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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?2011-04-18
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    @Jose: Don't you forget that your test functions have compact support?2011-04-18
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    as noted above, $\phi$ has compact support so the $-f\phi$ term is zero and you have $$ \frac{d}{dx}T_{\phi}=T_{\frac{df}{dx}} $$2011-04-18
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    Alright, I've edited the question according to your comments. The last equation now implies the original proposition?2011-04-18
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    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.2011-04-19
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    @Theo: Right. I've edited it again. If you care to post that as an answer I'll mark it as correct. Thanks!2011-04-19

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As you say, since $\phi$ has compact support, integration by parts yields

$$\frac{d}{dx} T_f (\phi) = - \int f \frac{d \phi}{dx} = \int \phi \frac{df}{dx} = T_{\frac{df}{dx} } (\phi)$$

since there are no boundary terms.

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. Indeed, if $\int g \phi = \int g' \phi$ for all $\phi \in \mathcal{C}_0^{\infty}$, by that lemma $g - g' = 0$ up to a null set. That is, $g = g'$ almost everywhere.

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.