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

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