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:


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


\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})…