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.
Distributional derivative coincides with classical derivative?
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0@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
1 Answers
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.