# Distributional derivative coincides with classical derivative?

I am having some trouble understanding the precise meaning of the following statement: "if $f \in C^1 (\Omega)$ for some $\Omega \subset \mathbb R^n$, then the distributional derivative of $f$ coincides with its classical derivative". I know that $f$ induces the following distribution: $$T_f (\phi) = \int f \phi ,$$ where the integral is taken over $\Omega$ $\phi \in \mathscr D (\Omega)$ is a test function (I'm assuming $\mathbb R^n = \mathbb R$ for simplicity), and then the distributional derivative of $f$ can be represented as $$\frac{d}{dx} T_f (\phi) = - \int f \frac{d \phi}{dx} .$$ What is then to be understood by saying that the classical derivative of $f$ is equal to its distributional derivative? Integration by parts shows that $$\frac{d}{dx} T_f (\phi) = - \int f \frac{d \phi}{dx} = - \left( - \int \phi f' \right) = T_{\frac{df}{dx} } (\phi) ,$$ therefore $$\frac{d}{dx} T_f (\phi) = T_{\frac{df}{dx} } (\phi) .$$ Since this last equation is valid for every test function $\phi$, does it then follow that both concepts of derivative are equivalent? Any comments would be much appreciated.

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@Jose: over what range are you integrating? –  Raskolnikov Apr 18 '11 at 23:07
@Jose: Don't you forget that your test functions have compact support? –  Jonas Teuwen Apr 18 '11 at 23:11
as noted above, $\phi$ has compact support so the $-f\phi$ term is zero and you have $$\frac{d}{dx}T_{\phi}=T_{\frac{df}{dx}}$$ –  yoyo Apr 18 '11 at 23:37
Alright, I've edited the question according to your comments. The last equation now implies the original proposition? –  Jose L. Lykón Apr 18 '11 at 23:44
Two things: first, the two terms $f\phi$ should not be there in the penultimate displayed equations. Second, if a distribution is represented by a $L_{\mathrm{loc}}^{1}$-function $g$ then this function is uniquely determined (a.e.). In particular, if this function $g$ can be chosen to be continuous, then it is unique. –  t.b. Apr 19 '11 at 1:00
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As you say, since $\phi$ has compact support, integration by parts yields
$$\frac{d}{dx} T_f (\phi) = - \int f \frac{d \phi}{dx} = \int \phi \frac{df}{dx} = T_{\frac{df}{dx} } (\phi)$$
Now if a distribution is represented by a locally integrable function $g$, then this function is unique up to null-sets by the fundamental lemma of calculus of variations. In particular, if this function $g$ can be chosen to be continuous, its continuous representative is unique. Therefore $\frac{d}{dx} T_f = T_{\frac{df}{dx}}$ as distributions. The higher-dimensional case is only more difficult notationally. Therefore differentiability in the sense of distributions generalizes classical differentiation.