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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.

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If $f$ is merely absolutely continuous, its classical and distributional derivatives still coincide, I believe? I don't have Rudin at hand, so I don't know what his example entails. –  Harald Hanche-Olsen Jul 4 '10 at 18:09
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If $f$ is merely absolutely continuous, doesn't its classical derivative only exist almost everywhere? –  Willie Wong Jul 4 '10 at 19:44
    
... my point being that even in the original question, the classical derivative of a $C^1$ function only coincides with the distributional derivative in the sense that the classical derivative is a (perhaps preferred) representative in a equivalence class; or in other words one can choose another function as the distributional derivative such that it differs from the "classical derivative" on a set of measure zero. –  Willie Wong Jul 4 '10 at 19:51
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@Harald: I hope I didn't cause any offence! I was just trying to observe that, granting all our comments are essentially contained in Rudin, unless the OP comes back and clarifies his thoughts, the question itself is not terribly meaningful. (In the sense that if our comments were on the mark, it is a case of "go back and re-read the book", and if our comments were off the mark, then god knows what the actual question is.) –  Willie Wong Jul 5 '10 at 9:38
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@Willie: No offence. We're in total agreement, using different words. –  Harald Hanche-Olsen Jul 5 '10 at 12:48

3 Answers 3

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 nth Sobolev space: closure of test functions under the nth 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 d/dx extends by continuity to map nth 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 nth Sobolev implies u is in the n+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...

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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.

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I think you somewhat misrepresented Lebesgue's decomposition theorem. There are “step functions” with an infinite number of steps, whose distributional derivatives are infinite linear combinations of delta functions – but more seriously, there are functions like the Cantor function en.wikipedia.org/wiki/Cantor_function whose distributional derivatives are singular but non-atomic measures. –  Harald Hanche-Olsen Jul 6 '10 at 15:49
    
Thank you for the comment. Well, the functions whose distributional derivatives have an atomic component are excluded by the condition $f'\in L_{loc}(\mathbb R)$. And the functions with singular non-atomic distributional derivatives are ruled out by the second bullet point: $f$ should be absolutely continuous. –  Andrey Rekalo Jul 6 '10 at 16:05

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.

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