Ngô Quốc Anh

July 24, 2021

A contraction type argument

Filed under: Uncategorized — Ngô Quốc Anh @ 10:47

Let f : [0,1] \to [0, +\infty) be a function. First, we have the following trivial result:

Observation. If a non-negative funtion f satisfies the following inequality

\displaystyle f(x) \leq \frac 12 f(x)

for all 0 \leq x \leq 1, then we must have \displaystyle f \equiv 0 in [0,1].

The proof of the above observation depends on the non-negativity of f. It is worth noting that we do not require the continuity of f. Here in this post, we are interested in the following

Main result. If the non-negative, continuous function f satisfies f=o(1) near zero and

\displaystyle f(x) \leq \frac 12 \sup_{0 \leq y \leq x} f(y)

for all 0 \leq x \leq 1, then there exists some \delta >0 in such a way that \displaystyle f \equiv 0 in [0,\delta].

The above result, in a special setting, appears in a recent work of Hyder and Sire. (See this if you cannot access the content.) Now we discuss a proof of the above main result, original due to one of my young colleagues.

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June 11, 2021

Second-order differentiability in terms of partial derivatives

Filed under: Uncategorized — Ngô Quốc Anh @ 0:28

Let U \subset \mathbf R^n be open, a \in U is an arbitrary point, and f : U  \to \mathbf R is a function. Recall that f is called differentiable at a if there is a linear map, denoted by A, such that

\displaystyle \lim_{\|h\| \searrow 0} \frac{ \| f(a+h)-f(a) - A (h) \| }{\|h\|} = 0.

The linear map A is called derivative of f at a, denoted by f'(a). Notice that in the nominator of the above quotient, the symbol \| \cdot \| is simply the absolute value function. But this is no longer true for higher-order derivatives that we are going to define.

The following theorem is well-known.

Theorem 1 (1st order differentiability). The function f : U \to \mathbf R is differentiable at a \in U if all partial derivatives \partial_i f exist in a neighborhood of a and are continuous at a.

When f'(a) exists, we must have

\displaystyle f'(a) (h) = \sum_{i=1}^n \partial_i f(a) h_i.

In this note, we want to extend the above theorem for higher-order derivatives. To be more precise, we prove the following

Theorem 2 (2nd order differentiability). The function f : U \to \mathbf R is twice differentiable at a \in U if all partial derivatives \partial_{ij} f exist in a neighborhood of a and are continuous at a.

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March 31, 2021

Monotonicity of (1+1/x)^(x+α)

Filed under: Uncategorized — Ngô Quốc Anh @ 1:11

This post concerns the monotonicity of

\displaystyle f: x \mapsto \Big(1+\frac 1x\Big)^{x+\alpha}

on (0,+\infty) with \alpha \geq 0. The two cases \alpha=0 and \alpha=1 are of special because these are always mentioned in many textbooks as

\displaystyle \lim_{n \to \infty}  \Big(1+\frac 1n\Big)^n= \lim_{n \to \infty}  \Big(1+\frac 1n\Big)^{n+1} =e.

Clearly, f is monotone increasing with respect to \alpha. Hence we are left with the monotonicity of f with respect to x. To study this problem, we examine f' with respect to x.

Derivative of f. It is easy to get

\displaystyle f'(x) =  \Big(1+\frac 1x\Big)^{x+\alpha} \Big[ \log \Big(1+\frac 1x\Big)- \frac{x+a}{x(x+1)}\Big].

Hence the sign of f' is determined by the sign of

\displaystyle g(x) =  \log \Big(1+\frac 1x\Big)- \frac{x+a}{x(x+1)} .

on (0, \infty).

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April 16, 2020

Restriction of gradient, Laplacian, etc on level sets and applications

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 18:10

This topic is devoted to proofs of several interesting identities involving derivatives on level sets. First, we start with the case of gradient. We shall prove

The first identity

\displaystyle \partial_\nu f = \pm |\nabla f|

on the level set

\displaystyle \big\{ x \in \mathbf R^n : f(x) =0 \big\}.

The above identity shows that while the right hand side involves the value of f in a neighborhood, however, the left hand side indicates that only the normal direction is affected. Heuristically, any change of f along the level set does not contribute to any derivative of f, namely, on the boundary of the level set, the norm of \nabla f is actually the normal derivative \partial_\nu f. Therefore, the only direction taking into derivatives of f is in the normal direction and this should be true for higher-order derivatives of f.

Next we prove the following

The second identity

\displaystyle \partial_\nu \big(x \cdot \nabla f \big)=(\partial_\nu^2 f) (x \cdot \nu)

on the level set

\displaystyle \big\{ x \in \mathbf R^n : \partial_1 f(x) = \cdots = \partial_n f(x) = 0 \big\}.

Combining the above two identities, we can prove

The third identity

\displaystyle \partial_\nu^2 f=\Delta f

on the level set

\displaystyle \big\{ x \in \mathbf R^n : \partial_1 f(x) = \cdots = \partial_n f(x) = 0 \big\}

which basically tells us how to compute the restriction of Laplacian on level sets. This note is devoted to a rigorous proof of the above facts together with a simple application of all these identities.

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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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