## 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).

$$int_{a}^{b} f(t) overline{g(t)} dt$$
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 $$mge 3$$.

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

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.”