For simplicity, given a real function $f\in L^1_{loc}(\Omega)$, we define both weak or distributional derivatives by $\int f'\phi = - \int f \phi'$ for all test functions $\phi$.

Now, take $\Omega = (-1,1)$, and $f(x) = I_{x>0}$ an indicator function.

Then, according to Example 2 of Section 5.2 of Evans' PDE book, there is no weak derivatives. see http://books.google.com.hk/books?id=Xnu0o_EJrCQC&lpg=PP1&vq=page%20243&pg=PA257#v=onepage&q&f=false

But, it is well known that $f' = \delta_0$ as a distribution. In fact, every distribution has its derivative according to Rudin's book on Functional Analysis, see Section 6.1.

At this far, can anybody clarify the following questions?

1) weak derivatives is stronger than distributional derivatives? How strong if yes?

2) is $\delta_0$ a $L^1_{loc}(-1,1)$, a locally integrable function? see also https://math.stackexchange.com/a/135866/79963

3) Most PDE books use weak derivatives, not distributional one?



(1) One usually wants to be weak derivatives to be functions, that is distributions represented by $\def\loc{\mathrm{loc}}$$L^1_\loc$-functions, that is we say that $f$ is weakly differentiable iff there is an $f' \in L^1_\loc$ such that $$ \int_\Omega f\phi' = -\int_\Omega f'\phi\quad \phi\in C_c^\infty(\Omega) $$ Now some functions, such as $I_{x>0}$ don't have a weak derivative. As every distribution $T \in \mathcal D'(\Omega)$ has derivatives of any order given by $T^{(n)}(\phi) = (-1)^n T(\phi^{(n)})$ in this case, the distributional derivatives cannot be represented by functions, see (2).

(2) No. $\delta$ cannot be represented by a function $g \in L^1_\loc(\Omega)$. To see this, suppose $g \in L^1_\loc(\Omega)$ were representing $\delta$, that is $$ \int_{-1}^1 g\phi = \delta(\phi) = \phi(0) $$ for all $\phi \in C^\infty_c(-1,1)$. Let $\phi_n$ be a sequence in $C^\infty_c$ such that $0\le \phi_n \le 1$, $\mathop{\rm supp} \phi_n \subseteq[-\frac 1n, \frac 1n]$, $\phi_n(0) = 1$. Then $\phi_n \to 0$ almost everywhere, hence $$ 1 = \phi_n(0) = \int_{-1}^1 \phi_n g \to 0 $$ contradiction.

(3) Also distributional derivatives which cannot be represented by functions play a role in PDE theory, so I cannot say, this holds for most books, for some, it surely does.


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