Difference between revisions of "Aufgaben:Problem 15"

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==Problem==
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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
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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): $$\sigma(z)=z\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\left(1-\frac{z}{\gamma}\right )e^{\frac{z}{\gamma}+\frac{z^{2}}{2\gamma^{2}}}$$
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== Exercise a) ==
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Show that \(\forall a\notin\Gamma_{\omega}\), $$\wp(z)-\wp(a)=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^{2}\sigma(a)^{2}}$$
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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 \(\frac{d}{dz}\log\theta(z+\frac{1+\omega}{2})=\frac{1}{z}+c_0+c_1z+\cdots\)
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'''''Hint''''': show that both sides of the equality are elliptic funtions with the same poles and zeros.
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$$ 1_B= \begin{cases} 1 \quad x \in B \\ 0\quad  x\notin B  \end{cases} $$
  
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=== Solution a) ===
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and a linear map
  
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\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}
  
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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)
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where \( dx' \) is the surface volume element on \( \partial B \). In particular, \(1 \cdot \mu_{\partial B} =: \mu_{\partial B} \)
  
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'''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 \(z\in\Gamma_\omega\)
 
  
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''Proof'':
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a) Show that \(\forall \phi \in \mathcal  D (\mathbb{R}^3) \)
  
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# $$\begin{align}\wp(z)-\wp(a)=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})+\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})|_{z=a}=$$ $$-\frac{d}{dz}\frac{\theta '(z+\frac{1+\omega}{2})}{\theta(z+\frac{1+\omega}{2})}+const=$$ $$-\frac{\theta" (z+\frac{1+\omega}{2})\theta(z+\frac{1+\omega}{2})-(\theta '(z+\frac{1+\omega}{2}))^2}{\theta(z+\frac{1+\omega}{2})^2}+const\end{align}$$
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\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}
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#\(\theta\) holomorphic \(\theta',\theta"\) holomorphic too, and then the uinique poles of \(\wp(z)-\wp(a)\equiv\) the zeros of \(\theta(z+\frac{1+\omega}{2})\)
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#'Proposition on lecture notes (p. 37 Fourier-Heat-ecc): \(\theta(\gamma+\frac{1+\omega}{2})=0\forall\gamma\in\Gamma_\omega\)
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#$$\wp(z)-\wp(a)=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})+\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})|_{z=a}=$$ $$-\frac{d}{dz}(\frac{1}{z}+c_0+c_1z+\cdots)+c_1+const=$$ $$-(-\frac{1}{z^2}+2c_1+\cdots)+const$$ \(\Rightarrow z=0\) poles of second order.
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#\(\wp(z)-\wp(a)\) elliptic on \(\Gamma_\omega\Rightarrow\) every \(\gamma\in\Gamma_\omega\)  is a pole of order 2.
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#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 counted. 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 4. we have \(N_\infty=2\Rightarrow N_0=2\)
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#\(\wp(z)-\wp(a)=0\iff\wp(z)=\wp(a)\overset{\wp\  is\ even}{=}\wp(-a)\Rightarrow\ z=\pm a\) 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).
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#Ellipticity \(\Rightarrow z\) zero \(\iff z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\)
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<p style="text-align:right;">\(\square\)</p>
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<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
 
  
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'''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 \(z\in\left\{\Gamma_\omega\right\}\)
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b) Prove that \(1_B, f \cdot 1_B, \mu_{\partial B}\) and \( f \cdot \mu_{\partial B}\) are distributions on \( \mathbb{R}^3 \).
  
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''Proof'':
 
  
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#Periodic in two directions'': Let \(\gamma\in\Gamma_\omega=\mathbb{Z}+\omega\mathbb{Z}\), then $$-\frac{\sigma(z+a+\gamma)\sigma(z-a+\gamma)}{\sigma(z+\gamma)^2\sigma(a)^2}\overset{prop\ \theta -fct}{=}$$ $$-\frac{e^{a_\gamma(z+a)+b_\gamma}\sigma(z+a)e^{a_\gamma(z-a)+b_\gamma}\sigma(z-a)}{(e^{a_\gamma z+b_\gamma}\sigma(z))^2\sigma(a)^2}=$$ $$-\frac{e^{2a_\gamma z+2b_\gamma}\sigma(z+a)\sigma(z-a)}{e^{2a_{\gamma} z+2b_\gamma}\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}$$.
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== Solution of part a) ==
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#''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
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<p style="text-align:right;">\(\square\)</p>
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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.
  
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<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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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.
  
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'''Claim''': \(f,g\) elliptic with same poles, zeros then \(f=Ag\), with \(A\) const
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=== (1) ===
  
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''Proof'': We know that a function \(k(z)\) with zeros \(a_1 ,\cdots ,a_m\), poles \(b_1, \cdots , b_r\) 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 \(q(a_i)\neq 0\ \forall a_i, q(b_i)\neq 0\ \forall b_i\).
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$$|\langle 1_B,\phi \rangle| = \left| \int_{\mathbb{R}^3} 1_B(x)\phi(x)dx \right| = \left| \int_{B}\phi(x)dx \right| \le \left| \, |B| \sup_{x \in B}(\phi (x)) \right| = |B| \, \left|\sup_{x \in B}(\phi (x)) \right| \le |B| \sup_{x \in B}|\phi (x)| \le  |B| \sup_{x \in \mathbb{R}^3}|\phi (x)| = |B| \, {||\phi||}_{L^\infty(\mathbb{R}^3)}$$
  
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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.
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=== (2) ===
  
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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 poles. But an elliptic function without poles is constant since \(q(z):=\frac{h(z)}{l(z)},\ q(\mathbb{C})=q(P)\) compact, where \(P\) fundamental parallelogram, so bounded and after Liouville is \(q(z)\) constant) \(\Rightarrow \ f(z)=\frac{h(z)}{l(z)}g(z)=Ag\)
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$$|\langle f \cdot 1_B,\phi \rangle| = \left| \int_{\mathbb{R}^3} (f \cdot 1_B)(x)\phi(x)dx \right| = \left| \int_{B} f(x)\phi(x)dx \right| \le \left| \, |B| \sup_{x \in B}(f(x)\phi(x)) \right| \le |B| \, \left|\sup_{x \in B}(f(x))\right| \, \left|\sup_{x \in B}(\phi (x))\right| \le $$
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<p style="text-align:right;">\(\square\)</p>
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$$|B| \, \sup_{x \in B}|f(x)| \, \sup_{x \in B}|\phi (x)| \le  |B| \, \sup_{x \in B}|f(x)| \, \sup_{x \in \mathbb{R}^3}|\phi (x)| = |B| \, {||f||}_{L^\infty(B)} \, {||\phi||}_{L^\infty(\mathbb{R}^3)} $$
  
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<hr style="border:0;border-top:thin dashed #ccc;background:none;"/>
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=== (3) ===
  
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'''Claim''': Assuming that $$\wp(z)-\wp(a)=K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}$$ we claim that \(K=-1\)
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$$ |\mu_{\partial B}(\phi)| = |(1 \cdot \mu_{\partial B})(\phi)| = \left| \int_{\partial B} 1 \cdot \phi(x) dx \right| \le \left| \, |\partial B| \sup_{x \in \partial B} (\phi(x)) \right| = |\partial B| \, \left|\sup_{x \in \partial B} (\phi(x)) \right| \le |\partial B|\sup_{x \in \partial B} |\phi(x)| \le |\partial B|\sup_{x \in \mathbb{R}^3} |\phi(x)| = |\partial B| \, {||\phi||}_{L^\infty(\mathbb{R}^3)} $$
  
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''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  \(\lim_{z\rightarrow 0}\wp(z)-\wp(a)=\lim_{z\rightarrow 0}K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\Rightarrow 1=\lim_{z\rightarrow 0}z^2 K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\)
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=== (4) ===
  
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Now we prove that \(\sigma\) is odd and that \(\lim_{z\rightarrow 0}\sigma(z)/z=1\):
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$$ | (f \cdot \mu_{\partial B})(\phi )| = |\int_{\partial B} f(x) \phi(x) dx| \le \left| \, |\partial B| \sup_{x \in \partial B}(f(x)\phi(x)) \right| = |\partial B| \, \left|\sup_{x \in \partial B}(f(x)\phi(x)) \right| \le |\partial B| \, \left|\sup_{x \in \partial B}(f(x)) \right| \, \left|\sup_{x \in \partial B}(\phi(x)) \right| \le $$
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$$ |\partial B| \sup_{x \in \partial B}|f(x)| \sup_{x \in \partial B}|\phi(x)| \le |\partial B| \sup_{x \in B}|f(x)| \sup_{x \in \mathbb{R}^3}|\phi(x)| = |\partial B| \, {||f||}_{L^\infty(B)}{||\phi||}_{L^\infty(\mathbb{R}^3)} $$
  
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\(\sigma(-z)=-z\prod(1+\frac{z}{\gamma})e^{\frac{-z}{\gamma}+\frac{z^2}{2\gamma^2}}\overset{\gamma '=-\gamma}{=}-z\prod(1-\frac{z}{\gamma '})e^{\frac{z}{\gamma '}+\frac{z^2}{2\gamma '^2}}=-\sigma(z)\) and \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=\lim_{z\rightarrow 0}\prod(1-\frac{z}{\gamma})e^{\frac{z}{\gamma}+\frac{z^2}{2\gamma ^2}}=\prod1=1\Rightarrow\) $$1=\lim_{z\rightarrow 0}z^2(K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2})=$$ $$\lim_{z\rightarrow 0}K\frac{z^2}{\sigma(z)^2}\frac{\sigma(z+a)\sigma(z-a)}{\sigma(a)^2}=$$ $$K\frac{\sigma(a)\sigma(-a)}{\sigma(a)^2}=$$ $$=K\frac{-\sigma(a)^2}{\sigma(a)^2}=-K\Rightarrow K=-1$$
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== Solution of part b) ==
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\(\Rightarrow \wp(z)-\wp(a)=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\)
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<p style="text-align:right;">\(\blacksquare\)</p>
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<hr style="border:0;border-top:dashed #a6a6a6;background:none;"/>
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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 \( \phi \in \mathcal D (\Omega)\).
  
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<span style="color:#2f2f2f;font-size:85%;">''Note'': In the lecture we saw that \(\vartheta\) functions are analytic and then \(\sigma\) analytic: when we want to show that \(-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\) is meromorphic, it's enough write that as a product / combination of analytic function it's analytic too except points for which the denominator is zero. This means that the zeros of the function are the ones of the numerator, while the poles the ones of the denominator.</span>
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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/ )).
  
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== Exercise b) ==
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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.
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Show that $$\wp'(a)=-\frac{\sigma(2a)}{\sigma(a)^4},\ \forall a\notin\Gamma_{\omega}$$
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=== Solution b) ===
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=== (1) ===  
  
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$$\wp '(a)=\lim_{z\rightarrow a} \frac{\wp(z)-\wp(a)}{z-a}\overset{a)}{=}$$ $$\lim_{z\rightarrow a}-\frac{\sigma(z+a)\sigma(z-a)}{(z-a)\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}\lim_{z\rightarrow a}\frac{\sigma(z-a)}{z-a}=$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}\overset{a)}{=}$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}$$
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\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}
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<p style="text-align:right;">\(\blacksquare\)</p>
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First we show linearity. Let \(a,b \in \mathbb{R}\) and \(\phi, \psi \in \mathcal D(\mathbb{R}^3)\).
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$$ 1_B(a\phi+b\psi) = \int_{B} (a\phi+b\psi)(x)dx = \int_{B} a\phi(x)+b\psi(x)dx = a\int_{B}\phi(x)dx + b\int_{B}\psi(x)dx = a 1_B(\phi) + b 1_B(\psi) $$
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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.
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 +
$$ \lim\limits_{j \to \infty}{1_B(\phi_j)} = \lim\limits_{j \to \infty}{\int_{B}\phi_j(x)dx} = \int_{B} \lim\limits_{j \to \infty}{\phi_j(x)}dx = \int_{B} \phi(x) dx = 1_B(\phi)$$
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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 \(\mathbb{R}^3\)).
 +
 
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=== (2) ===
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\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}
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 +
The linearity is exactly the same, as is the continuousness, for all the remaining parts.
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=== (3) ===
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\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}
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=== (4) ===
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\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}

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}

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