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4

After applying a change of variables to the integral on $M$, the given condition is that $$\frac1{|M|}\int_N |\det Df| u \circ f = \frac1{|N|}\int_Nu\circ f$$ for every $u$. Thus it seems this condition restricts you to exactly the diffeomorphisms with constant Jacobian determinant. When $|N|=|M|$ these are known as measure-preserving diffeomorphisms, which ...

3

Let $q = p/(p-1)$. Then $p,q$ satisfy the requirements of Youngs inequality, so that $$|u|^{p-1} |u_{x_n}| \leq C [|u|^{q(p-1)} + |u_{x_n}|^p]= C [|u|^p + |u_{x_n}|^p].$$ The other terms (the signum and terms involving $\zeta$ can be bounded by constants).

3

This counterexample is in terms of continuous but non-differentiable functions because it's easy to describe. At the points of non-differentiability one can smooth out the function to the degree required, e.g. by convolution with mollifiers etc. Take $f\equiv 1$. $J$ is a function whose graph is an infinite sequence of hats (by "hat" I mean like the graph ...

3

"Locally" is ambiguous here $f$ is locally Lipschitz in $\Omega$ if and only if $f \in W^{1,\infty}_{loc}(\Omega)$ The validity of this claim depends on interpretation of "locally Lipschitz". Does it mean every point of $\Omega$ has a neighborhood in which $f$ is Lipschitz, or there is $L$ such that every point of $\Omega$ has a neighborhood in ...

3

Step (B) isn't really by the induction hypothesis, it's the definition of weak $D^\gamma$ (a.k.a., "formal integration by parts"). Step (C) is by the induction hypothesis, distributing $D^\gamma$ according to the Leibniz rule. To understand (D), split the sum in (C) in two, express the first one in terms of $\rho$, and then rename the index $\rho$ as ...

3

For context, I "quoted" the beginning of proof below. The line "and consequently" does not involve integration by parts: that was already done. Instead, it involves the Peter-Paul inequality $$2ab\le \epsilon a^2+\epsilon^{-1}b^2$$ which allows us to absorb one term into the left hand side, which we are estimating. Common thing in PDE, by the way. ...

2

The way you asked the question, the answer is obviously negative, take any constant function different from 0. If M is not connected then even zero average is not enough to guarantee the inequality.

2

The trace theorem, as stated by Evans: Assume $U$ is bounded and $\partial U$ is $C^1$. Then there exists a bounded linear operator $T:W^{1,p}(U)\to L^p(\partial U)$ such that $Tu=u_{|\partial U}$ if $u\in W^{1,p}(U)\cap C(\overline{U})$ $\|Tu\|_{L^p(\partial U)}\le C\|u\|_{W^{1,p}(U)}$ for all $u\in W^{1,p}(U)$, with $C$ depending only on ...

2

First, your definition of $||f||_s$ is wrong: the square root on the right-hand side is missing. Now, if we use the correct definition $$||f||_s=\sqrt{\sum_m(1+m^2)^s|\hat f(m)|^2},$$ then everything is pretty straightforward: $$\sup_x|f(x)|=\sup_x\left|\sum_m\hat f(m)e^{imx}\right|\le\sum_m|\hat f(m)|=\sum_m |\hat f(m)|(1+m^2)^{s/2}(1+m^2)^{-s/2}$$ (by ...

2

Observe that $$\int_{\partial B(0,r)} u^2 \nu \cdot \frac x{|x|^2} \, dS = \frac{1}{r}\int_{\partial B(0,r)} u^2 \, dS$$ because $\nu$ is pointing in the direction of $x$, which makes $\nu\cdot x = |x|$. After you plug this into the equation before "therefore", all that's left to do is to rearrange the term. The integral with $u^2/|x|^2$ goes to the left ...

2

I'll use notation $\Delta u$ instead of $-\mu$, because it fits better in the formulas. Take a smooth domain $G$ such that $\operatorname{supp} \Delta u\subset G \Subset \Omega$. Using a smooth cutoff function, write $\psi$ as $\psi_1+\psi_2$ where $\psi_1$ has compact support in $\Omega$ and $\psi_2\equiv 0$ on $\overline{G}$. Apply Green's identity to ...

2

Integration in polar coordinates (or spherical, in higher dimensions). Write $z=x+\rho\theta$ where $\theta\in S^{n-1}$ is a unit vector and $\rho\in [0,r]$. Then for any integrable $f$ we have $$\int_{B(x,r)} f(z)\,dz = \int_0^r \int_{S^{n-1}} f(x+\rho \theta) \rho^{n-1}\,d\theta \,d\rho \tag{1}$$ (Compare with $n=2$ case, when this is the usual polar ...

2

It seems to me that one can easily extend the definition so that, for each $f\in L^1$, for almost all $a$, $\int\delta(x-a)f(x)\,dx$ is defined; just define it to be the limit in the Lebesgue density theorem. But you seem to want the quantifiers in the other order: For almost all $a$, for all $f\in L^1$, $\dots$. I see no reasonable way to get this. That ...

2

(Too long for a comment) In my opinion, this is an interesting curiosity and I think the correct answer is what Andreas has said - simply define $\langle \delta_a,f\rangle$ as the Lebesgue limit. This might save you some space if you need to say something like "the value of $f$ at $x$" in the context of an $L^1$ function (for which function values are not ...

2

I can show (1), but I don't know if $\Delta(\Phi\circ u)$ is a measure. The equality (1) has to be understood in the sense of distributions, so we must figure out how $(\Phi'\circ u)\Delta u$ acts on test functions. We know how $\Delta u$ acts: $T(\phi)=\int u\Delta \phi$. Since $u$ is a $W^{1,2}$ function, $T$ is a particularly nice distribution: its value ...

2

Concept 4 is very different from 1,2,3. It is just a different generalization of convexity which happens to have the same name (which, incidentally, makes for some ambiguous questions posted on this site). Since the definition 4 is due to Morrey, it is a good idea to call it Morrey quasiconvexity to avoid confusing with 1,2,3. Properties 1 and 3 are ...

2

Let $\phi_\epsilon$ denote a mollifier and let $u_\epsilon = u*\phi_\epsilon$ is a smooth function in $\Omega_\epsilon := \{x \in \Omega \mid \mathop{\rm dist}(x, \partial\Omega) > \epsilon\}$. As $Du_\epsilon = Du*\phi_\epsilon = 0$ in $\Omega_\epsilon$, $u_\epsilon$ is locally constant in $\Omega_\epsilon$. Hence, as $u_\epsilon \to u$ allmost ...

1

An element of $L^{1}$ is an equivalence class of functions which are equal a.e.. Elements of $L^{1}$ don't have pointwise values. Point values can make sense for specific cases, such as when there is a function in the equivalence class which happens to be continuous--that's because if there is such a function in the equivalence class, then there cannot be a ...

1

Because nobody is answering, I thought I'd offer a start. Your equation for the resolvent of $A$ is $$(\lambda I -A)\left(\begin{array}{c}f \\ g\end{array}\right)= \left(\begin{array}{cc}\lambda & -I \\ I-\Delta & \lambda\end{array}\right) \left(\begin{array}{c}f \\ ... 1 This is not true. Indeed, f\in H^1(R^{n+1})\Leftrightarrow f and all of its first derivatives (in the sense of distributions) are in L^2. But if f\in H^1(R,H^1(R^n)), then necessarily \frac{\partial^2 f}{\partial t\partial x_j} also lie in L^2, which is not the case for a general function in H^1(R^{n+1}). 1 I think you need \Omega of bounded measure (e.g. bounded). Fix t\in[0,T]. You have u_n(t)\to u(t) in H^{-1}. Furthermore, \sup_n \|u_n(t)\|_\infty=M<\infty by assumption. Therefore your sequence u_n is uniformly integrable, which means, that for every \epsilon>0, there exists \delta>0 such that for all measurable set ... 1 This is usually known as the Hardy-Littlewood-Sobolev inequality. By the way, there is no Sobolev space, in the statement! See also the classic paper by Elliot H. Lieb, Sharp constants in the Hardy-Littlewood-Sobolev and related inequalities. Most mathematicians call "Sobolev inequality" the main part of the Sobolev Embedding Theorem. 1 No, it's not nearly that simple. The constant C quantifies the connectivity of the manifold. It can be imagined as the severity of traffic jams that occur when all inhabitants of the manifold decide to drive to a random place at the same time. For example, let M be two unit spheres S^2 joined by a thin cylinder of radius r\ll 1 and length 1. There ... 1 We want to estimate \int_B |\nabla\bar u|^p  in terms of \int_{B^+} |\nabla u|^p. To begin with,$$\int_B |\nabla \bar u|^p = \int_{B^+} |\nabla u|^p +\int_{B^-} |\nabla \bar u|^p$$On B^-, the function \bar u is the sum of two functions:$$v(x) = -3u(x_1,\ldots,x_{n-1},-x_n)$$and$$w(x) = 4u(x_1,\ldots,x_{n-1},-\frac{x_n}{2})$$Compute the ... 1 Poincaré inequality holds for every subspace of W^{1,p}(\Omega) which has compact embedding in L^p and does not contain constants. Let me be more precise. Theorem. Let \Omega be an open, Lipschitz, bounded, connected set in \mathbb R^d and let p \in [1,+\infty). Let W \subset W^{1,p} be a subspace which has compact embedding in ... 1 Not exactly sure I am correct. At first, take a look at 8.8.1 which may be helpful. First substitute u(x)/x as u(x), you will get: ||(u(x)/x)/x||<=p/(p-1)||(u(x)/x)'|| =p/(p-1)||u'/x-u(x)/x^2||<=p/(p-1)(||u'(x)/x||+||u(x)/x^2||). It gives: ||u(x)/x^2|| <= Cp1 ||u'(x)/x||. It means we only need to show that: ||u'(x)/x||<=Cp2||u''||. Now ... 1 No, you can't have p<1 there. Take the constant function u\equiv \lambda. Your inequality becomes$$\|\lambda\|_{2^*}\le \|\lambda\|_1^p which (if $p< 1$) fails when $\lambda$ is large enough. When you imagine an inequality you'd like to be valid, consider how it scales when $u$ is multiply by a positive number, or (when working on vector ...

1

If $\Delta u$ is a signed measure, it can be decomposed into a positive and negative parts. Consequently, $u$ can be be written as a difference of two subharmonic functions. Such $u$ is called $\delta$-subharmonic. There are $W^{1,1}$ functions that are not $\delta$-subharmonic. I'll give an example in one dimension: $u(x)=\int_0^x W(t)\,dt$ where $W$ is ...

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