Definition of Sobolev space: distributional derivative or weak derivative?

Question

I am studying about the basic property of Sobolev spaces.

1. In the PDE textbook by Evans, it defines the Sobolev space $W^{1,p}$ by requiring the weak derivative (and itself) to belong to $L^p.$

2. In the book Infinite dimensional dynamical systems in mechanics and physics by Temam, it defines the Sobolev space $W^{m,p}$ by requiring all distributional derivatives with order $\leq m$ to be in $L^p$. In this question, we may take $m=1$ for simplicity.

As it is known to us all, the concept of the distributional derivative is much weaker than the weak derivative. For example, each $f\in L^1_{loc}(U)$ has a distributional derivative, but is not necessary to have a weak derivative.

My question: are the two definitions equivalent (at least in some sense, for example, isomorphism), and why?

Answer 1

You should edit your question and include the definition of those "weak" and "distributional" derivatives. But anyway, let me try and give a conceptual, higher-level answer. (I will consider the space $W^{1, p}(\Omega)$, where $p\in (1, \infty)$ and $\Omega$ is an open subset of $\mathbb R^n$).

The particular type of generalized derivative you choose is not that important. What is conceptually important is that your space must be the completion of $C^1(\Omega)$ (or even $C^\infty(\Omega)$) under the norm $$ \lVert f\rVert_{1,p}=\left( \int_\Omega \lvert f(x)\rvert^p \, dx + \int_\Omega \lvert \nabla f(x)\rvert^p\, dx\right)^\frac 1p .$$ This is why you need Sobolev spaces. You need to have a complete space based on the Lp norm that involves the first derivatives.

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EDIT. Let me expand on these points.

1. Why do we want first derivatives? Well, because we want to study operators such as $-\Delta$, the Laplacian, which we can write as a bilinear form via formulas such as $$-\int_{\Omega}\Delta u(x)v(x)\,dx= \int_{\Omega}\nabla u(x) \cdot \nabla v(x)\, dx, $$ (note that here we implicitly set the boundary conditions, but let us sweep that under the rug). So first derivatives are the bare minimum we need.
2. Why do we want $L^p$ norms? A first reason, especially when $p=2$, is that these may be dictated by the underlying physical model (see, e.g. here and here). From a purely mathematical standpoint, the $L^p$ norms are uniformly convex; this is a favorable algebraic feature which implies deep compactness properties. Ultimately, these translate into good existence theorems for solutions to PDEs or related problems. (There is also another point, which however is especially important when $\Omega=\mathbb R^n$. The $L^p(\mathbb R^n)$ norm is the simplest one that supports scaling symmetry: replacing $f=f(x)$ by $f_\lambda(x)=f(\lambda x)$, for a $\lambda >0$, produces $\lVert f_\lambda\rVert_p=\lambda^{n/p}\lVert f \rVert_p$, modelling the change in units of measurement).
3. Everything in points 1 and 2 fundamentally relies on completeness. Simply put, you can't do any analysis on a non-complete space. This is clear since the ancient Greeks, who bashed their heads valiantly against problems such as the squaring of the circle or the diagonal of the square.

EXAMPLE. See the Wikipedia page on the Lax--Milgram theorem for the model example of this program. Note that the space which is used there is $H^1_0$, which in our notation corresponds to a closed subspace of $W^{1,2}$. This space is Hilbert, a property that is fundamentally used in the proof. Now, a Hilbert space is the prototypical example of a complete and uniformly convex space, like in the points 3 and 2 above.

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Your question can this be reformulated as:

What kind of generalised derivative should we choose, in order to constructively build such completion of $C^1(\Omega)$?

My answer is that this is less important. Whatever type of generalised derivative you choose you will end up with a complete space that has $C^1(\Omega)$ as a dense subspace. Therefore, all generalised derivatives should produce the same Sobolev space. And if a notion of generalised derivative DOES NOT yield a completion of $C^1(\Omega)$, then that is a most serious flaw. Presumably, that notion is not right.

Answer 2

As already hinted by Dean Miller, the two definitions are not only equivalent - they're exactly the same definition with a different phrasing. This becomes evident once you unpack these definitions. $\newcommand{\pl}{\partial}$

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What does it mean for $u \in L^p$ to have a weak derivative in $L^p$?

Well, the definition is: there are functions $\pl_i u \in L^p$ ($i = 1,2,\ldots,n$) satisfying $$ \int \pl_i u \cdot \varphi = - \int u \cdot \pl_i \varphi \quad \text{for } \varphi \in C_c^\infty. $$ (note that $\pl_i \varphi$ is the classical derivative, while $\pl_i u$ slightly abuses the notation - one has to understand that it's just the weak derivative, and that it's unique)

What does it mean for $u \in L^p$ to have a distributional derivative in $L^p$?

The only difference here is that you don't need more assumptions on $u$ to make sense of the distributional derivative. By definition, the $i$-th partial distributional derivative is the linear operator $D_i u$ given by $$ D_i u [\varphi] := - \int u \cdot \pl_i \varphi \quad \text{for } \varphi \in C_c^\infty. $$ Now, $D_i$ a linear operator, so what do we mean when we say "it's in $L^p$"? Actually, it's another abuse of language, and we actually mean that "$D_i u$ can be represented by an $L^p$ function", or more precisely, there is a function $\pl_i u \in L^p$ satisfying $$ D_i u [\varphi] = \int \pl_i u \cdot \varphi \quad \text{for } \varphi \in C_c^\infty. $$ (it's not uncommon to see even more notational abuse here, with $\pl_i u$ denoting both the distribution and the related function)

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As you can see, it's one and the same thing. The notions of distributions and distributional derivatives give us a general framework with very little assumptions, which is sometimes handy. But if you're given some particular PDE, you only consider derivatives up to some order, and typically you also need a complete space in order for standard existence methods to work - that's where Sobolev spaces come in. If your PDE is not linear (or with rough coefficients), then you have to assume these derivatives are represented by functions, anyway. Thus, people most often talk about weak derivatives, even if it's the same thing in the end.