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Suppose that we have a sequence of functions $u_n \in W^{2,2}(\mathbb{R}^N)$, $N \geq 3$, such that $$ \|\nabla u_n\|_{L^2(\mathbb{R}^N)} = 1 \quad \text{and} \quad \|\Delta u_n\|_{L^2(\mathbb{R}^N)} = 1, \quad \forall n \in \mathbb{N}. $$

Are these assumptions enough to conclude that $\|u_n\|_{W^{2,2}(\mathbb{R}^N)} \leq C < +\infty$ for some $C > 0$ and all $n \in \mathbb{N}$?

Apart $\Delta u_n$, the norm $\|u_n\|_{W^{2,2}(\mathbb{R}^N)}$ contains also the second mixed derivatives, which probably could make $\|u_n\|_{W^{2,2}(\mathbb{R}^N)}$ unbounded. However, maybe the additional assumption $\|\nabla u_n\|_{L^2(\mathbb{R}^N)} = 1$ prevents this scenario?

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    The usual elliptic regularity estimate bounds the $W^{2,2}$ norm of $u$ in terms of the $L^2$ norms of $\Delta u$ and $u$. On a bounded domain you can get this latter bound from your gradient bound along with Poincaré's inequality, but I'm not sure how to handle the whole space (short of modifying your question by adding a decay estimate at infinity). – 2017-02-08
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    @AnthonyCarapetis Thanks for the reply! Indeed, in the case of a bounded domain we can say more (especially if we have zero traces). However, I'm also not aware about the equivalence of norms in the case of entire space... (you are right that $L^2$-norm of $u$ have to be included, but it is not clear about mixed derivatives). – 2017-02-08

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Take any $u \in C_c^\infty(\mathbb{R}^N)$ and integrate by parts twice: $$ \int |\Delta u|^2 = \int \sum_{i,j} u_{ii} u_{jj} = - \int \sum_{i,j} u_{iij} u_j = \int \sum_{i,j} u_{ij} u_{ij} = \int |Du|^2. $$ That is, $\|\Delta u\|_{L^2(\mathbb{R}^N)} = \|D^2 u\|_{L^2(\mathbb{R}^N)}$. By density of $C_c^\infty(\mathbb{R}^N)$ in $W^{2,2}(\mathbb{R}^N)$, the same equality holds for any $u \in W^{2,2}(\mathbb{R}^N)$.

We can apply Sobolev-Poincare inequality $\|u\|_{L^p(\mathbb{R}^N)} \le C_N \| Du \|_{L^2(\mathbb{R}^N)}$ with $p = \frac{2N}{N-2}$ to bound the $L^p$-norm, but the $L^2$-norm seems out of reach.

Edit. Previous version of this answer referred to Poincare's inequality in $\mathbb{R}^n$ with $p=2$, which is false. It can be seen by considering the family of functions $u_\lambda(x) := u(\lambda x)$ with different values of $\lambda \in \mathbb{R}$ that the only possible value for $p$ is $p = \frac{2N}{N-2}$.

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    Integration by parts - this is what I really missed) Thnx! But, just to be more precise, the Poincare inequality cannot be applied in $\mathbb{R}^N$. (However, for my aims it is not necessary, since I realised that I was needed to work not in $W^{2,2}(\mathbb{R}^3)$, but in the space $\{u:~ u \in L^6(\mathbb{R}^3),~ \nabla u \in W^{1,2}(\mathbb{R}^3)\}$, and $u \in L^6(\mathbb{R}^3)$ is achieved by assumption $\|\nabla u\|_{L^2} = 1$, due to the Sobolev inequality.) – 2017-02-09
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    Thanks. I corrected the answer accordingly - this mistake could be quite misleading. – 2017-02-09