Difference between revisions of "Aufgaben:Problem 15"

Revision as of 12:57, 24 June 2015

Let \( f \) be a continuous real-valued function on \( \mathbb{R}^n \). 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)

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

(2)

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

(3)

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

(4)

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

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}