# How to define Biharmonic operator for second order Sobolev spaces and show that it is continuous?

I am studying an article where the authors assume that $\Omega \subset \mathbb{R}^N$, $N>4$ . Somewhere in the paper we have

$$\Delta^2 (\cdot) : W^{2,2}(\Omega) \to W^{-2,2}(\Omega)$$

(This operator is at the begining of the Page 11-th of article)

My question is this: I can not understand how does these linear operator has been defined? Because biharmonic operator is fourth-order but how can we define biharmonic operator for functions of second order Sobolev spaces?

Also author has said this is a continuous linear operator. How can I prove this is a continuous operator?

• The obvious guess for the definition is that $f\in W^{2, 2}_0(\Omega)$ satisfies $\Delta^2 g = 0$ weakly if $\int_\Omega \Delta f \Delta g = 0$ for all $f\in W^{2, ,2}(\Omega)$. – user99914 Sep 20 '15 at 4:41

Expanding on John Ma's comment: consider the bilinear form on $W^{2,2}$ defined by $$B(u,v)=\int_\Omega \Delta u\Delta v$$ Only two derivatives of each function are involved, so this form is bounded on $W^{2,2}$: $$|B(u,v)|\le C\|u\|_{W^{2,2}}\|v\|_{W^{2,2}}\tag{1}$$
Hence, for every $u\in W^{2,2}$ the map $v\mapsto B(u,v)$ is a bounded linear functional on $W^{2,2}$. Restrict it to $W_0^{2,2}$ and you have an element of $W^{-2,2}$ (which by definition is the dual of $W_0^{2,2}$). This element of $W^{-2,2}$ is what we call $\Delta^2 u$. This definition agrees with the classical bi-Laplacian for smooth functions, since for those we can integrate by parts, getting $B(u,v)=\int (\Delta^2 u )v$.
The continuity of $\Delta^2u$ with respect to $u$ follows from (1) as well, since $$\|v\mapsto B(u,v)\|_{W^{-2,2}}\le C\|u\|$$