| CARVIEW |
In this note, we shall discuss the weak form of several ODEs via integration by parts. To be more precise, we are interested in the following two simple ODEs
and
where the given function 
1. THE CASE OF
Initially, if 
holds for any test function ![\phi \in C^1([a,b])](https://s0.wp.com/latex.php?latex=%5Cphi+%5Cin+C%5E1%28%5Ba%2Cb%5D%29&bg=ffffff&fg=333333&s=0&c=20201002)
(more…)Theorem 1. Assume that
. Then any weak solution to
A classical version of the inverse function theorem for 1D asserts the following.
Suppose that
is an invertible function with inverse
. If
with
and
is continuous at
, then
and
Very often, some of the above hypotheses for
In this note, we show by constructing counter-examples that in general the continuity of 
1. The case of continuity
We start with the simplest case, namely if we only assume that 
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
is a non-empty subset of
, and if
of
.
The second property, call C, is the completeness of reals.
(C, completeness axiom): If
and
are non-empty subsets of
for any
and
, then there is some
such that
for any
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
and
with
, there exists some natural number
such that
.
Cantor’s intersection theorem: A decreasing nested sequence of non-empty, closed intervals in
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).
(more…)Let ![f : [0,1] \to [0, +\infty)](https://s0.wp.com/latex.php?latex=f+%3A+%5B0%2C1%5D+%5Cto+%5B0%2C+%2B%5Cinfty%29&bg=ffffff&fg=333333&s=0&c=20201002)
Observation. If a non-negative funtion
for all
, then we must have
in
.
The proof of the above observation depends on the non-negativity of
Main result. If the non-negative, continuous function
near zero and
for all
in such a way that
.
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.
(more…)Let 
The linear map 
The following theorem is well-known.
Theorem 1 (1st order differentiability). The function
is differentiable at
exist in a neighborhood of
When
In this note, we want to extend the above theorem for higher-order derivatives. To be more precise, we prove the following
(more…)Theorem 2 (2nd order differentiability). The function
exist in a neighborhood of
This post concerns the monotonicity of
on 
Clearly, 
Derivative of
![\displaystyle f'(x) = \Big(1+\frac 1x\Big)^{x+a } \Big[ \log \Big(1+\frac 1x\Big)- \frac{x+a }{x(x+1)}\Big].](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+f%27%28x%29+%3D++%5CBig%281%2B%5Cfrac+1x%5CBig%29%5E%7Bx%2Ba+%7D+%5CBig%5B+%5Clog+%5CBig%281%2B%5Cfrac+1x%5CBig%29-+%5Cfrac%7Bx%2Ba+%7D%7Bx%28x%2B1%29%7D%5CBig%5D.&bg=ffffff&fg=333333&s=0&c=20201002)
Hence the sign of
on 
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
on the level set
The above identity shows that while the right hand side involves the value of 
Next we prove the following
The second identity
on the level set
Combining the above two identities, we can prove
The third identity
on the level set
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.
This topic concerns a very classical question: extend of a function 
Throughout this topic, by 
CONTINUITY IS NOT ENOUGH. Let us consider the first situation where the given function
Let ![X = [0,\frac 12 ) \cup (\frac 12, 1]](https://s0.wp.com/latex.php?latex=X+%3D+%5B0%2C%5Cfrac+12+%29+%5Ccup+%28%5Cfrac+12%2C+1%5D&bg=ffffff&fg=333333&s=0&c=20201002)
Since
Hence any extension 
SIMPLE OBSERVATIONS. We start with the following basic results.
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
be a diffeomorphism in
. Fix a point
where
denotes the open ball in
.
As a remark and to be more exact, we require 
We now proceed with the proof whose proof is divide into a few steps.
Step 1. First we use
where the error vector-valued function 
The following Leibniz integral rule is well-known
Theorem. Let
be a function such that both
are continuous in
and
-plane, including
,
. Also suppose that the functions
and
are both continuous and both have continuous derivatives for
The purpose of this note is to show that, in fact, it is not is not necessary to assume the function 
As mentioned above, we want to show that this is also the case for integrals of the form above.
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![[a,b]](https://s0.wp.com/latex.php?latex=%5Ba%2Cb%5D&bg=ffffff&fg=333333&s=0&c=20201002)
Ngô Quốc Anh