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.

  • $\begingroup$ 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 $p > 1$ $\endgroup$
    – reuns
    Dec 17, 2016 at 22:46
  • $\begingroup$ @user1952009 I don't understand which part of the question you are trying to address. $\endgroup$
    – Ryan Unger
    Dec 17, 2016 at 22:50
  • $\begingroup$ 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) $\endgroup$
    – reuns
    Dec 17, 2016 at 22:54

1 Answer 1


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: $$ \int \nabla f \varphi = - \int f \operatorname{div} \varphi \quad \text{for } f \in W^{1,p}(\Omega). $$

  • $\begingroup$ 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. $\endgroup$
    – Ryan Unger
    Dec 18, 2016 at 20:27
  • $\begingroup$ 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. $\endgroup$
    – Ryan Unger
    Dec 18, 2016 at 20:30
  • $\begingroup$ 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. $\endgroup$
    – Ryan Unger
    Dec 19, 2016 at 11:34
  • $\begingroup$ 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. $\endgroup$ Dec 19, 2016 at 17:04
  • $\begingroup$ 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. $\endgroup$ Dec 19, 2016 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.