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

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In summary: H^3(0,1)$. Then, we have:$$\|j_n - v\|_{H^2(0,1)} = \|v_n + \phi_n - v\|_{H^2(0,1)} \leq \|v_n - v\|_{H^2(0,1)} + \|\phi_n\|_{H^2(0,1)}$$Since $v_n \rightarrow v$ in $H^2(0,1)$, the first term on the right-hand side goes to $0$ as $n \rightarrow \infty$. And since $\phi_n \rightarrow 0
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mathematix89
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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
 
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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
 

1. What is Sobolev space H^2(0,1)?

Sobolev space H^2(0,1) is a function space that contains all functions whose first and second derivatives are square integrable on the interval (0,1). It is a useful tool in mathematical analysis for studying the regularity of functions and solving differential equations.

2. How is Sobolev space H^2(0,1) different from H^3(0,1)?

While both Sobolev spaces H^2(0,1) and H^3(0,1) contain functions with square integrable first and second derivatives, H^3(0,1) also includes functions whose third derivatives are square integrable on the interval (0,1). This means that H^3(0,1) is a larger function space than H^2(0,1).

3. What are the applications of Sobolev space H^2(0,1)?

Sobolev space H^2(0,1) has many applications in mathematics and physics, including the study of elliptic partial differential equations, optimal control theory, and shape optimization problems. It is also used in the finite element method for numerical solutions of differential equations.

4. How is the norm defined in Sobolev space H^2(0,1)?

The norm in Sobolev space H^2(0,1) is defined as the sum of the squared norms of the function and its first and second derivatives. This norm measures the smoothness of a function and is used to characterize the regularity of functions in the space.

5. Can functions in Sobolev space H^2(0,1) be extended beyond the interval (0,1)?

Functions in Sobolev space H^2(0,1) can be extended to larger intervals, but the regularity of the function may decrease. For example, a function in H^2(0,1) may only have a continuous first derivative when extended to a larger interval, instead of a square integrable first derivative. This is known as the trace theorem in Sobolev spaces.

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