Limit hulls?

In the course of trying to explain \liminf and \limsup “visually” I was led to this variation on the notion of convex hull. Given a set A in the plane, the convex hull is constructed by considering all the half-planes H that contain A and taking their intersection. The requirement A\subset H can be relaxed for instance by saying that A\setminus H has measure zero. But what happens if we instead ask that A\setminus H be bounded? Suppose for instance that A is the first quadrant. Then any half-plane of the form H_t=\{v: v\cdot (1,1)>t\}, for t>0, is a candidate and we see that their intersection is empty. But it isn’t empty if we pass to the projective plane. In that case we would get an arc of the line at infinity, which could maybe be called the “limit hull” of A. It seems the case, but I haven’t sat down to check, that whenever A is unbounded, its limit hull is non-empty.

The connection with \liminf was as follow. Let y=f(x) be the graph of your favorite function. First define the infimum of f “visually” by saying that “we draw a horizontal line under the graph and then raise it as far as we can until it bumps into the graph of f. The \liminf_{x\rightarrow+\infty} f(x) can be defined analogously by drawing a horizontal line under the graph and continue raising it as long as the set of x‘s where f is under the line forms a bounded set.

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Lectures about 5, 8 and 24

These three videos of John Baez talking about his favorite numbers are a lot of fun to watch.

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Benford’s law and Banach limits

In this nice short paper (gated), R. Raimi makes a connection between Benford’s law and Banach limits.

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An inequality useful for the p-Laplacian

Fix p>1. Let u,v\in{\mathbb R}^n, then

\displaystyle (u-v)\cdot(|u|^{p-2}u-|v|^{p-2}v)\geq 0.

Proof: Unfold and rewrite what we want to show as:

\displaystyle  (|u|^{p-2}+|v|^{p-2})(u\cdot v)\leq |u|^{p}+|v|^{p}.

By Cauchy-Schwarz the left hand-side is less than

\displaystyle |u|^{p-1}|v|+|v|^{p-1}|u|,

which can be written as

\displaystyle |u|^{p}+|v|^{p}+(|u|-|v|)(|v|^{p-1}-|u|^{p-1}).

But, since p>1,

\displaystyle (|u|-|v|)(|v|^{p-1}-|u|^{p-1})\leq 0.

Posted in Analysis | 2 Comments

Recreational question about factorials

Walking out of my Finite Math class I wondered if one can determine all possible triples of non-negative integers (a,b,c) such that

\displaystyle a!\ b!=c!

There are some trivial solutions. Whenever c=p! for some other non-negative integer p, then (c-1,p,c) is a solution.

Are there any other? (I haven’t put much thought to this question).

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A little-known definition of Hardy spaces

In the paper by Sedleckiń≠, A. M.
An equivalent definition of the H\sp{p} spaces in the half-plane, and some applications. (Russian)
Mat. Sb. (N.S.) 96(138) (1975), 75–82, 167, the following is proved: given an analytic function f on the upper-half plane {\mathbb H}=\{z: {\mathcal Im} z>0\} consider the following two quantities for 0<p<\infty:

\displaystyle \|f\|_{H_p}^p=\sup_{y>0}\int_{-\infty}^{+\infty}|f(x+iy)|^pdx

and

\displaystyle \|f\|_{H_p^\star}^p=\sup_{0<\theta<\pi}\int_0^\infty |f(re^{i\theta})|^pdr.

Then

\displaystyle A_p\|f\|_{H_p}\leq \|f\|_{H_p^\star}\leq B_p\|f\|_{H_p}

for some constants 0<A_p<B_p<\infty.

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Global harmonic miracle

That harmonic functions would satisfy the mean value property for infinitesimally small circles is intuitively somewhat clear. But that this property should persist on circles of arbitrary radius is quite miraculous. To recall, if h is harmonic on a domain \Omega\subset{\mathbb C} and one considers a disk D(z_0,r)\subset\Omega, then

\displaystyle h(z_0)=\frac{1}{\pi r^2}\int_{D(z_0,r)}h(w)dA(w).

One might wonder whether such a global property occurs for other shapes, say ellipses. The answer is no.

Fact: Suppose U\subset {\mathbb C} is open and has finite area A, z_0\in U, and

\displaystyle h(z_0)=\frac{1}{A}\int_U h(w)dA(w)

for every integrable harmonic function h on U. Then U is a disk centered at z_0.

To see why, pick a point z_1\not\in U that is closest to z_0. Then play with the function

\displaystyle h(z)=2{\mathcal Re}\left(\frac{z-z_0}{z-z_1}\right).

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