Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
@Jose: over what range are you integrating? –  Raskolnikov Apr 18 '11 at 23:07
1  
@Jose: Don't you forget that your test functions have compact support? –  Jonas Teuwen Apr 18 '11 at 23:11
1  
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
show 1 more comment

1 Answer

up vote 2 down vote accepted

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. 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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.