Intuition behind the definition of the space $H_{0}^{1}(I) (=W_{0}^{1,2})$

Question

I am reading Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim Brezis. It is defined the Sobolev space $W^{1,p}(I)$ as:

$$W^{1,p}(I)= \{ u\in L^{p}(I) | \exists g \in L^p(I) :\int_{I}^{}u \varphi'= - \int_{I}^{}g \varphi \}$$

Where I is an open interval in $\mathbb{R}$.

Then it is stated that every function $u \in W^{1,p}(I) $ admits one (only one) continuous representative on I, i.e:

For every $u \in W^{1,p}(I) $ there exists $ \hat{u} \in C(\bar{I})$ such that $u= \hat{u}$ almost everywhere.

Then it is said that the book is not going to distinguish between $u$ and its continuous representative $\hat{u}$.

So I am a bit confused about the definition of $H_{0}^{1} (I)$. It is defined as:

$H_{0}^1 $ (Also denoted $W_{0}^{1,2}$) is the closure of $C^{1}_{c}(I)$ in $W^{1,2}(I)$.

And then used to solve some boundary value problems in one dimension. So the questions are:

• It doesn't matter whether the functions in $H_{0}^{1}(I)$ are continuous or not, as u and its continuous representative are the same in $L^p $, so they are the same in this context. Is this correct?
• I don't really get why is this space ($H_{0}^{1}(I)$) defined as a closure and how do its elements look like. Is it for the sole purpose that these elements are characterized by $u=0$ on $\delta u$?

Answer 1

By construction, $H_0^1(I)$ is a closed subspace of $H^1(I)$, hence for every $u\in H^1_0(I)$ there is $\hat u\in C(\bar I)$ such that $u=\hat u$ a.e.

The power of the definition of $H_0^1(\Omega)$ to be the closure of $C_c^1(\Omega)$ (traditionally it is defined as the closure of $C_c^\infty(\Omega)$!) is that it allows to approximate every function $u\in H^1_0(\Omega)$ by smooth functions.

This is usually exploited by the so-called density argument: First one proves a statement for smooth functions (usually involving complicated integrals that at first sight are difficult to justify for $H^1$ functions), then by density of the smooth functions in $H^1_0(\Omega)$ the statement holds true for the whole $H^1_0(\Omega)$.

This is in particular useful for $\Omega\subset \mathbb R^n$ with $n>1$: regardless of the regularity of the boundary of $\Omega$, the approximation stuff works for $H^1_0(\Omega)$ functions. However, it is much harder to justify such results for $H^1(\Omega)$-functions.

Answer 2

For Sobolev spaces You don´t want the restriction to continuous functions, because You need more general spaces to apply tricks from functional analysis. It´s just that in one dimension it turns out there is a continuous representative. Usually the closure of $C^{\infty}_c(U)$ in $W^2(U)$ for $U\subset \mathbb{R}^n$ is denoted $H_0(U)$(where $H$ suggests a Hilbert space) and these are the functions, that in a certain sense vanish on $\partial U$, i.e. for the Trace operator $T:W^2(U)\rightarrow L^2(\partial U)$You have $Tu\equiv 0$ if and only if $u\in H_0(U)$. That´s why they are of interest for certain boundary problems. Actually it makes no sense to say the function $u$ vanishes on the boundary if it is just defined almost everywhere. However in one dimension You can argue via the continuous representative.