calculus and analysis – How do I suppress an automatic sign change? II

Let me be try to be highly specific, as my previous attempt How do I suppress an automatic sign change? to pose the question initially had a sign error, and perhaps became a little muddled.



In the course of pursuing the question Evaluate a certain three-dimensional constrained integral, the term (one of 694)

r = (202338335476512488921084723200 x^6 Sqrt(-(-1 + 2 x) (2 - x +  2 Sqrt(1 - x - 2 x^2)))Boole(1/38 (10 - Sqrt(5)) < x <= 1/4))/(319794090309 (723 + 17 Sqrt(5)))

is generated.

My attempt, r/.c, to apply (so the term becomes integrable) the rule

c := Sqrt(-(-1 + 2 x) (2 - x + 2 Sqrt(1 - x - 2 x^2))) -> Sqrt((1 - 2 x) (2 - x + 2 Sqrt(1 - x - 2 x^2)))

fails (because apparently the expression -(-1 + 2 x) is ab initio converted to (1-2 x)).

What needs to be done, so that the intended conversion takes place?

Unfortunately, it would seem the apparent “automatic” conversion of $-(-1 + 2 x)$ to $(1-2 x)$ is not so “automatic” that it is performed in the formula for $r$ itself, which would obviate the apparent dilemma.

complex analysis – An inner product on $mathcal{C}[a,b]$

I’ve to prove that the functional
$$langle f,grangle = int_{a}^{b} int_{a}^{b} frac{sin(pi(t-s))}{pi (t-s)} f(s) overline{g(t)}dsdt$$
is an inner product on $mathcal{C}(a,b)$ (complex continuous functions).

I’ve already made a similar exercise where I shown that
$$int_{a}^{b} f(t) overline{g(t)} dt$$
is an inner product. And most of the properties follow the same reason. But I can’t see how
$$langle f,frangle = int_{a}^{b} int_{a}^{b} frac{sin(pi(t-s))}{pi (t-s)} f(s) overline{f(t)}dsdt = 0 iff f equiv 0.$$
The implication $f=0 Longrightarrow langle f,frangle=0$ is clear from of the definition of the functional, but since $frac{sin(pi(t-s))}{pi (t-s)} = 0$ for $pi(t-s) = npi$ I can’t get the other implication. Same with $langle f,frangle geq 0.$

real analysis – Band limited initial data : Regularity for Navier Stokes Equation defined on a Torus $mathbb{T}^m$

Consider the Navier stokes equation and the Euler equation defined on a Torus(periodic solutions).
Let the dimensionality of the space $mathbb{T}^m$ be $mge 3$.

Link to the Problem.

Has it been investigated partially or conclusively, the regularity of the solutions when the initial data $u_0(x) = u(x,0)$ is a trigonometric polynomial of a certain degree?

References to any closely related research is also appreciated.

functional analysis – Convolution and weak $L^p$ spaces

Let $chi$ a non negative continuous function whose support is included in $)-1,1($ with $intchi=1$. Let $varphi$ a function in $L^{p’}$ non negative with compact support in $mathbb{R}setminus{0}$. Let $q,q’in)1,+infty($ such that $frac1q+frac1{q’}=1$.

  1. Show that $$limlimits_{ntoinfty}nint_{mathbb{R}^2}chi(nt’)varphi(t)frac{1}{vert t-t’vert^{frac{1}{q}}}dt’dt=int_{mathbb{R}}varphi(t)frac{1}{vert tvert^{frac{1}{q}}}dt$$
  1. Deduce that there does not exist a positive constant $C$ such that
    $$forall(f,g)in L^1times L_omega^q,~Vert fstar gVert_{L^q}leqslant CVert fVert_{L^1}Vert gVert_{L_omega^q}$$
    where we denote $L_omega^p$ the weak $L^q$ space endowed with the norm $Vert gVert_{L_omega^q}=suplimits_{lambda>0}lambda^qmu((vert gvert >lambda))$.

  2. Deduce that there does not exist a positive constant $C$ such that
    $$forall(f,g)in L^{q’}times L_omega^q,~Vert fstar gVert_{L^q}leqslant CVert fVert_{L^{q’}}Vert gVert_{L_omega^q}$$

My work

  1. Suppose $Supp(varphi)subset(-K,-varepsilon)cup(varepsilon,K)$ for $varepsilon,K>0$. Let $chi_n$ defined as $chi_n(x)=nchi(nx)$. Then $Supp(chi_n)subset)-frac1n,frac1n($. $chi_n$ is an approximation of unity derived from $chi$. Let $n>frac2varepsilon$ so that for any $(t,t’)in Supp(phi)times Supp(chi_n)$, $vert t-t’vert>fracvarepsilon2$. Note that if $mgeqslant n$, this is still true. Define $g:tmapstofrac{1}{vert tvert^{frac1q}}$ for $tin(-K-fracvarepsilon2,-fracvarepsilon2)cup(fracvarepsilon2,K+fracvarepsilon2)$ and $0$ everywhere else. Then $g$ is in every $L^p(mathbb{R})$, $1leqslant pleqslant +infty$.

In this way, the integral has a sense and we have:
$$nint_{mathbb{R}^2}chi(nt’)varphi(t)frac{1}{vert t-t’vert^{frac{1}{q}}}dt’dt=int_{mathbb{R}}varphi(t)mathbf{1}_{(-K,-varepsilon)cup(varepsilon,K)}(t)int_{mathbb{R}}chi_n(t)mathbf{1}_{(-fracvarepsilon2,fracvarepsilon2)}(t’)frac1{vert t-t’vert^{frac1q}}dt’dt=int_{mathbb{R}}varphi(t)int_{mathbb{R}}chi_n(t)g(t-t’)dt’dt=int_{mathbb{R}}varphi(t)(chi_nstar g)(t)dt$$

Because $gin L^1(mathbb{R})$ and $chi_n$ is an approximation of unity, $chi_nstar gxrightarrow(nto+infty){L^1}g$. This means that, up to an extraction, $chi_nstar gxrightarrow(nto+infty){} g$ almost everywhere.

In this way, we can apply the dominated convergence theorem: $varphi(t)(chi_nstar g)(t)$ converges almost everywhere to $varphi(t) g(t)$. And for almost all $t$, we have thanks to the Young inequality:
$$vertvarphi(t)(chi_nstar g)(t)vertleqslantvertvarphi(t)vertVertchi_nstar gVert_{infty}leqslantvertvarphi(t)vertunderbrace{Vertchi_nVert_{L^1}}_{=1}Vert gVert_{L^infty}$$
and $vertvarphivertVert gVert_{L^infty}$ is integrable thanks to the Hölder inequality:
$$int_{R}vertvarphi(t)Vert gVert_{L^infty}leqslantVert gVert_{L^infty}VertvarphiVert_{L^{q’}}Bigl(mu(Supp(varphi)Bigr)^{frac1q}$$

So we get the wanted convergence.

  1. First we show that $vertcdotvert^{frac{-1}q}in L_omega^q(mathbb{R})$. Indeed, let $lambda>0$.
    $$lambda^qmu((vertcdotvert>lambda))=lambda^qmu((vertcdotvert<frac{1}{lambda^q}))=2$$

Then, suppose this constant $C$ exists. Then from the first question we have $chi_nstar gin L^q$ as $chi_nin L^1$ and
$$int_{mathbb{R}}varphi(t)(gstarchi_n)(t)dtleqslant CVertvarphi Vert_{L^{q’}}Vert gVert_{L_omega^q}$$ which means that, when $n$ goes to infinity,
$$int_mathbb{R}varphi(t)g(t)dtleqslant CVertvarphi Vert_{L^{q’}}Vert gVert_{L_omega^q}$$

I tried using some well thought $varphi$ to find a contradiction but could not find one. For example, I tried characteristic functions, but it does not lead to a contradiction. Any hints ?

fa.functional analysis – Is a specific product function orthogonal to all harmonic functions

Suppose $Omega=(-1,1)^3$. Let $f:(-1,1)to mathbb R$ and $g:(-1,1)^2to mathbb R$ be smooth functions and suppose that given any harmonic function on $Omega$ (i.e. $Delta u =0$ on $Omega$), with $u in L^2(Omega)$, there holds:
$$ int_{Omega} u(x^1,x^2,x^2) f(x^1)g(x^2,x^3),dx=0.$$

Does it follow that $f$ and $g$ are identically zero?

Real Analysis – I don’t have any examples on proving that the infimum of a set is equal to a specific number.

let x be a positive real number and let A = {m+nx: m,n ∈ Z and m+nx>0}. Prove that infA=0

So far I have the following:
Let a = m+nx. A is a set of positive real numbers so 0<a ∀ a ∈ A, thus 0 is a lower bound for A. Since 0 is a lower bound of A, then we say A is bounded by definition. We know that inf A exists by the completeness axiom.

real analysis – First order PDE in complex variables?

You can start by looking at the chain rule for wirtinger derivatives, from which you deduce that

$$
partial_{bar z} exp(h(z)) = exp(h(z)) cdot partial_{bar z} h(z)
$$

Therefore, if you find a function $h$ such that $partial_{bar z} h = – g(z)$ (I think you forgot a “$-$” sign in your solution for the real case!) taking $f(z) = exp(h(z)) $ will solve your problem. In general, this is known as the d-bar problem (or $barpartial-$problem).

real analysis – Does there exist an injective Lipschitz map on the disk whose gradient switches between two given matrices?

While solving a problem in calculus of variations, I came to the following question:

Let $A,B$ be two real $2 times 2$ matrices with positive determinants, and suppose that $operatorname{rank}(A-B)=1$.

Let $D subseteq mathbb{R}^2$ be the closed unit disk.
Does there exist a Lipschitz map $u:D
to D$
such that:

  1. $nabla u in {A,B}$ a.e. in $D$
  2. $u$ is surjective.
  3. $u^{-1}(y)$ is a singleton for almost every $y in D$.

By a theorem by Ball and James, any $u$ satisfying condition $(1)$ is of the following form:

Write $B-A=an^T$ for some $a,n inmathbb{R}^2$, $n$ a unit vector. Then
$$
u(x)=Ax+h(xcdot n)a+v_0,
$$

for some $v_0 in mathbb{R}^2$, and a Lipschitz function $h:mathbb{R} to mathbb{R}$ with $h’ in {0,1}$ a.e..

So, it “remains” to examine maps $u$ of the form above.


The assumptions $(1)$ and $(2)$ imply that the average integral of the Jacobian $Ju$ is $1$, so the measure of the set where $nabla u=A$ is predetermined by $det A,det B$.


If it matters, then in my case of application, $A in alpha text{SO}(2)$, where $alpha>1$, and the singular values of $B$ satisfy:
$sigma_1+sigma_2>beta>1$ and $sigma_1 sigma_2 <<1$ very close to zero. Here $alpha, beta$ are some fixed parameters.

real analysis – Proving that, if $y_1geq y_2$ at some interval $[x_0, x_1]$, there exists a point where $y_1’>y_2’$

I am working on an exercise on ODE and I need the following lemma to solve it.
Lemma: Let two functions $y_1, y_2$: $(x_0, x_1)tomathbb{R}$, such that $y_1(x_0) = y_2(x_0)$ and $y_1(x_1) = y_2(x_1)$. Moreover $y_1(x)>y_2(x) ; forall xin (x_0, x_1)$. Prove that there exists some point $zin(x_0,x_1)$, for which $y_1′(z) > y_2′(z)$.

It is geometrically obvious, as every point near enough $x_0$ satisfies the condition, but I couldn’t prove it formally. I tried working with local maximums and minimums, but I think there is a more obvious way than considering all options for $max$ and $min$.

calculus and analysis – How do I suppress an automatic sign change?

In a comment on the answer of user JimB to Evaluate a certain three-dimensional constrained integral , I remarked

“I have a integration limit transformation rule of the form Sqrt((1 – 2 x) (2 – x + 2 Sqrt(1 – x – 2 x^2))) -> 1 – 2 x + Sqrt(1 – x – 2 x^2), which Mathematica implements finely, but I also need Sqrt((-1 + 2 x) (2 – x + 2 Sqrt(1 – x – 2 x^2))) -> 1 – 2 x + Sqrt(1 – x – 2 x^2),–note changes of sign in first Sqrt–which Mathematica immediately converts to the previous rule, so the second rule doesn’t get enforced. So, I need to suppress the immediate conversion.”