Distributional derivative of non continuously differentiable functions

Question

Hello,

let $f$ be a continuously differentiable function on $R^n$. Then its classical derivative and its distributional derivative coincide.

It is known (cf. Rudin, Functional Analysis, Sect. 6.13) that this correspondence breaks down if $f$ is merely differentiable. Whereas in the sense of distributions $f$ is infinitely often differentiable, and obeys Schwartz' theorem, we usually can't investigate much more than the first derivatives in the classical meaning.

I want to train my intution with respect to different notions of derivation, and wonder how there is still a correspondence, or whether there is a point when both derivatives become virtually incommensurable.

Answer 1

There is a certain refinement of the question which turns out to have useful, interesting content, namely, talking about ($L^2$) Sobolev spaces, both local and global. For simplicity, on the real line, the 0th Sobolev space is just $L^2$. For positive integer $n$, there are three characterizations of the $n$th Sobolev space: closure of test functions under the $n$th Sobolev norm-squared:

$$|f|^2_n = |f|^2 + |f'|^2 + |f''|^2 + ...+|f^{(n)}|^2$$

closure of smooth functions whose n derivatives are in $L^2$, under the same norm, and the collection of distributions in $L^2$, whose n derivatives are in $L^2$.

The $\frac{d}{dx}$ extends by continuity to map $n$th to ($n-1$)th Sobolev space, and is "$L^2$ differentiation". It is not classical.

And/but this is not just a bunch of definitions: Sobolev's imbedding theorem shows that the nth Sobolev space is inside ($n-1$)-times continuously differentiable functions. (In dimension $N$, the discrepancy is $N/2 + \epsilon$.)

In higher dimensions, "elliptic regularity" is the assertion (proven decades ago) that operators $D$ such as the Laplacian have the property that $Du=f$ with $f$ in $n$th Sobolev implies $u$ is in the $n+\operatorname{deg}(D)$ Sobolev space. Part of the technical point here is that the proof really proves something about $L^2$ differentiability, not classical.

In fact, I would claim that this circle of ideas deserves to be part of every mathematician's worldview...

Answer 2

Let me try to answer the question how I understand it (basically, just to expand a bit on the comments by Harald and Willie).

Let $Df$ be a distributional derivative of a differentiable function $f:\mathbb R\to\mathbb R$. This implies, in particular, that $Df$ is a linear continuous functional on the space of test functions $\mathcal{D}(\mathbb R)$. By definition, $Df$ can be identified with a function $g:\mathbb R\to\mathbb R$ iff $$\langle Df,\phi\rangle=(g,\phi)\equiv\int_{\mathbb R} g(x)\phi(x)dx\qquad\qquad\qquad\qquad \quad(1)$$ for all $\phi\in\mathcal{D}(\mathbb R)$. If such $g$ exists, it is unique (up to modifications on a measure-zero set).

So the question is: given a differentiable function $f$, when does (1) hold with $g=f'$?

• First, in order for the integral in (1) to be finite for all $\phi\in\mathcal{D}(\mathbb R)$, the function $g=f'$ must be locally integrable on $\mathbb R$ in the sense of Lebesgue, i.e. $$\int\limits_{a}^{b}|g(x)|dx=\int\limits_{a}^{b}|f'(x)|dx < \infty, \quad \forall a,b\in\mathbb R.\qquad(2)$$ In other words, the function $f$ should have bounded variation on all finite intervals $[a,b]\subset\mathbb R$. Note, that $f'$ is always measurable so (2) simply means that the derivative is not wildly unbounded.

• Condition (2) is necessary but not sufficient. If $f$ is of bounded variation then by Lebesgue's decomposition theorem it can be written as $$f=f_{ac}+f_{sing}$$ where $f_{ac}$ is absolutely continuous and $f_{sing}$ is a step function. The example in Rudin's book shows that if $f_{sing}$ does not vanish then we cannot integrate by parts to get the identity $$\int_{\mathbb R} f'(x)\phi(x)dx=-\int_{\mathbb R} f(x)\phi'(x)dx\qquad\qquad\qquad(3)$$ and (1) also fails.

• Finally, if $f$ is absolutely continuous on $\mathbb R$ (i.e. $f=f_{ac}$) then $Df$ can be identified with $f'$ (one can justify the integration by parts in (3) and get (1) with $g=f'$ in this case). This is a necessary and sufficient condition.

Answer 3

G.B.Folland, Introduction to Partial Differential Equations, has a Proposition (0.33): Let Omega be an open domain in Rn. If u belongs to C(Omega) and the distributional derivatives D(superscript)alpha)u for all absolute(alpha) < or equal to k can be identified with C(Omega) functions (not yet identified as the continuous derivatives, just continuous functions), then u belongs to C superscript(k)(Omega). (i.e., the space of functions on the domain Omega with continuous derivatives of all orders less than or equal to k.

The proof proceeds by induction once it is established for k = 1. So he establishes the identity of distribution derivatives with classical continuous derivatives in the event that the distributional derivatives can be identified with or associated with some continuous functions as hinted at in equation (1) above. I found that his proof is difficult even though he says it's easy. Of course, some distributional derivatives don't correspond to an L1(loc) let alone a C function. In such an event the distributional derivative is not associated with a continuous function as in eq(1) above.