Sobolev space H^2(0,1) and H^3(0,1)

[mathematix89]

Hi there. I want to show (or have a reference that proves) that the Sobolev space \[ H^3(0,1) \] equipped with the inner product \[ (j ,v)_{H^3 (0,1) } = \int_0^1 j_{xxx} \; v_{xxx} \] is dense in the space \[ H^2(0,1) \] endowed with the scalar product \[ (j,v)_{H^2 (0 ,1 ) }= \int_0^1 j_{xx} \; v _{xx} \].
Thank you already

 

[jvicens]

for your help!Hi there,

Thank you for your question. The Sobolev space $H^3(0,1)$ is a space of functions that have three weak derivatives in $L^2(0,1)$. This means that the functions in $H^3(0,1)$ are not necessarily continuous, but they have well-defined weak derivatives. The inner product defined in this space is given by:

$$(j,v)_{H^3(0,1)} = \int_0^1 j_{xxx}v_{xxx} dx$$

Similarly, the Sobolev space $H^2(0,1)$ is a space of functions that have two weak derivatives in $L^2(0,1)$ and the inner product in this space is given by:

$$(j,v)_{H^2(0,1)} = \int_0^1 j_{xx}v_{xx} dx$$

To show that $H^3(0,1)$ is dense in $H^2(0,1)$, we need to show that for any function $v \in H^2(0,1)$, there exists a sequence of functions $\{j_n\} \subset H^3(0,1)$ such that $j_n \rightarrow v$ in $H^2(0,1)$. In other words, we need to show that for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, we have $\|j_n - v\|_{H^2(0,1)} < \epsilon$.

To prove this, we will use the fact that the space of smooth functions with compact support is dense in $H^2(0,1)$. Let $C^\infty_c(0,1)$ denote this space. Then, for any $v \in H^2(0,1)$, there exists a sequence $\{v_n\} \subset C^\infty_c(0,1)$ such that $v_n \rightarrow v$ in $H^2(0,1)$.

Now, we will construct the sequence $\{j_n\}$ as follows. Let $j_n = v_n + \phi_n$, where $\phi_n$ is a smooth function with compact support such that