I was looking through some old course work and found an exercise which I am not able to fully solve. The task is the following:
Prove that the differential equation $$ u + \partial^4_x u - \partial^2_x u = f $$ for any given $f \in L^2(\mathbb{R})$ has a weak solution $u \in H^4(\mathbb{R})$ (fourth order Sobolev space). Do not explicitly calculate a solution.
So far I have the following approach to a solution:
Multiply both sides of the differential equation with $\varphi \in H^4_0(\mathbb{R})$ (closure of $C_c^\infty(\mathbb{R})$ in the norm of $H^4(\mathbb{R})$) and integrate over $\mathbb{R}$, such that
$$
\int_\mathbb{R} u \varphi \, dx + \int_\mathbb{R} \partial^4_x u \varphi \, dx - \int_\mathbb{R} \partial^2_x u \varphi \, dx = \int_\mathbb{R} f \varphi \, dx \,.
$$
Then by partial integration we get
$$
\underbrace{\int_\mathbb{R} u \varphi \, dx + \int_\mathbb{R} \partial^2_x u \partial^2_x \varphi \, dx + \int_\mathbb{R} \partial_x u \partial_x\varphi \, dx}_{=\langle u, \varphi\rangle_{H^2(\mathbb{R})}} = \underbrace{\int_\mathbb{R} f \varphi \, dx}_{=F(\varphi)} \,.
$$
where the left side is the scalar product of the second order Sobolev space and the right side is a functional of $\varphi$, meaning $F:H^4_0(\mathbb{R}) \rightarrow \mathbb{R}$. One can show that this functional is linear and bounded. Then by the Riesz representation theorem there exists a function $u \in H^4_0(\mathbb{R})$, such that
$$
F(\varphi) = \langle u, \varphi\rangle_{H^2(\mathbb{R})} \ \ \forall \varphi \in H^4_0(\mathbb{R}) \,.
$$
Now I am not sure if my reasoning is correct. Does this implicate that also $u \in H^4(\mathbb{R})$, or is it only $u \in H^2(\mathbb{R})$ since we only have the second order scalar product. I would be nice if someone could clear this up for me. Maybe my approach is flawed and there is a better way to show it. Many thanks in advance.