Difference between revisions of "Aufgaben:Problem 15"

Latest revision as of 13:08, 3 August 2015

Let $ f $ be a continuous real-valued function on $ \mathbb{R}^3 $. Let $B = B_1(0) $ denote the unit ball in $ \mathbb{R}^3 $ . We define

$$ 1_B= \begin{cases} 1 \quad x \in B \\ 0\quad x\notin B \end{cases} $$

and a linear map

\begin{split} f \cdot \mu_{\partial B} : \mathcal D (\mathbb{R}^3) &\rightarrow \mathbb{R}^3 \\ \phi &\mapsto (f \cdot \mu_{\partial B})(\phi) = \int_{\partial B} f(x') \phi(x') dx' \end{split}

where $ dx' $ is the surface volume element on $ \partial B $. In particular, $1 \cdot \mu_{\partial B} =: \mu_{\partial B} $

a) Show that $\forall \phi \in \mathcal D (\mathbb{R}^3) $

\begin{split} |\langle 1_B,\phi \rangle| &\le |B| {||\phi||}_{L^\infty(\mathbb{R}^3)} \qquad \qquad \quad &(1) \\ |\langle f \cdot 1_B,\phi \rangle| &\le |B| {||f||}_{L^\infty(B)}{||\phi||}_{L^\infty(\mathbb{R}^3)} &(2) \\ |\mu_{\partial B}(\phi)| &\le |\partial B| {||\phi||}_{L^\infty(\mathbb{R}^3)} &(3) \\ | (f \cdot \mu_{\partial B})(\phi )| &\le |\partial B| {||f||}_{L^\infty(B)}{||\phi||}_{L^\infty(\mathbb{R}^3)} &(4) \end{split}

b) Prove that $1_B, f \cdot 1_B, \mu_{\partial B}$ and $ f \cdot \mu_{\partial B}$ are distributions on $ \mathbb{R}^3 $.

Solution of part a)

First we note, that the space of test functions $\mathcal D (\mathbb{R}^3) \subset \mathcal S (\mathbb{R}^3) \subset L^\infty (\mathbb{R}^3) $ and because $\phi$ is in the space of testfunctions, it follows that $ \phi $ is integrable.

The extreme value theorem (Weierstrass) states, that a real valued function which is continuous on a compact subset of $ \mathbb{R}^n $ is bounded and takes its minimum and maximum in this subset. This will be used when talking about the function $ f $. It follows that $ f $ is bounded on the unit ball.

(1)

$$|\langle 1_B,\phi \rangle| = \left| \int_{\mathbb{R}^3} 1_B(x)\phi(x)dx \right| = \left| \int_{B}\phi(x)dx \right| \le \left| \, |B| \sup_{x \in B}(\phi (x)) \right| = |B| \, \left|\sup_{x \in B}(\phi (x)) \right| \le |B| \sup_{x \in B}|\phi (x)| \le |B| \sup_{x \in \mathbb{R}^3}|\phi (x)| = |B| \, {||\phi||}_{L^\infty(\mathbb{R}^3)}$$

(2)

$$|\langle f \cdot 1_B,\phi \rangle| = \left| \int_{\mathbb{R}^3} (f \cdot 1_B)(x)\phi(x)dx \right| = \left| \int_{B} f(x)\phi(x)dx \right| \le \left| \, |B| \sup_{x \in B}(f(x)\phi(x)) \right| \le |B| \, \left|\sup_{x \in B}(f(x))\right| \, \left|\sup_{x \in B}(\phi (x))\right| \le $$ $$|B| \, \sup_{x \in B}|f(x)| \, \sup_{x \in B}|\phi (x)| \le |B| \, \sup_{x \in B}|f(x)| \, \sup_{x \in \mathbb{R}^3}|\phi (x)| = |B| \, {||f||}_{L^\infty(B)} \, {||\phi||}_{L^\infty(\mathbb{R}^3)} $$

(3)

$$ |\mu_{\partial B}(\phi)| = |(1 \cdot \mu_{\partial B})(\phi)| = \left| \int_{\partial B} 1 \cdot \phi(x) dx \right| \le \left| \, |\partial B| \sup_{x \in \partial B} (\phi(x)) \right| = |\partial B| \, \left|\sup_{x \in \partial B} (\phi(x)) \right| \le |\partial B|\sup_{x \in \partial B} |\phi(x)| \le |\partial B|\sup_{x \in \mathbb{R}^3} |\phi(x)| = |\partial B| \, {||\phi||}_{L^\infty(\mathbb{R}^3)} $$

(4)

$$ | (f \cdot \mu_{\partial B})(\phi )| = |\int_{\partial B} f(x) \phi(x) dx| \le \left| \, |\partial B| \sup_{x \in \partial B}(f(x)\phi(x)) \right| = |\partial B| \, \left|\sup_{x \in \partial B}(f(x)\phi(x)) \right| \le |\partial B| \, \left|\sup_{x \in \partial B}(f(x)) \right| \, \left|\sup_{x \in \partial B}(\phi(x)) \right| \le $$ $$ |\partial B| \sup_{x \in \partial B}|f(x)| \sup_{x \in \partial B}|\phi(x)| \le |\partial B| \sup_{x \in B}|f(x)| \sup_{x \in \mathbb{R}^3}|\phi(x)| = |\partial B| \, {||f||}_{L^\infty(B)}{||\phi||}_{L^\infty(\mathbb{R}^3)} $$

Solution of part b)

A distribution is (by definition from the MMP II script "NewtonianPotential" page 8) a continuous linear functional on $ \mathcal D (\Omega)$, in our case $ \mathcal D (\mathbb{R}^3)$. The linear functional $u(\phi) $ is continuous when $\lim\limits_{j\rightarrow \infty}{u(\phi_j)} = u(\phi) $ for all sequences $ \phi_j$ , $j \ge 1$, in $ \mathcal D (\Omega)$ that converge to some $ \phi \in \mathcal D (\Omega)$.

Because our functionals have the form of an integral and we always integrate over a bounded subset of $\mathbb{R}^3 $ ($B$ or $\partial B$), we can take the limes inside the integral (we don't even have to apply the dominated convergence theorem (cheeer \o/ )).

Important to note is the definition of the functionals. While $\mu_{\partial B}$ and $ f \cdot \mu_{\partial B}$ are already defined as functionals, $1_B$ and $f \cdot 1_B$ are functions on $\mathbb{R}^3$. But we can easily define associated functionals over the scalar product, which is also a hint in the MMP II script "NewtonianPotential" page 9.

(1)

\begin{split} 1_B: \mathcal D(\mathbb{R}^3) &\to \mathbb{R} \\ \phi &\mapsto 1_B(\phi) = \langle 1_B , \phi \rangle = \int_{\mathbb{R}^3} 1_B(x)\phi(x)dx = \int_{B} \phi(x)dx \end{split}

First we show linearity. Let $a,b \in \mathbb{R}$ and $\phi, \psi \in \mathcal D(\mathbb{R}^3)$. $$ 1_B(a\phi+b\psi) = \int_{B} (a\phi+b\psi)(x)dx = \int_{B} a\phi(x)+b\psi(x)dx = a\int_{B}\phi(x)dx + b\int_{B}\psi(x)dx = a 1_B(\phi) + b 1_B(\psi) $$

Now the continuousness: Let $\phi_j \in \mathcal D(\mathbb{R}^3), j \ge 1$ be an arbitrary sequence in $\mathcal D(\mathbb{R}^3)$ with $\lim\limits_{j\to\infty}{\phi_j} = \phi$. We check the continuousness with the definition from the script.

$$ \lim\limits_{j \to \infty}{1_B(\phi_j)} = \lim\limits_{j \to \infty}{\int_{B}\phi_j(x)dx} = \int_{B} \lim\limits_{j \to \infty}{\phi_j(x)}dx = \int_{B} \phi(x) dx = 1_B(\phi)$$ Where we used that the $\phi_j(x)$ are bounded and therefore we can always find a function that is integrable on B and is bigger than all $\phi_j(x) \forall j,x$ (we can even choose a constant for that, because B is a compact subset of $\mathbb{R}^3$).

(2)

\begin{split} f \cdot 1_B: \mathcal D(\mathbb{R}^3) &\to \mathbb{R} \\ \phi &\mapsto (f \cdot 1_B)(\phi) = \langle f \cdot 1_B , \phi \rangle = \int_{\mathbb{R}^3} 1_B(x)f(x)\phi(x)dx = \int_{B} f(x)\phi(x)dx \end{split}

The linearity is exactly the same, as is the continuousness, for all the remaining parts.

(3)

\begin{split} \mu_{\partial B}: \mathcal D(\mathbb{R}^3) &\to \mathbb{R} \\ \phi &\mapsto \mu_{\partial B}(\phi) = \int_{\partial B} \phi(x)dx \end{split}

(4)

\begin{split} f \cdot \mu_{\partial B}: \mathcal D(\mathbb{R}^3) &\to \mathbb{R} \\ \phi &\mapsto (f \cdot \mu_{\partial B})(\phi) = \int_{\partial B} f(x)\phi(x)dx \end{split}

@@ Line 1: / Line 1: @@
-==Problem==
+Let \( f \) be a continuous real-valued function on \( \mathbb{R}^3 \). Let \(B = B_1(0) \) denote the unit ball in \( \mathbb{R}^3 \) . We define
-Recall, \(\vartheta\) is a ''theta function relative to a lattice'' \(\Gamma_{\omega}=\mathbb{Z}+\mathbb{Z}\omega\), \(\omega\in\mathbb{H}\), if \(\vartheta\) is entire and \(\forall\gamma\in\Gamma_{\omega}\ \exists a_{\gamma},b_{\gamma}\in\mathbb{C}\) such that \(\vartheta\left(z+\gamma\right)=e^{a_{\gamma}z+b_{\gamma}}\vartheta(z),\ \forall z\in\mathbb{C}\). Consider the following theta function relative to \(\Gamma_{\omega}\) (you may assume that the product converges and that \(\sigma\) satisfies the above definition): $$\sigma(z)=z\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\left(1-\frac{z}{\gamma}\right )e^{\frac{z}{\gamma}+\frac{z^{2}}{2\gamma^{2}}}$$
-''a)''  Show that \(\forall a\notin\Gamma_{\omega}\), $$\wp(z)-\wp(a)=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^{2}\sigma(a)^{2}}$$
+$$ 1_B= \begin{cases} 1 \quad x \in B \\ 0\quad  x\notin B  \end{cases} $$
-where \(\wp=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2}+c_1)\)   is the Weiertrass \(\wp\)-function, and \(\theta(z)=\sum_{n=-\infty}^{\infty}e^{2\pi inz}e^{\pi i\omega z^2}\)  is the ''Riemann theta function'' and \(c_1\) is the coefficient in the expansion \(\frac{d}{dz}\log\theta(z+\frac{1+\omega}{2})=\frac{1}{z}+c_0+c_1z+\cdots\)
-'''''Hint''''': show that both sides of the equality are elliptic funtions with the same poles and zeros.
+and a linear map
-''b)'' Show that $$\wp'(a)=-\frac{\sigma(2a)}{\sigma(a)^4},\ \forall a\notin\Gamma_{\omega}$$
+\begin{split} f \cdot \mu_{\partial B} : \mathcal D (\mathbb{R}^3) &\rightarrow \mathbb{R}^3 \\ \phi &\mapsto (f \cdot \mu_{\partial B})(\phi) = \int_{\partial B} f(x') \phi(x') dx' \end{split}
-== Solutions ==
+where \( dx' \) is the surface volume element on \( \partial B \). In particular, \(1 \cdot \mu_{\partial B} =: \mu_{\partial B} \)
-From '''''Freitag, Busam - Funktionentheorie''''' [Springer], Problem V.6.4 is exactly the same problem (the hint is related too): only to add some proofs.
-It says: "Auf beiden Seiten steht bei festem \(a\) eine elliptische Funktion mit denselben Nullstellen \((\pm a)\) und Polstellen (to prove 1). Daher stimmen sie bis auf einen konstanten Faktor überein (to prove 2). Für die Normierung benutzt man die Beziehung \(\lim_{z\rightarrow 0}z^2(\wp(z)-\wp(a))=1\). dass auf der rechten Seite dasselbe herauskommt, folgt aus den Relationen \(\sigma(a)=-\sigma(-a)\) und \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=1\), welche unmittelbar aus der Definition folgen (better see the passages)."
+a) Show that \(\forall \phi \in \mathcal  D (\mathbb{R}^3) \)
-From Serie 9, ex 1 we know that \(\wp\) is elliptic and then \(\wp(z)-\wp(a)\) is elliptic too (elliptic fct \(\pm\) const: still meromorphic and 2-per)
-'''Claim''': \(\wp(z)-\wp(a)\) has zeros first order in every \(z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\) and poles second order in every \(z\in\left\{\Gamma_\omega\right\}\)
+\begin{split} |\langle 1_B,\phi \rangle| &\le |B| {||\phi||}_{L^\infty(\mathbb{R}^3)} \qquad \qquad \quad &(1) \\  |\langle f \cdot 1_B,\phi \rangle| &\le |B| {||f||}_{L^\infty(B)}{||\phi||}_{L^\infty(\mathbb{R}^3)}  &(2) \\ |\mu_{\partial B}(\phi)| &\le |\partial B| {||\phi||}_{L^\infty(\mathbb{R}^3)} &(3) \\ | (f \cdot \mu_{\partial B})(\phi )| &\le |\partial B| {||f||}_{L^\infty(B)}{||\phi||}_{L^\infty(\mathbb{R}^3)} &(4)  \end{split}
-''Proof'':
-''i)'': $$\wp(z)-\wp(a)=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})+\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})|_{z=a}=$$ $$-\frac{d}{dz}\frac{\theta '(z+\frac{1+\omega}{2})}{\theta(z+\frac{1+\omega}{2})}+const=$$ $$-\frac{\theta" (z+\frac{1+\omega}{2})\theta(z+\frac{1+\omega}{2})-(\theta '(z+\frac{1+\omega}{2}))^2}{\theta(z+\frac{1+\omega}{2})^2}+const$$
+b) Prove that \(1_B, f \cdot 1_B, \mu_{\partial B}\) and  \( f \cdot \mu_{\partial B}\) are distributions on \( \mathbb{R}^3 \).
-''ii)'': \(\theta\) holomorphic \(\theta',\theta"\) holomorphic too, and then the uinique poles of \(\wp(z)-\wp(a)\equiv\) the zeros of \(\theta(z+\frac{1+\omega}{2})\)
-''iii)'': Proposition on lecture notes (p. 37 Fourier-Heat-ecc): \(\theta(\gamma+\frac{1+\omega}{2})=0\forall\gamma\in\Gamma_\omega\)
+== Solution of part a) ==
+First we note, that the space of test functions \(\mathcal D (\mathbb{R}^3) \subset \mathcal S (\mathbb{R}^3) \subset L^\infty (\mathbb{R}^3) \) and because \(\phi\) is in the space of testfunctions, it follows that \( \phi \) is integrable.
-''iv)'': $$\wp(z)-\wp(a)=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})+\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})|_{z=a}=$$ $$-\frac{d}{dz}(\frac{1}{z}+c_0+c_1z+\cdots)+const=$$ $$-(-\frac{1}{z^2}+c_1+\cdots)+const$$ \(\Rightarrow z=0\) pole of second order.
+The extreme value theorem (Weierstrass) states, that a real valued function which is continuous on a compact subset of \( \mathbb{R}^n \) is bounded and takes its minimum and maximum in this subset. This will be used when talking about the function \( f \). It follows that \( f \) is bounded on the unit ball.
-''v)'': \(\wp(z)-\wp(a)\) elliptic on \(\Gamma_\omega\Rightarrow\) every \(\gamma\in\Gamma_\omega\)  is a pole of order 2.
+=== (1) ===
-''vi)'': NB: thanks to ellipticity (double periodicity) we can "move" the parallelogram such that there are no poles / zeros on the boundary. Consider a new parallelogram \(P\) with \(z=0\) inside it. From complex analysis: Prop 1.6: \(f\ elliptic\Rightarrow \int_{\partial P} fdz=0\) and argument principle: \(\int_{\partial P} \frac{f'}{f}dz=2\pi i[N_0-N_infty]\). Where \(N_0,N_\infty\) are the number of zeros respectively poles with multiplicity conted. If \(f\ elliptic\Rightarrow \frac{f'}{f}\) is elliptic and then, from prop 1.6 \(\int_{\partial P} \frac{f'}{f}=0\Rightarrow N_0=N_\infty\). From ''iv)'' we have \(N_\infty=2\Rightarrow N_0=2\)
+$$|\langle 1_B,\phi \rangle| = \left| \int_{\mathbb{R}^3} 1_B(x)\phi(x)dx \right| = \left| \int_{B}\phi(x)dx \right| \le \left| \, |B| \sup_{x \in B}(\phi (x)) \right| = |B| \, \left|\sup_{x \in B}(\phi (x)) \right| \le |B| \sup_{x \in B}|\phi (x)| \le  |B| \sup_{x \in \mathbb{R}^3}|\phi (x)| = |B| \, {||\phi||}_{L^\infty(\mathbb{R}^3)}$$
-''vii)'': \(\wp(z)-\wp(a)=0\iff\wp(z)=\wp(a)\overset{\wp\  is\ even}{=}\wp(-a)\Rightarrow\z=\pm a\) are zeros of \(\wp(z)-\wp(a)\). From ''vi'' we know that they are the only zeros in \(P\) (W.l.o.g \(a\in P\) as we can move the parallelogram).
+=== (2) ===
-''viii)'': Ellipticity \(\Rightarrow z\) zero \(\iff z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\square\)
+$$|\langle f \cdot 1_B,\phi \rangle| = \left| \int_{\mathbb{R}^3} (f \cdot 1_B)(x)\phi(x)dx \right| = \left| \int_{B} f(x)\phi(x)dx \right| \le \left| \, |B| \sup_{x \in B}(f(x)\phi(x)) \right| \le |B| \, \left|\sup_{x \in B}(f(x))\right| \, \left|\sup_{x \in B}(\phi (x))\right| \le $$
+$$|B| \, \sup_{x \in B}|f(x)| \, \sup_{x \in B}|\phi (x)| \le  |B| \, \sup_{x \in B}|f(x)| \, \sup_{x \in \mathbb{R}^3}|\phi (x)| = |B| \, {||f||}_{L^\infty(B)} \, {||\phi||}_{L^\infty(\mathbb{R}^3)} $$
-'''Claim''': \(-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\) is elliptic and has zeros first order in every \(z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\) and poles second order in every \(z\in\left\{\Gamma_\omega\right\}\)
+=== (3) ===
-''Proof'':
+$$ |\mu_{\partial B}(\phi)| = |(1 \cdot \mu_{\partial B})(\phi)| = \left| \int_{\partial B} 1 \cdot \phi(x) dx \right| \le \left| \, |\partial B| \sup_{x \in \partial B} (\phi(x)) \right| = |\partial B| \, \left|\sup_{x \in \partial B} (\phi(x)) \right| \le |\partial B|\sup_{x \in \partial B} |\phi(x)| \le |\partial B|\sup_{x \in \mathbb{R}^3} |\phi(x)| = |\partial B| \, {||\phi||}_{L^\infty(\mathbb{R}^3)} $$
-''i) Periodic in two directions'': Let \(\gamma\in\Gamma_\omega=\mathbb{Z}+\omega\mathbb{Z}\), then $$-\frac{\sigma(z+a+\gamma)\sigma(z-a+\gamma)}{\sigma(z+\gamma)^2\sigma(a)^2}\overset{prop\ \theta -fct}{=}$$ $$-\frac{e^{a_\gamma(z+a)+b_\gamma}\sigma(z+a)e^{a_\gamma(z-a)+b_\gamma}\sigma(z-a)}{(e^{a_\gamma z+b_\gamma}\sigma(z))^2\sigma(a)^2}=$$ $$-\frac{e^{2a_\gamma z+2b_\gamma}\sigma(z+a)\sigma(z-a)}{e^{2a_{\gamma} z+2b_\gamma}\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}$$.
+=== (4) ===
-''ii) Meromorphic'' $$-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{(z+a)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z+a}{\gamma})e^{\frac{z+a}{\gamma}+\frac{(z+a)^2}{2\gamma ^2}}](z-a)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z-a}{\gamma})e^{\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}]}{(z\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}(1-\frac{z}{\gamma})e^{\frac{z}{\gamma}+\frac{z^2}{2\gamma ^2}})^2(a\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{a}{\gamma})e^{\frac{a}{\gamma}+\frac{a^2}{2\gamma ^2}})^2}=$$ $$-\frac{(z^2-a^2)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z+a}{\gamma})(1-\frac{z-a}{\gamma})e^{\frac{z+a}{\gamma}+\frac{(z+a)^2}{2\gamma ^2}+\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}]}{z^2a^2\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z}{\gamma})^2 (1-\frac{a}{\gamma})^2 e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{2a}{\gamma}+\frac{a^2}{\gamma ^2}}]}=$$ $$-\frac{(z^2-a^2)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z+a}{\gamma})(1-\frac{z-a}{\gamma})e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{a^2}{\gamma ^2}}]}{z^2a^2\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}[(1-\frac{z}{\gamma})^2 (1-\frac{a}{\gamma})^2 e^{\frac{2z}{\gamma}+\frac{z^2}{\gamma ^2}+\frac{2a}{\gamma}+\frac{a^2}{\gamma ^2}}]}=$$ $$-\frac{(z^2-a^2)\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\gamma^2(\gamma-z-a)(\gamma-z+a)e^{\frac{-2a}{\gamma}}}{z^2a^2\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}(\gamma-z)^2 (\gamma-a)^2}=$$ $$-\frac{(z+a)(z-a)}{z^2a^2}\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\gamma^2 e^{-\frac{2a}{\gamma}}\frac{(z+a-\gamma)(z-a-\gamma)}{(\gamma-z)^2(\gamma-a)^2}$$ \(\Rightarrow\) it's meromorphic and we observe that zeros and poles are as claimed. Meromorphic + 2-periodic = elliptical \(\square\)
+$$ | (f \cdot \mu_{\partial B})(\phi )| = |\int_{\partial B} f(x) \phi(x) dx| \le \left| \, |\partial B| \sup_{x \in \partial B}(f(x)\phi(x)) \right| = |\partial B| \, \left|\sup_{x \in \partial B}(f(x)\phi(x)) \right| \le |\partial B| \, \left|\sup_{x \in \partial B}(f(x)) \right| \, \left|\sup_{x \in \partial B}(\phi(x)) \right| \le $$
+$$ |\partial B| \sup_{x \in \partial B}|f(x)| \sup_{x \in \partial B}|\phi(x)| \le |\partial B| \sup_{x \in B}|f(x)| \sup_{x \in \mathbb{R}^3}|\phi(x)| = |\partial B| \, {||f||}_{L^\infty(B)}{||\phi||}_{L^\infty(\mathbb{R}^3)} $$
-'''Claim''': \(f,g\) elliptic with same pole, zeros then \(f=Ag\), with \(A\) const
+== Solution of part b) ==
-''Proof'': We know that a function \(k(z)\) with zeros \(a_1 ,\cdots ,a_m\), pole \(b_1, \cdots , b_r\) can be written as \(k(z)=q(z)\frac{\prod_{a_i\  zero}(z-a_i)}{\prod_{b_i\  pole}(z-b_i)}\) with \(q(z)\) analitic and \(q(a_i)\neq 0\ \forall a_i, q(b_i)\neq 0\ \forall b_i\).
+A distribution is (by definition from the MMP II script "NewtonianPotential" page 8) a continuous linear functional on \( \mathcal D (\Omega)\), in our case \( \mathcal D (\mathbb{R}^3)\). The linear functional \(u(\phi) \) is continuous when \(\lim\limits_{j\rightarrow \infty}{u(\phi_j)} = u(\phi) \) for all sequences \( \phi_j\) ,  \(j \ge 1\),  in \( \mathcal D (\Omega)\) that converge to some \( \phi \in \mathcal D (\Omega)\).
-Then we write $$f(z)=h(z)\frac{\prod_{a_i zero}(z-a_i)}{\prod_{b_i pole}(z-b_i)},\ \ g(z)=l(z)\frac{\prod_{a_i\  zero}(z-a_i)}{\prod_{b_i\ pole}(z-b_i)}$$ with \(h,l\) analytic and with prop. of above.
+Because our functionals have the form of an integral and we always integrate over a bounded subset of \(\mathbb{R}^3 \) (\(B\) or \(\partial B\)), we can take the limes inside the integral (we don't even have to apply the dominated convergence theorem (cheeer \o/ )).
-If \(f,g\) elliptic, then \(\frac{f}{g}\)  is elliptic too: \(\frac{f}{g}=\frac{h(z)}{l(z)}\) analitic as a quotient of analitic functions, then \(\frac{h(z)}{l(z)}\) has no singularities and then no pole. But an elliptic function without pole is constant \(q(z):=\frac{h(z)}{l(z)},\ q(\mathbb{C})=q(\Gamma_\omega)\) compact, so bounded and after Liouville is \(q(z)\) constant) \(\Rightarrow \ f(z)=\frac{h(z)}{l(z)}g(z)=Ag\ \square\)
+Important to note is the definition of the functionals. While \(\mu_{\partial B}\) and \( f \cdot \mu_{\partial B}\) are already defined as functionals, \(1_B\) and \(f \cdot 1_B\) are functions on \(\mathbb{R}^3\). But we can easily define associated functionals over the scalar product, which is also a hint in the MMP II script "NewtonianPotential" page 9.
-'''Claim''': Assuming that $$\wp(z)-\wp(a)=K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}$$ we claim that \(K=-1\)
+=== (1) ===
-''Proof'': $$\lim_{z\rightarrow 0} z^2(\wp(z)-\wp(a))=$$ $$\lim_{z\rightarrow 0}z^2\wp(z)-\overset{=0}{\overbrace{\lim_{z\rightarrow 0}z^2\wp(a)}}=$$ $$\lim_{z\rightarrow 0}z^2(-\frac{d}{dz}(\frac{d}{dz}\log\theta(z+\frac{1+\omega}{2})+c_1))=$$ $$-\lim_{z\rightarrow 0}z^2(\frac{d}{dz}(\frac{1}{z}+c_0+zc_1+\cdots))=$$ $$-\lim_{z\rightarrow 0}z^2(-\frac{1}{z^2}+c_1+\cdots)=1$$ but  \(\lim_{z\rightarrow 0}\wp(z)-\wp(a)=\lim_{z\rightarrow 0}K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\Rightarrow 1=\lim_{z\rightarrow 0}K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\)
+\begin{split} 1_B: \mathcal D(\mathbb{R}^3) &\to \mathbb{R} \\ \phi &\mapsto 1_B(\phi) = \langle 1_B , \phi \rangle = \int_{\mathbb{R}^3} 1_B(x)\phi(x)dx = \int_{B} \phi(x)dx \end{split}
-Now we prove that \(\sigma\) is odd and that \(\lim_{z\rightarrow 0}\sigma(z)/z=1\): \(\sigma(-z)=-z\prod(1+\frac{z}{\gamma})e^{\frac{-z}{\gamma}+\frac{z^2}{2\gamma^2}}\overset{\gamma '=-\gamma}{=}-z\prod(1-\frac{z}{\gamma '})e^{\frac{z}{\gamma '}+\frac{z^2}{2\gamma '^2}}=-\sigma(z)\) and \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=\lim_{z\rightarrow 0}\prod(1-\frac{z}{\gamma})e^{\frac{z}{\gamma}+\frac{z^2}{2\gamma ^2}}=\prod1=1\Rightarrow\) $$1=\lim_{z\rightarrow 0}z^2(K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2})=$$ $$\lim_{z\rightarrow 0}K\frac{z^2}{\sigma(z)^2}\frac{\sigma(z+a)\sigma(z-a)}{\sigma(a)^2}=$$ $$K\frac{\sigma(a)\sigma(-a)}{\sigma(a)^2}=$$ $$=K\frac{-\sigma(a)^2}{\sigma(a)^2}=-K\Rightarrow K=-1$$
+First we show linearity. Let \(a,b \in \mathbb{R}\) and \(\phi, \psi \in \mathcal D(\mathbb{R}^3)\).
-\(\Rightarrow \wp(z)-\wp(a)=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\blacksquare\)
+$$ 1_B(a\phi+b\psi) = \int_{B} (a\phi+b\psi)(x)dx = \int_{B} a\phi(x)+b\psi(x)dx = a\int_{B}\phi(x)dx + b\int_{B}\psi(x)dx = a 1_B(\phi) + b 1_B(\psi) $$
+Now the continuousness: Let \(\phi_j \in \mathcal D(\mathbb{R}^3), j \ge 1\) be an arbitrary sequence in \(\mathcal D(\mathbb{R}^3)\) with \(\lim\limits_{j\to\infty}{\phi_j} = \phi\). We check the continuousness with the definition from the script.
+$$ \lim\limits_{j \to \infty}{1_B(\phi_j)} = \lim\limits_{j \to \infty}{\int_{B}\phi_j(x)dx} = \int_{B} \lim\limits_{j \to \infty}{\phi_j(x)}dx = \int_{B} \phi(x) dx = 1_B(\phi)$$
+Where we used that the \(\phi_j(x)\) are bounded and therefore we can always find a function that is integrable on B and is bigger than all \(\phi_j(x) \forall j,x\) (we can even choose a constant for that, because B is a compact subset of \(\mathbb{R}^3\)).
+=== (2) ===
+\begin{split} f \cdot 1_B: \mathcal D(\mathbb{R}^3) &\to \mathbb{R} \\ \phi &\mapsto (f \cdot 1_B)(\phi) = \langle f \cdot 1_B , \phi \rangle = \int_{\mathbb{R}^3} 1_B(x)f(x)\phi(x)dx = \int_{B} f(x)\phi(x)dx \end{split}
+The linearity is exactly the same, as is the continuousness, for all the remaining parts.
+=== (3) ===
+\begin{split} \mu_{\partial B}: \mathcal D(\mathbb{R}^3) &\to \mathbb{R} \\ \phi &\mapsto \mu_{\partial B}(\phi) = \int_{\partial B} \phi(x)dx  \end{split}
+=== (4) ===
+\begin{split} f \cdot \mu_{\partial B}: \mathcal D(\mathbb{R}^3) &\to \mathbb{R} \\ \phi &\mapsto (f \cdot \mu_{\partial B})(\phi) = \int_{\partial B} f(x)\phi(x)dx  \end{split}

Difference between revisions of "Aufgaben:Problem 15"

Latest revision as of 13:08, 3 August 2015

Contents

Solution of part a)

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(2)

(3)

(4)

Solution of part b)

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