# Ngô Quốc Anh

## June 3, 2022

### Non-zero derivative does not imply the continuity of the inverse

Filed under: Uncategorized — Ngô Quốc Anh @ 0:20

A classical version of the inverse function theorem for 1D asserts the following.

Suppose that $f : U \to V$ is an invertible function with inverse $f^{-1} : V \to U$. If $f$ is differentiable at $x_0$ with $f'(x_0) \ne 0$ and $f^{-1}$ is continuous at $y_0 = f(x_0)$, then $f^{-1}$ is differentiable at $y_0$ and

$\displaystyle \big(f^{-1}\big)'(y_0) = \frac 1{f'(x_0)}.$

Very often, some of the above hypotheses for $f$ is replaced by, say, $f$ is continuous “in a neighborhood” and one-to-one. But in the proof, while the one-to-one property immediately gives the existence of the inverse function $f^{-1}$, the “non-local” continuity yields the continuity of the inverse $f^{-1}$. Here we are interested in the “local” continuity/differentiablity.

In this note, we show by constructing counter-examples that in general the continuity of $f^{-1}$ at $y_0$ cannot be dropped. In the sequel, we denote by $\mathbb N$ the set set of positive integers, namely $\mathbb N = \{1,2,3,...\}$.

1. The case of continuity

We start with the simplest case, namely if we only assume that $f$ is continuous at $x_0$. In this scenario, the following example due to R.E. Megginson is well-known. Consider the function $f : \mathbf R \to \mathbf R$ defined by

\displaystyle f(x)=\left\{\begin{aligned} & \frac 1{2n} & & \text{if } x= \frac 1n \text{ for some } n \in \mathbb N \\ & \frac 1n & & \text{if } x= n \text{ for some odd } n \in \mathbb N \setminus \{1\} \\ & \frac n2 & & \text{if } x= n \text{ for some even } n \in \mathbb N \setminus \{1\} \\ & x & & \text{otherwise}.\end{aligned}\right.

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## October 15, 2021

### Least upper bound axiom, completeness axiom, and Archimedean property and Cantor’s intersection theorem are equivalent

Filed under: Giải Tích 1 — Ngô Quốc Anh @ 17:03

This note concerns the equivalence between the three properties usually taken as an axiom in synthetic constructions of the real numbers. We start with the least upper bound property, call L, which is usually appeared in construction of the real numbers.

(L, least upper bound axiom): If $A$ is a non-empty subset of $\mathbf R$, and if $A$ has an upper bound, then $A$ has a least upper bound $u$, such that for every upper bound $v$ of $A$, there holds $u \leq v$.

The second property, call C, is the completeness of reals.

(C, completeness axiom): If $X$ and $Y$ are non-empty subsets of $\mathbf R$ with the property $x \leq y$ for any $x \in X$ and $y \in Y$, then there is some $c \in \mathbf R$ such that $x \leq c \leq y$ for any $x \in X$ and $y \in Y$.

The third, also last, property, call AC, is the set of two results: the Archimedean property and Cantor’s intersection theorem. These two results often appear as consequences of the construction of reals.

Archimedean property: For any real numbers $x$ and $y$ with $x>0$, there exists some natural number $n$ such that $nx > y$.

Cantor’s intersection theorem: A decreasing nested sequence of non-empty, closed intervals in $\mathbf R$ has a non-empty intersection.

Our aim is to prove that in fact the above three properties (L), (C), and (AC) are equivalent. Our strategy is to show the following direction:

(L) ⟶ (C) ⟶ (AC) ⟶ (L).

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## 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

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

on $(0,+\infty)$ with $a \in \mathbb R$. The two cases $a =0$ and $a =1$ 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 $a$. Hence we are left with the monotonicity of $f$ with respect to $x$. From the above discussion, the function $f$ is monotone increasing with respect to $x$ when $a = 0$. When $a<0$, as the function $(1+1/x)^{-|a|}$ is monotone increasing with respect to $x$, we deduce that the function $f$ is monotone increasing with respect to $x$. Hence, we are left with the case $a > 0$. To study this problem, we examine $f'$ with respect to $x$.

Derivative of $f$. It is easy to get

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

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.

## 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.

## 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

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

## 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.

## 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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