# Singular support of distributions

I've read the following definition:

Let $u \in \mathcal{D}^\prime(\mathbb{R}^n)$. We say that $y_0 \notin \mathrm{sing} \ \mathrm{supp} \ u$ if there exists $\phi \in \mathcal{C}_c^\infty(\mathbb{R}^n)$ such that $\phi(y_0) \neq 0$ and $\phi u \in \mathcal{C}^\infty(\mathbb{R}^n)$.

However, after thinking for a while, I don't understand it very well.

Firstly, in general, the support of a function is a subset of its domain, isn't it? In this case, I guess the singular support is just a type of support -that of singularities-, but the domain of an element in $\mathcal{D}^\prime(\mathbb{R}^n)$ is not $\mathbb{R}^n$ but $\mathcal{C}_c^\infty(\mathbb{R}^n)$. However, $y_0 \in \mathbb{R}^n$. What if $u$ is not a distribution induced by some $L^1_{\text{loc}}(\mathbb{R}^n)$ function?

On the other hand, what's the meaning of $\phi u \in \mathcal{C}^\infty(\mathbb{R}^n)$? I mean, by definition, given $u \in \mathcal{D}^\prime(\mathbb{R}^n)$, we have that for every $\phi \in \mathcal{C}_c^\infty(\mathbb{R}^n)$ and $j = 1, \dots, n$ $$\partial_j u (\phi) = - u(\partial_j \phi).$$ So, it doesn't follow that $u$ is infinitely differentiable?

Thanks a lot for your help!

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It should be pointed out that $\partial_j u(\phi)=-u(\partial_j\phi)$ is taken to be the definition of the derivative of a distribution. The formula certainly doesn't imply that distributions are everywhere smooth. – icurays1 Nov 23 '12 at 5:17

Let us review some idea. If $f \in L^{1}_{\mathrm{loc}}(\Bbb{R}^n)$, then we can define an element $T_{f} \in \mathcal{D}'(\Bbb{R}^n)$ by

$$\left< T_{f}, \phi \right> = \int_{\Bbb{R}^n} f\phi, \quad \forall \phi \in C_{c}^{\infty}(\Bbb{R}^n).$$

Then the embedding $f \mapsto T_{f}$ allows us to identify $L^{1}_{\mathrm{loc}}(\Bbb{R}^n)$ as a subspace of $\mathcal{D}'(\Bbb{R}^n)$.

Of course we may not write $u = u(x) = \cdots$ for arbitrary $u \in \mathcal{D}'(\Bbb{R}^n)$. But let us take a deeper look on the identification above. We observe that the value of $f$ at $x = x_0$ is completely determined by the family of functions $f \phi$ such that $\phi$ is supported on a neighborhood of $x_0$. Then every claim on local property of $f$ can be translated in terms of the local data $\{ f \phi \}$. For example, the value of $f$ at $x = x_0$ is $y_0$ whenever

$$\lim_{\varepsilon \to 0} \left< T_{f}, \phi_{\varepsilon} \right> = y_0$$

for $\phi_{\varepsilon}(x) = \varepsilon^{-n} \phi\left( \varepsilon^{-1} (x - x_0) \right)$, where $\phi \in C^{\infty}_{c}(\Bbb{R}^n)$ supported on a neighborhood of $0$ and its total mass satisfies $\int_{\Bbb{R}^n} \phi = 1$. In this way, we can regard $f = T_{f}$ as a function-like object such that at each point $x$ there corresponds a local data of $f$. The term generalized function is used for this point of view.

The key point is that, this idea extends naturally to distributions as well. In particular, $u$ is completely determined by its local data at each point $x \in \Bbb{R}^n$. But in this case, the local data no longer need not yield a value of $u$ at $x_0$. This is why we define the support of a distribution to be a subset of $\Bbb{R}^n$.

Of course, it can happen that the local data give rise to the function value of $u$ at some points. Indeed, we can speak of the value of $u \in \mathcal{D}'(\Bbb{R}^n)$ at a point $x_0$ whenever the condition above holds with $u$ instead of $T_f$. Thus it may also happen that $u$ reduces to a function on some open set. A good example is the Dirac delta $\delta_0$, where $\delta_0 = 0$ on $\Bbb{R}^n \setminus \{0\}$.

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