The Poincaré-constant of a domain ${\Omega\subset\mathbb{R}^{n}}$ is the smallest constant ${C}$ such that the estimate

$\displaystyle \|u- \bar u\|_{p}\leq C\|\nabla u\|_{p}$

holds (where ${\bar u = |\Omega|^{-1}\int_{\Omega} u}$ is the mean value of ${u}$).

These constants are known for some classes of domains and some values of ${p}$: E.g. Payne and Weinberger showed in 1960 that for ${p=2}$ and convex ${\Omega}$ the constant is ${\tfrac{\mathrm{diam}(\Omega)}{\pi}}$ and Acosta and Duran showed in 2004 that for ${p=1}$ and convex ${\Omega}$ one gets ${\tfrac{\mathrm{diam}(\Omega)}2}$.

I do not know of any other results on these constants, and while playing around with these kind of things, I derived another one: Since I often work with images ${u:\Omega\rightarrow [0,1]}$ I will assume that ${0\leq u\leq 1}$ in the following.

Using the co-area formula we can express the ${1}$-norm of the gradient via the length of the level sets: Denoting by ${\mathcal{H}^{n-1}}$ the ${n-1}$-dimensional Hausdorff measure (which is, roughly speaking, the length in the case of ${n=2}$) and with ${E_{t}}$ the level set of ${u}$ at level ${t}$, the co-area formula states that

$\displaystyle \int_{\Omega}|\nabla u| = \int_{0}^{1}\mathcal{H}^{n-1}(\partial E_{t}) dt. \ \ \ \ \ (1)$

Now we combine this with the isoperimetric inequality, which is

$\displaystyle \mathcal{H}^{n-1}(\partial E_{t})\geq n |B_{1}|^{\tfrac1n}|E_{t}|^{\tfrac{n-1}{n}}, \ \ \ \ \ (2)$

where ${B_{1}}$ is unit ball and ${|\cdot|}$ denotes the Lebesgue measure. Combining~(1) and~(2) we get

$\displaystyle \int_{\Omega}|\nabla u| \geq n |B_{1}|^{\tfrac1n}\int_{0}^{1}|E_{t}|^{\tfrac{n-1}{n}} dt. \ \ \ \ \ (3)$

Now we use the trivial fact that

$\displaystyle u(x)^{2} = \int_{0}^{u(x)}2t dt = \int_{0}^{1}2t\chi_{\{x\mid u(x)\geq t\}}(x)dt$

combined with Fubini to get

$\displaystyle \int_{\Omega}u^{2} = \int_{0}^{1}2t\int_{\Omega}\chi_{\{x\mid u(x)\geq t\}}(x)dx\, dt = 2\int_{0}^{1}t |E_{t}|dt. \ \ \ \ \ (4)$

Since ${|E_{t}|/|\Omega|\leq 1}$ and ${(n-1)/n<1}$ it holds that ${|E_{t}|\leq |\Omega|^{\tfrac1n}|E_{t}|^{\tfrac{n-1}n}}$. Combining this with~(4) we get

$\displaystyle \int_{\Omega}u^{2}\leq 2|\Omega|^{\tfrac1n}\int_{0}^{1}t |E_{t}|^{\tfrac{n-1}n}dt \leq 2|\Omega|^{\tfrac1n}\int_{0}^{1} |E_{t}|^{\tfrac{n-1}n}dt.$

Finally, we use~(3) to obtain

$\displaystyle \int_{\Omega}u^{2}\leq \frac{2|\Omega|^{\tfrac1n}}{n|B_{1}|^{\tfrac1n}}\int_{\Omega}|\nabla u|$

in other words

$\displaystyle \|u\|_{2}^{2}\leq \frac{2|\Omega|^{\tfrac1n}}{n|B_{1}|^{\tfrac1n}}\|\nabla u\|_{1}.$

This estimate is quite explicit, does not need the subtraction of the mean value, does not need convexity of ${\Omega}$, but also does not obey the scaling ${u\mapsto au}$ (which is of no surprise since we used the condition ${0\leq u\leq 1}$ which also does not obey this scaling).

In dimension ${n=2}$ the estimate takes the simpler form

$\displaystyle \|u\|_{2}^{2}\leq \sqrt{\tfrac{|\Omega|}{\pi}}\|\nabla u\|_{1}.$

Im Wintersemester 2015/2016 habe ich die Vorlesung “Analysis 3” gehalten und dazu dieses Skript verfasst:

Diese Seite dient dazu, in den Kommentaren gefundene Fehler zu sammlen und hier zu dokumentieren.

Errata zur gedruckten Version:

• Seite 16: “$|\det(U)|$ und $\int_{\mathbb{R}}$ statt $\int_{\mathbb{-R}}$
• Seite 17, zweiter Absatz: “wobei $G$ eine Teilmenge des $\mathbb{R}^{n+1}$ ist”
• Seite 34 Beispiel 18.5 ii) Argument leicht umformuliert da vorher nicht ganz korrekt
• Seite 45: $\varphi_2 = \dots = \varphi''$
• Seite 58 Satz 19.3 Schritt2. die letzte Zeile. ” Dies heißt aber, dass $\varphi^1,\dots,\varphi^n$ linear abhängig sind. Widerspruch!”
Korrektur: “Dies heißt aber, dass $\psi^1,\dots,\psi^n$ linear abhängig sind. Widerspruch!”
• Seite 59 die erste Bemerkung (zweite Zeile): Lösungen $\varphi^k$ statt $\psi^k$;
• Seite 61 Korollar 19.7 “Fundamentalmatrix zu $y'=A(x)y$” statt “$y'=A(x)$“.
• Seite 70, Beweis von Satz 19.17: $+b(x)$ statt $-b(x)$.
• Seite 80, Beipsiel 19.29, 2.: $\mathrm{i}\omega$ ist eine Nullstelle.
• Seite 107, Satz 21.4: $\tilde\phi:\tilde T\to M$.
• Seite 116: $\omega_5 = \tfrac{8\pi^2}{3}$.
• Seite 137 unten: “aus $f_\nu\to f$ glm. folgt…”
• Seite 140, Beweis von Lemma 23.11: $\partial^\alpha\phi_\nu\stackrel{\mathcal{D}}{\to}\partial^\alpha\phi$ .
• Seite 141: $[x\phi(x)]_{-\infty}^0$
• Seite 152: $(LE)*\rho = \delta_0*\rho$

Das Skript zur zugehörigen Vorlesung “Analysis 2” ist hier zu finden.

Im Sommersemester 2015 habe ich die Vorlesung “Analysis 2” gehalten und dazu dieses Skript verfasst:

Diese Seite dient dazu, in den Kommentaren gefundene Fehler zu sammlen und hier zu dokumentieren.

Errata zur gedruckten Version:

• p. 54, Beweis zu Satz 13.32: In der letzten Zeile müsste $D_i f(x)$ statt $D_if(0)$ stehen.
• p. 61, Beweis zu Satz 14.8: Nach dem ersten Absatz müsste es heißen $v-y=A(g(v))(g(v)-g(y))$.
• p. 63, Beweis zu Satz14.11: In der Zeile unter (*) ist ein „gilt“ zu viel.
• p. 66, Beweis zu Satz14.13: In der vorletzten Zeile fehlt das „d“ von „passend“.
• p. 82. Lemma 15.16: In der ersten Zeile muss es  $j\neq k$ heißen.
• p. 93, Zeile über Beispiel 16.1: Es muss heißen „..allerdings keine brauchbaren Sätze”.
• p. 106, Beweis zu Satz 16.24: Im ersten Absatz, dritte Zeile: „Somit ist $x\mapsto D_t f(x,t)$…“
• p. 108, Beispiel 16.26: dritte Gleichung, zweites Integral muss heißen
$\int_0^\infty x^3\exp(-tx)\mathrm{d}x = \tfrac{6}{t^4}$.
• p. 108, Beispiel 16.26: vorletztes Integral muss heißen $\int_0^\infty x^n\exp(-x)\mathrm{d}x = n!$.
• p. 109,  zweiter Absatz muss lauten “… bei jedem Integral um einen Grenzwert…”.
• p. 109 letzter Satz muss lauten “…und den Größen…”.
• p. 110, Satz 16.27: Letzter Satz muss lauten “Ist $\lambda(M)<\infty$, so ist $f\in L^1(\mathbb{R})$.

Zur ergänzenden Vorlesung zur Analysis 2 für Physiker mit der knappen “Hands-on” Einführung in die Vektoranalysis gibt es hier ebenfalls das Skript:

Errata:

In my Analysis class today I defined the trigonometric functions ${\sin}$ and ${\cos}$ by means of the complex exponential. As usual I noted that for real ${x}$ we have ${|\mathrm{e}^{\mathrm{i} x}| = 1}$, i.e. ${\mathrm{e}^{\mathrm{i} x}}$ lies on the complex unit circle. Then I drew the following picture:

This was meant to show that the real part and the imaginary part of ${\mathrm{e}^{\mathrm{i} x}}$ are what is known as ${\cos(x)}$ and ${\sin(x)}$, respectively.

After the lecture a student came to me and noted that we could have started with ${a>1}$ and note that ${|a^{\mathrm{i} x}|=1}$ and could do the same thing. The question is: Does this work out? My initial reaction was: Yeah, that works, but you’ll get a different ${\pi}$

But then I wondered, if this would lead to something useful. At least for the logarithm one does a similar thing. We define ${a^x}$ for ${a>0}$ and real ${x}$ as ${a^x = \exp_a(x) = \exp(x\ln(a))}$, notes that this gives a bijection between ${{\mathbb R}}$ and ${]0,\infty[}$ and defines the inverse function as

$\displaystyle \log_a = \exp_a^{-1}.$

So, nothing stops us from defining

$\displaystyle \cos_a(x) = \Re(a^{\mathrm{i} x}),\qquad \sin_a(x) = \Im(a^{\mathrm{i} x}).$

Many identities are still valid, e.g.

$\displaystyle \sin_a(x)^2 + \cos_a(x)^2 = 1$

or

$\displaystyle \cos_a(x)^2 = \tfrac12(1 + \cos_a(2x)).$

For the derivative one has to be a bit more careful as it holds

$\displaystyle \sin_a'(x) = \ln(a)\cos_a(x),\qquad \cos_a'(x) = -\ln(a)\sin_a(x).$

Coming back to “you’ll get a different ${\pi}$”: In the next lecture I am going to define ${\pi}$ by saying that ${\pi/2}$ is the smallest positive root of the functions ${\cos}$. Naturally this leads to a definition of “${\pi}$ in base ${a}$” as follows:

Definition 1 ${\pi_a/2}$ is the smallest positive root of ${\cos_a}$.

How is this related to the area of the unit circle (which is another definition for ${\pi}$)?

The usual analysis proof goes by calculating the area of a quarter the unit circle by integral ${\int_0^1 \sqrt{1-x^2} dx}$.

Doing this in base ${a}$ goes by substituting ${x = \sin_a(\theta)}$:

$\displaystyle \begin{array}{rcl} \int\limits_0^1\sqrt{1-x^2}\, dx & = & \int\limits_0^{\pi_a/2}\sqrt{1-\sin_a(\theta)^2}\, \ln(a)\cos_a(\theta)\, d\theta\\ & = & \ln(a) \int\limits_0^{\pi_a/2} \cos_a(\theta)^2\, d\theta\\ & = & \ln(a) \frac12 \int\limits_0^{\pi_a/2}(1 + \cos_a(2\theta))\, d\theta\\ & = & \frac{\ln(a)}{2} \Big( \frac{\pi_a}{2} + \int\limits_0^{\pi_a/2}\cos_a(2\theta)\, d\theta\\ & = & \frac{\ln(a)\pi_a}{4} + 0. \end{array}$

Thus, the area of the unit circle is now ${\ln(a)\pi_a}$

Oh, and by the way, you’ll get the nice identity

$\displaystyle \pi_{\mathrm{e}^\pi} = 1$

(and hence, the area of the unit circle is indeed ${\ln(\mathrm{e}^\pi)\pi_{\mathrm{e}^\pi} = \pi}$)…