Ngô Quốc Anh

April 14, 2019

Extending functions between metric spaces: Continuity, uniform continuity, and uniform equicontinuity

Filed under: Giải Tích 3, Giải tích 8 (MA5206) — Tags: — Ngô Quốc Anh @ 15:02

This topic concerns a very classical question: extend of a function f : X \to Y between two metric spaces to obtain a new function \widetilde f : \overline X \to Y enjoying certain properties. I am interested in the following three properties:

  • Continuity,
  • Uniformly continuity,
  • Pointwise equi-continuity, and
  • Uniformly equi-continuity.

Throughout this topic, by X and Y we mean metric spaces with metrics d_X and d_Y respectively.

CONTINUITY IS NOT ENOUGH. Let us consider the first situation where the given function f : X \to Y is only assumed to be continuous. In this scenario, there is no hope that we can extend such a continuous function f to obtain a new continuous function \widetilde f : \overline X \to Y. The following counter-example demonstrates this:

Let X = [0,\frac 12 ) \cup (\frac 12, 1] and let f be any continuous function on X such that there is a positive gap between f(\frac 12+) and f(\frac12-). For example, we can choose

\displaystyle f(x)=\begin{cases}x^2&\text{ if } x<\frac 12,\\x^3 & \text{ if } x>\frac 12.\end{cases}

Since f is monotone increasing, we clearly have

\displaystyle f(\frac12-)-f(\frac 12+)=\frac18.

Hence any extension \widetilde f of f cannot be continuous because \widetilde f will be discontinuous at x =\frac 12. Thus, we have just shown that continuity is not enough. For this reason, we require f to be uniformly continuous.

SIMPLE OBSERVATIONS. We start with the following basic results.

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October 18, 2018

Jacobian determinant of diffeomorphisms measures the quotient of area of small balls

Filed under: Uncategorized — Ngô Quốc Anh @ 23:50

This post concerns a widely mentioned feature of the Jacobian determinant of diffeomorphisms whose proof is not easy to find. The precise statement of the result is as follows:

Geometric meaning of the Jacobian determinant: Let U \subset \mathbf R^n be open and \phi : U \to \phi (U) be a diffeomorphism in \mathbf R^n. Fix a point a \in U. Then

\displaystyle |\det J_\phi (a) | = \lim_{r \searrow 0} \frac{{\rm vol}(\phi(B(a,r)))}{{\rm vol}(B(a,r))},

where B(a, r) denotes the open ball in \mathbf R^n centered at a with radius r.

As a remark and to be more exact, we require \phi to be a C^1-diffeomorphism. Before proving the above result, it is worth noting that it is true for linear maps, whose proof is not hard. One way to realize this is to make use of the change of variable formula for multiple integrals. The proof presented here is inspired by the proof of Lemma 5.1.12 in this book.

We now proceed with the proof whose proof is divide into a few steps.

Step 1. First we use \| \cdot \| to denote a norm on \mathbf R^n. Clearly, because \phi is a C^1-diffeomorphism we can write

\displaystyle \phi (x) = \phi (a) + J_\phi (a) \cdot (x- a) + \vec \varepsilon (x) \| x- a\|,

where the error vector-valued function \vec \varepsilon enjoys the following properties

\displaystyle \| \vec \varepsilon (x) \| \to 0

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September 14, 2018

Leibniz rule for proper integral with parameter whose limits also depends on the parameter

Filed under: Uncategorized — Ngô Quốc Anh @ 21:24

The following Leibniz integral rule is well-known

Theorem. Let f(x, t) be a function such that both f(x, t) and its partial derivative f_x(x, t) are continuous in t and x in some region of the (x, t)-plane, including a(x) \leqslant t \leqslant b(x), x_0 \leqslant x \leqslant x_1. Also suppose that the functions a(x) and b(x) are both continuous and both have continuous derivatives for x_0 \leqslant x \leqslant x_1. Then, for x_0 \leqslant x \leqslant x_1,

\displaystyle \frac {d}{dx}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)=f\big (x,b(x)\big )b'(x)-f\big (x,a(x)\big) a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial f }{\partial x}}(x,t)\,dt.

The purpose of this note is to show that, in fact, it is not  is not necessary to assume the function f to be continuous. We note that this is indeed the case in which the limits of the integral \int_{a(x)}^{b(x)}f(x,t)\,dt do not depend on the parameter x. For convenience, it is routine to assume the continuity, which immediately implies that all integrals are well-defined.

As mentioned above, we want to show that this is also the case for integrals of the form above.

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October 9, 2017

Jacobian of the stereographic projection not at the North pole

Filed under: Uncategorized — Ngô Quốc Anh @ 18:57

Denote \pi_P : \mathbb S^n \to \mathbb R^n the stereographic projection performed with P as the north pole to the equatorial plane of \mathbb S^n. Clearly when P is the north pole N, i.e. N = (0,...,0,1), then \pi_N is the usual stereographic projection.

As we have already known that, for arbitrary \xi \in \mathbb S^n, the image of \xi is

\displaystyle \pi_P : \xi \mapsto x = P+\frac{\xi-P}{1-\xi \cdot P}.

Here the point x \in \mathbb R^n is being understood as a point in \mathbb R^{n+1} by adding zero in the last coordinate. For the inverse map, it is not hard to see that

\displaystyle \pi_P^{-1} : x \mapsto \xi =\frac{|x|^2-1}{|x|^2+1}P+\frac 2{|x|^2+1}x.

The purpose of this entry is to compute the Jacobian of the, for example, \pi_P^{-1} by comparing the ratio of volumes.

First pick two arbitrary points x, y \in \mathbb R^n and denote \xi = \pi_P^{-1}(x) and \eta = \pi_P^{-1}(y). The Euclidean distance between \xi and \eta is

\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.

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October 3, 2017

Stereographic projection not at the North pole and an example of conformal transformation on S^n

Filed under: Riemannian geometry — Tags: , — Ngô Quốc Anh @ 12:09

Denote \pi_P : \mathbb S^n \to \mathbb R^n the stereographic projection performed with P as the north pole to the equatorial plane of \mathbb S^n. Clearly when P is the north pole N, i.e. N = (0,...,0,1), then \pi_N is the usual stereographic projection.

Clearly, for arbitrary x \in \mathbb S^n, the image of x is

\displaystyle \pi_P : x \mapsto y = P+\frac{x-P}{1-x \cdot P}.

For the inverse map, it is not hard to see that

\displaystyle \pi_P^{-1} : y \mapsto x =\frac{|y|^2-1}{|y|^2+1}P+\frac 2{|y|^2+1}y.

Derivation of \pi_P and \pi_P^{-1} are easy, for interested reader, I refer to an answer in . Let us now define the usual conformal transformation \varphi_{P,t} : \mathbb S^n \to \mathbb S^n given by

\displaystyle \varphi_{P,t} : x \mapsto \pi_P^{-1} \big( t \pi_P ( x) \big)

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February 23, 2017

In a normed space, finite linearly independent systems are stable under small perturbations

Filed under: Giải tích 8 (MA5206) — Ngô Quốc Anh @ 23:21

In this topic, we show that in a normed space, any finite linearly independent system is stable under small perturbations. To be exact, here is the statement.

Suppose (X, \|\cdot\|) is a normed space and \{x_1,...,x_n\} is a set of linearly independent elements in X. Then \{x_1,...,x_n\} is stable under a small perturbation in the sense that there exists some small number \varepsilon>0 such that for any \|y_i\| < \varepsilon with 1 \leqslant i \leqslant n, the all elements of \{x_1+y_1,...,x_n+y_n\} are also linearly independent.

We prove this result by way of contradiction. Indeed, for any \varepsilon>0, there exist n elements y_i \in X with \|y_i\| < \varepsilon such that all elements of \{x_1+y_1,...,x_n+y_n\} are linearly dependent, that is, there exist real numbers \alpha_i with 1 \leqslant i \leqslant n such that

\displaystyle \alpha_1 (x_1+y_1) + \cdots + \alpha_n (x_n+y_n) =0

with

\displaystyle |\alpha_1| + \cdots + |\alpha_n| >0.

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December 11, 2016

Why the equation Δu+1/u^α=0 in R^n has no positive solution?

Filed under: Uncategorized — Ngô Quốc Anh @ 23:50

Following the idea due to Brezis in his note that we had already mentioned once, it seems that we can show that the following equation

\displaystyle \Delta u + \frac 1{u^\alpha}=0

in \mathbb R^n with n\geqslant 2 has no positive C^2-solution. Here we require \alpha>0.

Indeed, to see this, we first denote

f(t)=-t^{-\alpha}

with t>0. Clearly the function f is monotone increasing in its domain. Using this, we rewrite our equation as follows

\Delta u =f(u).

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September 9, 2016

Benefits of “complete” and “compact” for analysis on Riemannian manifolds

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 10:08

When working on Riemannian manifolds, it is commonly assumed that the manifold is complete and compact. (The case of non-compactness is also of interest too.) In this entry, let us review the role of completeness and compactness in this setting.

How important the completeness is? Let us recall that for given a Riemannian manifold (M,g), what we have is a nice structure as well as an appropriate analysis on any tangent space T_pM. For a C^1-curve \gamma : [a,b] \to M on M, the length of \gamma is

\displaystyle L(\gamma) = \int_a^b \sqrt{g(\gamma (t)) \langle \partial_t \gamma \big|_t, \partial_t \gamma \big|_t\rangle} dt

where \partial_t \gamma\big|_t \in T_{\gamma (t)}M a tangent vector. (Note that by using curves, the tangent vector \partial_t \gamma\big|_t is being understood as follows

\displaystyle  \partial_t \gamma\big|_t (f) = (f \circ \gamma)'(t)

for any differentiable function f at \gamma(t).) Length of piecewise C^1 curves can be defined as the sum of the lengths of its pieces. From this a distance on M whose topology coincides with the old one on M is given as follows

\displaystyle d_g(x,y) = \inf_\gamma L(\gamma)

where the infimum is taken on all over the set of all piecewise C^1 curves connecting x and y.

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April 20, 2016

Stereographic projection, 6

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:08

I want to propose an alternative way to calculate the Jacobian of the stereographic projection \mathcal S. In Cartesian coordinates  \xi=(\xi_1, \xi_2,...,\xi_{n+1}) on the sphere \mathbb S^n and x=(x_1,x_2,...,x_n) on the plane, the projection and its inverse are given by the formulas

\displaystyle\xi _i = \begin{cases} \dfrac{{2{x_i}}}{{1 + {{\left| x \right|}^2}}},&1 \leqslant i \leqslant n, \hfill \\ \dfrac{{{{\left| x \right|}^2} - 1}}{{1 + {{\left| x \right|}^2}}},&i = n + 1. \hfill \\ \end{cases}

and

\displaystyle {x_i} = \frac{{{\xi _i}}}{{1 - {\xi _{n + 1}}}}, \quad 1 \leqslant i \leqslant n.

It is well-known that the Jacobian of the stereographic projection \mathcal S: \xi \mapsto x is

\displaystyle \frac{\partial \xi}{\partial x} = {\left( {\frac{2}{{1 + {{\left| x \right|}^2}}}} \right)^n}.

The way to calculate its Jacobian is to compare the ratio of volumes. First pick two arbitrary points x, y \in \mathbb R^n and denote \xi = \mathcal S(x) and \eta = \mathcal S(y).

The Euclidean distance between \xi and \eta is

\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.

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August 9, 2015

The third fundamental lemma in the method of moving spheres

Filed under: Uncategorized — Ngô Quốc Anh @ 4:03

In this post, we proved the following result (appeared in a paper by Y.Y. Li published in J. Eur. Math. Soc. (2004))

Lemma 1. For n \geqslant 1 and \nu \in \mathbb R, let f be a function defined on \mathbb R^n and valued in [-\infty, +\infty] satisfying

\displaystyle {\left( {\frac{\lambda }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \leqslant f(y), \quad \forall |y - x| > \lambda > 0.

Then f is constant or \pm \infty.

Later, we considered the equality case in this post and proved the following result:

Lemma 2. Let n\geqslant 1, \nu \in \mathbb R and f \in C^0(\mathbb R^n). Suppose that for every x \in \mathbb R^n there exists \lambda(x)>0 such that

\displaystyle {\left( {\frac{\lambda(x) }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda(x) ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) =f(y), \quad \forall |y - x| > 0.

Then for some a \geqslant 0, d>0 and \overline x \in \mathbb R^n

\displaystyle f(x) = \pm a{\left( {\frac{1}{{d + {{\left| {x - \overline x } \right|}^2}}}} \right)^{\frac{\nu }{2}}}.

In this post, we consider the third result which can be stated as follows:

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