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.

Is it possible to characterize the distributional derivative as some sort of "best linear approximation" of a distribution (a la the Fréchet/Gâteaux derivatives), if viewed in the appropriate spaces?

For example, let $f \in D'$, $\phi \in D$, and let $\psi_x \in D$ be a mollified ramp function so that $\psi_x(x,y,z,...)=x$ on the support of $\phi$. What can we say about the quantity:

$$\frac{f(\phi)-f(\phi+s\psi_x)}{s}-\frac{\partial f}{\partial x}(\phi)$$

where $\frac{\partial f}{\partial x}$ is the usual distributional derivative?

Going further, if we relax possibilities for "derivative direction", so that simply $\psi \in D$, does it even make sense to talk about the quantity "$Df:D'\rightarrow D'$":

$$Df(\phi)(\psi):=\lim_{s\rightarrow 0} \frac{f(\phi)-f(\phi+s\psi)}{s}$$ ?

In the space of distributions we don't have a norm, but we still have a vector space structure and a notion of convergence, so it seems that something like this is at least plausible.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

(I use below $f$ for the distribution and $\phi$ for the test function...)

Let $\phi\in D'$ and $f\in D$. Let $\tau_x:D\to D$ be translation, so that $\tau_x(f)(y)=f(y+x)$, and let $\tau^t_x:D'\to D'$ be the transpose of $\tau_x$. Then $\frac{\tau_{h}f-f}h\to f'$ in $D$ when $h\to0$, so $$\langle\frac{\tau^t_h\phi-\phi}h,f\rangle=\langle\phi,\frac{\tau_{h}f-f}h\rangle\to\langle\phi,f'\rangle=\langle \phi',f\rangle$$ as $h\to 0$. Since the topology in $D'$ is that of pointwise convergence on $D$, it follows that $$\lim_{h\to0}\frac{\tau^t_h\phi-\phi}h=\phi',$$ which is something along the lines of what you wanted.

NB: Differentialting the actual map $\phi:D\to\mathbb R$ is not going to produce anything interesting, because the map is linear itself.

share|improve this answer
    
Thanks, this is the sort of thing I was looking for. That $\phi:D\rightarrow \Re$ is already linear explains why I was going around and around in circles getting nowhere with my original approach. (: –  Nick Alger Dec 11 '10 at 15:19

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.