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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''i)'': $$\wp(z)-\wp(a)=-\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})+\frac{d^2}{dz^2}\log\theta(z+\frac{1+\omega}{2})|_{z=a}=$$ $$-\frac{d}{dz}\frac{\theta '(z+\frac{1+\omega}{2})}{\theta(z+\frac{1+\omega}{2})}+const=$$ $$-\frac{\theta" (z+\frac{1+\omega}{2})\theta(z+\frac{1+\omega}{2})-(\theta '(z+\frac{1+\omega}{2}))^2}{\theta(z+\frac{1+\omega}{2})^2}+const$$
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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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''ii)'': \(\theta\) holomorphic \(\theta',\theta"\) holomorphic too, and then the uinique poles of \(\wp(z)-\wp(a)\equiv\) the zeros of \(\theta(z+\frac{1+\omega}{2})\)
 
  
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''iii)'': 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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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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''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.
 
  
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''v)'': \(\wp(z)-\wp(a)\) elliptic on \(\Gamma_\omega\Rightarrow\) every \(\gamma\in\Gamma_\omega\) is a pole of order 2.  
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== Solution of part a) ==
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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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''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 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 ''iv)'' we have \(N_\infty=2\Rightarrow N_0=2\)
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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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''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).
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=== (1) ===
  
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''viii)'': Ellipticity \(\Rightarrow z\) zero \(\iff z\in\left\{\pm a+\gamma | \gamma\in\Gamma_\omega\right\}\square\)
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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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'''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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=== (2) ===
  
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''Proof'':
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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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$$|B| \, \sup_{x \in B}|f(x)| \, \sup_{x \in B}|\phi (x)| \le  |B| \, \sup_{x \in B}|f(x)| \, \sup_{x \in \mathbb{R}^3}|\phi (x)| = |B| \, {||f||}_{L^\infty(B)} \, {||\phi||}_{L^\infty(\mathbb{R}^3)} $$
  
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''i) Periodic in two directions'': Let \(\gamma\in\Gamma_\omega=\mathbb{Z}+\omega\mathbb{Z}\), then $$-\frac{\sigma(z+a+\gamma)\sigma(z-a+\gamma)}{\sigma(z+\gamma)^2\sigma(a)^2}\overset{prop\ \theta -fct}{=}$$ $$-\frac{e^{a_\gamma(z+a)+b_\gamma}\sigma(z+a)e^{a_\gamma(z-a)+b_\gamma}\sigma(z-a)}{(e^{a_\gamma z+b_\gamma}\sigma(z))^2\sigma(a)^2}=$$ $$-\frac{e^{2a_\gamma z+2b_\gamma}\sigma(z+a)\sigma(z-a)}{e^{2a_{\gamma} z+2b_\gamma}\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}$$.
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=== (3) ===
  
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''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 \(\square\)
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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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'''Claim''': \(f,g\) elliptic with same pole, zeros then \(f=Ag\), with \(A\) const
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=== (4) ===
  
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''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 \(q(a_i)\neq 0\ \forall a_i, q(b_i)\neq 0\ \forall b_i\).
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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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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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== Solution of part b) ==
  
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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 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) \(\Rightarrow \ f(z)=\frac{h(z)}{l(z)}g(z)=Ag\ \square\)
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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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'''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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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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''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  \(\lim_{z\rightarrow 0}\wp(z)-\wp(a)=\lim_{z\rightarrow 0}K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\Rightarrow 1=\lim_{z\rightarrow 0}K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\)
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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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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 \(\lim_{z\rightarrow 0}\frac{\sigma(z)}{z}=\lim_{z\rightarrow 0}\prod(1-\frac{z}{\gamma})e^{\frac{z}{\gamma}+\frac{z^2}{2\gamma ^2}}=\prod1=1\Rightarrow\) $$1=\lim_{z\rightarrow 0}z^2(K\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2})=$$ $$\lim_{z\rightarrow 0}K\frac{z^2}{\sigma(z)^2}\frac{\sigma(z+a)\sigma(z-a)}{\sigma(a)^2}=$$ $$K\frac{\sigma(a)\sigma(-a)}{\sigma(a)^2}=$$ $$=K\frac{-\sigma(a)^2}{\sigma(a)^2}=-K\Rightarrow K=-1$$
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=== (1) ===  
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\(\Rightarrow \wp(z)-\wp(a)=-\frac{\sigma(z+a)\sigma(z-a)}{\sigma(z)^2\sigma(a)^2}\blacksquare\)
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== Exercise b) ==
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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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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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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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$$\wp '(a)=\lim_{z\rightarrow a} \frac{\wp(z)-\wp(a)}{z-a}\overset{a)}{=}$$ $$\lim_{z\rightarrow a}-\frac{\sigma(z+a)\sigma(z-a)}{(z-a)\sigma(z)^2\sigma(a)^2}=$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}\lim_{z\rightarrow a}\frac{\sigma(z-a)}{z-a}=$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}\lim_{z\rightarrow a}\frac{z-a}{z-a}\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}(1-\frac{z-a}{\gamma})e^{\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}\overset{Assume\ \prod<\infty}{=}$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}\prod_{\gamma\in\Gamma_{\omega},\gamma\neq 0}\lim_{z\rightarrow a}(1-\frac{z-a}{\gamma})e^{\frac{z-a}{\gamma}+\frac{(z-a)^2}{2\gamma ^2}}=$$ $$-\frac{\sigma(2a)}{\sigma(a)^4}\blacksquare$$
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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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