On distribution
Can you prove:
If $S$ is a distribution and $\phi$ is a test function then the function $t\rarr \in C^\infty$
If $S$ is a distribution and $\phi$ is a test function then the function $t\rarr
Risposte
What if we equipped $ccD'(Omega)$ with the weak* topology? In this case, it suffices to prove that
${psi in C_0^(oo)(Omega)" "|" " forall phi in C_0^(oo)(Omega): =0}={0}$
for the subspace $C_0^(oo)(Omega)$ to be weak* dense in $ccD'(Omega)$ (this is a consequence of the Hahn-Banach theorem).
Therefore we have to show that $0$ is the only function $psi in C_0^(oo)(Omega)$ for which
$ =int_(Omega) phi * psi " d"x = 0$, $forall phi in C_0^(oo)(Omega)$.
This follows from the Du Bois-Reymond lemma.
${psi in C_0^(oo)(Omega)" "|" " forall phi in C_0^(oo)(Omega):
for the subspace $C_0^(oo)(Omega)$ to be weak* dense in $ccD'(Omega)$ (this is a consequence of the Hahn-Banach theorem).
Therefore we have to show that $0$ is the only function $psi in C_0^(oo)(Omega)$ for which
$
This follows from the Du Bois-Reymond lemma.
Are you sure?
What I would expect is that such a result is true in the following weaker form (the same as for $L^p$-spaces):
$\Omega \subset RR^n$, $\omega \subset\subset \Omega$. Then for any $u \in ccD'$ there exists a sequence of functions $u_j \in C_0^\infty$ such that $u_j \to u$ in $ccD'(\omega)$ for $j \to \infty$.
In other words, I don't see how you can reach the boundary of $\Omega$.
What I would expect is that such a result is true in the following weaker form (the same as for $L^p$-spaces):
$\Omega \subset RR^n$, $\omega \subset\subset \Omega$. Then for any $u \in ccD'$ there exists a sequence of functions $u_j \in C_0^\infty$ such that $u_j \to u$ in $ccD'(\omega)$ for $j \to \infty$.
In other words, I don't see how you can reach the boundary of $\Omega$.
"irenze":
Ehm, $C_0^\infty$ is not dense in $ccD'$!!!
Simply because it is not dense in $L^\infty \subset ccD'$...
I knew the density result to be true for $Omega$ open $sub RR^n$. In fact, one can show that for any $u in ccD'(Omega)$ there exists a sequence of functions $u_j in C_0^(oo)(Omega)$ so that $u_j to u$ in $ccD'(Omega)$ for $j to oo$.
Ehm, $C_0^\infty$ is not dense in $ccD'$!!!
Simply because it is not dense in $L^\infty \subset ccD'$...
Simply because it is not dense in $L^\infty \subset ccD'$...
Consider a sequence $S_k$ of $C_0^(oo)$ functions converging to $S$ in $ccD'$. Such a sequence exists because $C_0^(oo)$ is dense in $ccD'$.
Then we're able to write
$ =int S_k psi$, $forall psi in C_0^(oo)$.
So, taking the $j$-th derivative with respect to $t$,
$(del^j)/(del t^j) =(del^j)/(del t^j)int S_k(x) phi(x+t)"d"x=int S_k(x) *(del^j)/(del t^j)phi(x+t)"d"x$.
But $(del^jphi(x+t))/(del t^j)=eta_j(x+t)$ is again $C_0^(oo)forall j in NN$, so, in the limit $k to oo$
$int S_k(x) *(del^j)/(del t^j)phi(x+t)"d"x=int S_k(x) eta_j(x+t) "d"x= to forall j$,
so the derivatives exist up to infinite order.
Then we're able to write
$
So, taking the $j$-th derivative with respect to $t$,
$(del^j)/(del t^j)
But $(del^jphi(x+t))/(del t^j)=eta_j(x+t)$ is again $C_0^(oo)forall j in NN$, so, in the limit $k to oo$
$int S_k(x) *(del^j)/(del t^j)phi(x+t)"d"x=int S_k(x) eta_j(x+t) "d"x=
so the derivatives exist up to infinite order.
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