Difference between revisions of "Talk:Aufgaben:Problem 15"
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--[[User:Djanine|Djanine]] ([[User talk:Djanine|talk]]) 14:25, 11 June 2015 (CEST) | --[[User:Djanine|Djanine]] ([[User talk:Djanine|talk]]) 14:25, 11 June 2015 (CEST) | ||
Besides, we wanted to solve it as easily and short as possible. | Besides, we wanted to solve it as easily and short as possible. | ||
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| + | ---- | ||
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| + | === Alternative solution of the continuousness in part b) === | ||
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| + | 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 \( |T(\phi)| \le C \|\phi\|_{C_b^k(K)} := C \sum\limits_{|\alpha| \leq k}^{} \sup\limits_{x \in K} |\partial^\alpha \phi(x)|\). | ||
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| + | For \( k = 0\) this becomes \( C \sup\limits_{x \in K} |\phi(x)|\). | ||
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| + | In part a) we have shown that \( |1_B(\phi)| = |\langle 1_B,\phi \rangle| \le |B| \sup\limits_{x \in B}|\phi (x)| \enspace(\le |B| \sup\limits_{x \in \mathbb{R}^3}|\phi (x)| )\). | ||
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| + | 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. | ||
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| + | $$ \Rightarrow C \ge^! \frac{|B| \sup\limits_{x \in B}|\phi (x)|}{\sup\limits_{x \in K}|\phi (x)|} $$ | ||
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| + | 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. | ||
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| + | Thus it follow that \( 1_B\) is a distribution. | ||
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| + | \(\sup\limits_{x \in K}|\phi (x)| \neq 0\) not sure about that. | ||
Revision as of 12:58, 24 June 2015
--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.
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 follow that \( 1_B\) is a distribution.
\(\sup\limits_{x \in K}|\phi (x)| \neq 0\) not sure about that.
