# When viewing Sobolev spaces as completions, how does the notion of weak derivative arise?

Suppose instead of defining the Sobolev spaces $W^{k,p}(\Omega), \Omega\subset\Bbb R^n$ as the space of functions whose Sobolev norm (with weak derivatives) is finite, we define it as the completion of the subset of $C^\infty(\Omega)$ functions whose Sobolev norm is finite (call it $C^{k,p}(\Omega)$). By a theorem of topology, these spaces are homeomorphic, since $C^{k,p}(\Omega)$ is dense in both $W^{k,p}(\Omega)$ and $\text{comp}(C^{k,p}(\Omega))$. So while these constructions are equivalent...that's not very clear.

For instance, it is not clear what $Df$ is supposed to mean, when $f$ is an equivalence class of Cauchy sequences in $C^{k,p}(\Omega)$. (I also don't know how to interpret the equivalent class as a function $f:\Omega\to\Bbb R$ in the first place. I imagine it is an equivalence class of functions that agree a.e. somehow.) In the "standard" viewpoint, we have that $Df$ is just the weak derivative. So how does one interpret what $Df$ means in the completion viewpoint, and can one show that $\int Df\varphi=-\int fD\varphi$, $\forall \varphi\in C^1_0(\Omega)$, just like for weak derivatives?

If this makes sense for $\Bbb R^n$, then I would like to understand it on manifolds. The completion viewpoint seems to be dominant in the geometric analysis literature, but no one explains what $\nabla f$ is actually supposed to be. In Chavel (Eigenvales in Riem. Geo.), we find:

Given a function $f\in L^2(M)$, we say that $Y\in\mathscr L^2(M)$ is a weak derivative of $f$ if $$\int_M\langle Y,X\rangle=-\int_M f\operatorname{div}(X)$$ for all compactly supported $C^1$ vector fields $X$.

(Here $\mathscr L^2(M)$ are the square integrable vector fields.)

Now, this viewpoint is somewhat different from the usual PDE one since we use compactly supported vector fields...but I suppose this is just a reflection of the fact that $\partial f/\partial x^i$ has no intrinsic meaning on a manifold.

I am either looking for someone to clear up my questions here, or give a good reference on this subject.

• I don't understand your question. If the weak derivative $g_i = \partial_{x_i} f$ is $L^1$ then $f = \int g_i dx_i$ and $f$ is continuous in $x_i$. So if $\nabla f \in L^1$ then $f$ is continuous (or there is a continuous function in its $L^p$ equivalence class). And locally $L^p \subset L^1$ for $p > 1$ Dec 17, 2016 at 22:46
• @user1952009 I don't understand which part of the question you are trying to address. Dec 17, 2016 at 22:50
• It seems you want a course on the weak-derivative and the (closed) unbounded operators in $L^{p}$ spaces. And the finite difference operator $T_h f(x)= \frac{f(x+h)-f(x)}{h}$ converges in the $W^{1,p} \to L^p$ operator norm to the weak-derivative operator (as usual by density of the smooth functions, a consequence of the convolution) Dec 17, 2016 at 22:54

First, you need to separate the concept of completion from the proof of its existence (construction via Cauchy sequences). A completion is just a complete metric space $$W^{k,p}$$ that contains (an isometric copy of) $$C^{k,p}$$ as a dense subset. Alternatively, you could use the following universal property to define it:

any uniformly continuous function $$f \colon C^{k,p} \to N$$ to any complete metric space $$N$$ has a unique uniformly continuous extension $$\bar{f} \colon W^{k,p} \to N$$.

For example, the completion of $$\mathbb{Q}$$ is $$\mathbb{R}$$, but we usually do not have problems with intepreting elements of $$\mathbb{R}$$ as numbers. Why? Because we can extend the algebraic operations on $$\mathbb{Q}$$ to operations on $$\mathbb{R}$$.

Technically speaking, elements of $$W^{k,p}$$ in general are not functions, just as it is the case for $$L^p$$. The reason for this is that there is no meaningful way to define $$f(x)$$ for chosen $$x \in \Omega$$. If $$f$$ is fixed, Lebesgue differentiation theorem states that $$f$$ is approximately continuous at a.e. $$x \in \Omega$$ and thus we can make sense of $$f(x)$$, but the set of admissible points $$x$$ depends on $$f$$. It is also worth mentioning that if the product $$k \cdot p$$ is greater than the dimension of the domain (or if $$k$$ is equal to the dimension), elements of $$W^{k,p}$$ can be represented by continuous functions and $$f(x)$$ makes perfect sense: the evaluation map $$W^{k,p} \ni f \mapsto f(x)$$ is continuous for every $$x \in \Omega$$.

Still, some other useful operations on functions are well-defined on $$L^p$$, such as integration: $$L^p(\Omega) \ni f \mapsto \int_D f, \quad \text{if } D \subseteq \Omega,$$ or multiplication by bounded functions. Of course, this can also be done for $$W^{k,p}$$.

Now what are weak derivatives? Note that the operation of taking classical gradient $$C^{1,p} \ni f \mapsto \nabla f \in L^p$$ is uniformly continuous; remember that $$C^{1,p}$$ is considered with Sobolev norm. Hence we can extend it continuously to $$W^{1,p}$$ as its completion, defining the weak gradient $$\nabla f$$. This should answer one of your questions.

You can easily see that this coincides with the definition you mentioned. Take any $$\varphi \in C_c^\infty(\Omega,\mathbb{R}^n)$$ and define the linear functional $$S_\varphi \colon W^{1,p}(\Omega) \to \mathbb{R}$$ by $$W^{1,p}(\Omega) \ni f \mapsto \int \nabla f \varphi + \int f \operatorname{div} \varphi.$$ Since the operations of taking weak gradient, multiplying by a bounded function and integrating are well-defined and continuous (in respective spaces and norms), $$S_\varphi$$ is continuous. On the other hand, $$S_\varphi(f) = 0$$ for all $$f \in C^{1,p}$$, which is a dense subset. Hence $$S_\varphi \equiv 0$$ and we obtain the other definition: $$\int \nabla f \varphi = - \int f \operatorname{div} \varphi \quad \text{for } f \in W^{1,p}(\Omega).$$

• I think it is meaningful to ask what $f(x)$ means for a representative of a point in $L^p(\Omega)$/$W^{k,p}(\Omega)$/$W^{k,p}_0(\Omega)$. For instance, we can ask when a class in some Sobolev space contains a continuous or $C^k$ function, which does depend on pointwise estimates. It's not clear to me what $\nabla f$ really is in the completion viewpoint, especially on manifolds. Dec 18, 2016 at 20:27
• I would like for $\nabla f$ to be interpreted as some sort of equivalence class of $L^p$ rough vector fields, so that I can actually "choose" one of them for pointwise estimates that hold a.e. Dec 18, 2016 at 20:30
• Pointwise estimates are used all the time, for instance in the dominated convergence theorem. I don't know how you can say $f(x)$ has no meaning. We also have theorems on mollificiation or Lebesgue points that have $f(x)$ in them. Also, on a manifold the gradient is not a function, it's a vector field, and it's not clear to me at all that the $L^p$ space of vector fields is complete - the usual proof fails miserably. So it's not clear that the gradient can be extended since the target might not be a Banach space. Dec 19, 2016 at 11:34
• I added some explanations in the answer. The point is that in general (i.e. for every possible $f,x$) $f(x)$ doesn't make sense. One can always choose a function $f$ from the corresponding equivalence class and certain kinds of operations don't depend on the choice of $f$, but evaluation at a chosen point is not one of them. Dec 19, 2016 at 17:04
• As for Sobolev spaces on manifolds - can you edit your question and give a more precise description of your problem? I have trouble trying to understand your point. Vector fields are also functions (they just don't take values in $\mathbb{R}$) and it seems to me that the standard proof of completeness of $L^p$ applies also to vector fields on manifolds. The problem whether the two definitions of Sobolev spaces coincide is a deeper one. Dec 19, 2016 at 17:11