Talk:Aufgaben:Problem 15
--Brynerm (talk) 11:38, 11 June 2015 (CEST) Just a remark. Wikipedia says \(T(\phi)\) continous \( \Leftrightarrow \exists k\in \mathbb{N}_0 \;|T(\phi)| \leq C \|\phi\|_{C_b^k(K)} := C \sum_{|\alpha| \leq k}^{} \sup_{x \in K}\left| \partial^\alpha \phi(x) \right|\)
That would probably relay more on part a) than your solution, but what you did, looks fine too to me.
--yzemp (talk) 14:19, 11 June 2015 (CEST) Yeah, we were aware of this definition, the problem is that we can't rely on wikipedia definitions when we solve these tasks. We found it better to stick to the script, this way we can be sure that we are allowed to use the definitions.
--Djanine (talk) 14:25, 11 June 2015 (CEST) Besides, we wanted to solve it as easily and short as possible.
In series 4 we had a similar exercise as in (b )there we needed to use the uniform convergence of \(\phi_j\) to show that \(\delta\) is a distribution. I remember using pointwise convergence which was marked as incorrect. Anyone know why pointwise convergence is not enough? I would think we need to do the same here. Edit: I think especially we need to consider the case where \(B\cap K = \emptyset\) separately. I don't think using only pointwise convergence is a problem. But it is convenient to use uniform convergence considering (a).
In both parts it is probably enough to prove (2) and (4) and then set f equal to one.
Carl (talk) 16:31, 30 July 2015 (CEST)
Alternative solution of the continuousness in part b)
Wikipedia says: a linear functional \(T: \mathcal D (\Omega) \to \mathbb{C} \) is a distribution \( \Leftrightarrow \forall \) compact subspaces \( K \subset \Omega \exists k\in \mathbb{N}_0 \) and \( \exists C>0 \) such that \( |T(\phi)| \le C \|\phi\|_{C_b^k(K)} := C \sum\limits_{|\alpha| \leq k}^{} \sup\limits_{x \in K} |\partial^\alpha \phi(x)|\).
For \( k = 0\) this becomes \( C \sup\limits_{x \in K} |\phi(x)|\).
In part a) we have shown that \( |1_B(\phi)| = |\langle 1_B,\phi \rangle| \le |B| \sup\limits_{x \in B}|\phi (x)| \enspace(\le |B| \sup\limits_{x \in \mathbb{R}^3}|\phi (x)| )\).
If we can show that \(|B| \sup\limits_{x \in B}|\phi (x)| \le C \sup\limits_{x \in K} |\phi(x)| \enspace \forall \) compact subspaces \( K \subset \mathbb{R}^3 \), the exercise will be done.
$$ \Rightarrow C \ge^! \frac{|B| \sup\limits_{x \in B}|\phi (x)|}{\sup\limits_{x \in K}|\phi (x)|} $$
As \(\sup\limits_{x \in K}|\phi (x)| \neq 0\) and \(|B| \sup\limits_{x \in B}|\phi (x)| < \infty \), we can find a finite \(C\) to fullfil this equation.
Thus it follows that \( 1_B\) is a distribution.
\(\sup\limits_{x \in K}|\phi (x)| \neq 0\) not sure about that.
Correct me, but I think this step is wrong. The absolute value should be taken inside the supremum.
$$ \left| \int_{B}\phi(x)dx \right| \le \left| |B| \sup_{x \in B}(\phi (x)) \right| $$
