$ 
Let $U \subset \Bbb R^2$ be an open set, $f \in C^4(U), g \in C_c^2(U)$ non-negative functions and
$$\Delta^2 : C^4(U) \rightarrow C(U),$$
$$\Delta^2 := {d^4 \over dx^4} + 2 {d^4 \over dx^2dy^2} + {d^4 \over dy^4},$$
be the biharmonic operator. Show that
$$\int_U (\Delta^2 f) g \ d\lambda^2 = \int_U (\Delta f) (\Delta g) d\lambda^2.$$

We can figure out that
$$\int_U (\Delta f) (\Delta g) d\lambda^2 = \int_U {d^2f \over dx^2} {d^2g \over dx^2} + {d^2f \over dx^2} {d^2g \over dy^2} + {d^2f \over dy^2} {d^2g \over dx^2} + {d^2f \over dy^2} {d^2 g \over dy^2}.$$
If it is necessary, I'll add more steps, but for the sake of clarity, just let me say that for the left side, I used the linearity of the integral and then pushed a part of the differential operators that stuck to $f$ at first on $g$, and by using the linearity again, I receive:
$$\int_U {d^2f \over dx^2} {d^2g \over dx^2} + {d^2f \over dx^2} {d^2g \over dy^2} + {d^2f \over dx^2} {d^2g \over dy^2} + {d^2f \over dy^2} {d^2 g \over dy^2}.$$
Everything seems correct except the third term. I received ${d^2f \over dx^2} {d^2g \over dy^2}$, but I need ${d^2f \over dy^2} {d^2g \over dx^2}$.
Does anyone see "directly" where I am wrong here?
 A: As your function g is compactly supported, we can find a ball $B=B_R(0)$ such that $supp(g) \subset B$. As a consequence, we have $g|_{\partial B}=0$ including its derivatives. Also, note that $supp(fg) \subset supp(g)$. We know want to consider the divergence theorem on $B$, which has a sufficiently smooth and compact boundary. Note that the right and the left hand side of the statement you are trying to proof both become $0$ outside of $B$.
It is a consequence of the divergence theorem, which has Greens identity as a consquence. To see that:
$$
div(g \nabla \Delta f)=<\nabla g, \nabla \Delta f>+g\Delta^2f
$$
Now that g is $C^2_C$, we can use the divergence theorem and the the usual greens identity (note that $g=0$ on the boundary as well as close to it):
$$
0=\int_{\partial B} fg d\vec{S}=\int_B div(g \nabla \Delta f)= \\\int_B<\nabla g, \nabla \Delta f>+g\Delta^2f= \\
-\int_B \Delta f \Delta g + \int_{\partial B}  (\partial_v g) \Delta f +\int_B g \Delta^2 f
$$
Then, we see that for any function $\phi$ s.t. the follwing expresion is integrable:
$$
\int_U \phi g=\int_{supp(g)} \phi g=\int_B \phi g
$$
Rearranging and using that the support of $g$ is within B and U gives you:
$$
\int_U \Delta g \Delta f=\int_U g \Delta^2f
$$
Please doublecheck if I did make a mistake as well.
Edit: I just noted that you should set $f,g=0$ outside of $U$, so everything you do is well defined 
