Given a real function $f\in L^1_{\text{loc}}(\Omega)$, 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.
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?
- weak derivatives is stronger than distributional derivatives? If
yes, how strong? - Is $\delta_0$ a $L^1_{loc}(-1,1)$, a locally integrable function?
see also this question. - Most PDE books use weak derivatives, not distributional one?
Best Answer
(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.