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I know the definitions of both weak and classical derivative. But I am trying to see the classical derivative as a weak derivative. When we have $\int f' \varphi = -\int f\varphi'$ for all $\varphi\in C^\infty$ with compact support. Is this definition consistent with the one for classical derivative, I mean, if $f\in C^1$ how can we obtain $f'$(classical) from the def. above?

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  • $\begingroup$ Have you tried integration by parts? Boundary conditions are zero if $\varphi\in C^{\infty}_c$. $\endgroup$ – David Jun 3 '13 at 11:01
  • $\begingroup$ And for the uniqueness you further need something like this $\endgroup$ – Ilya Jun 3 '13 at 11:02
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Integration by parts:

$$ \int_{-\infty}^\infty f\phi'\,dx = \lim_{\substack{a\to-\infty\\b\to\infty}} \left[ f\phi \right]_a^b - \int_{-\infty}^\infty f'\phi\,dx = - \int_{-\infty}^\infty f'\phi\,dx $$

since $\phi(a) = \phi(b) = 0$ if $|a|$ and $|b|$ are sufficiently large. (Remember that $\phi$ has compact support.)

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  • $\begingroup$ Yes sure thanks, but my question is, how do I have to choose $\varphi$ so that I obtain $f'$ when $f\in C^1$? $\endgroup$ – Mark Jun 3 '13 at 11:31
  • $\begingroup$ I am not asking to prove the integration by parts identity, but how to compute a classical derivative out of it when $f\in C^1$. In the formula above, it should say $f'\phi$ in the first integral btw. $\endgroup$ – Mark Jun 3 '13 at 11:59
  • $\begingroup$ Thanks for spotting the typo. I guess I don't understand your question though. The computation shows that the weak derivative of $f$ is $f'$, i.e. that the weak and classical derivatives coincide (when $f\in C^1$). $\endgroup$ – mrf Jun 3 '13 at 12:08
  • $\begingroup$ @Mark You can compute a classical derivative, then prove it like this that it is also the weak derivative. $\endgroup$ – Shuhao Cao Jun 3 '13 at 20:26
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I am guessing Mark's question is that how we know that the classical derivative is the weak derivative as functions. Or, more precisely, if there's a function $g$ (say in $L^p$) such that $\int g \phi = \int f' \phi = - \int f \phi'$ for all test functions $\phi$, is it necessary that $f' = g$ (almost everywhere)?

The question boils down to whether two functions that coincide when viewed as distributions are in fact the same as functions. Being the same as distributions is a priori a weaker notion than pointwise equality as functions. But one might resort to an approximation argument by taking the test function sequence $\{ \phi_n \}$ approaching the Dirac function in some sense.

Given a test function $\phi$, we can construct a convolution approximate unit $\{ \phi_t\}$ by defining $\phi_t(x) = \frac{1}{t} \phi(\frac{x}{t})$.

Fact: If $|\phi(x)| \leq \frac{C}{(1+|x|)^{1+\epsilon}}$ for some $C, \epsilon >0$, $\int \phi = 1$, and $f \in L^p$ ($1 \leq p \leq \infty$), then $f * \phi_t \rightarrow f$ pointwise on the Lebesgue set of $f$. In particular, we have pointwise convergence almost everywhere and where $f$ is continuous.

Such a $\phi$ with suitable decay can always be found. So if $f' \in L^p$ for some $p$ then any $g \in L^p$ that acts as the weak derivative of $f$ must be equal to $f'$ almost everywhere.

As for "computation", you can compute $f'$ in the usual way and quote uniqueness above.

Hope this helps.

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