New answers tagged

1

Adding to @WORDS's answer, I thought up a way to prove $W^{1,1}\subseteq AC$. Here it is. Let $u\in W^{1,1}(a,b)$. Let $\rho_n$ be the standard mollifiers. We have: \begin{align*} \rho_n\ast u,\rho_n\ast u'\in{}&\mathcal{C}^\infty(a,b)\,\,\forall n; \\ \rho_n\ast u\to{}&u\,\,\text{in }L^1(a,b); \\ \rho_n\ast u'\to{}&u'\,\,\text{in }L^1(a,b); \\ \...


1

No, the converse is not true. Let $\Omega$ be a slit disk in the plane, that is the unit disk minus the radius from $(0,0)$ to $(1,0)$. The Sobolev inequalities hold in this domain, as one can see by applying them in the top and bottom half-disks (which are Lipschitz domains). However, $\Omega$ is not a Sobolev extension domain. For example, the function ...


2

The space $\tilde{C}^k$ is not complete with respect to the norm $||.||_{k,p}$ -- assuming the norm $||.||_p$ is the $L^p$ norm $$||f||_p^p := \int_\Omega |f|^p dx$$ The reason is that convergence in this kind of norm does not preserve differentiability. Edit: provide simple example to reply to a comment: look, for example, at the functions $$f_n(x) = |x|^{...


2

Strictly speaking, the answer is no. $f \in W^{2,p}$ is really an equivalence class of functions which are equal a.e. It doesn't even exactly make sense to speak of their pointwise regularity. But even if you "open up" the equivalence class and look at the functions inside, not all of them have pointwise regularity. (For instance, you can do whatever you ...


1

There is a difference in that the elements of $W^{1,1}$ are equivalence classes of functions (up to equality a.e.), while the elements of AC and BV may (or may not) be understood as actual functions, for which changing the value at a point matters. That said, there is a close relation: $u\in W^{1,1}(I)$ if and only if $u$ has an absolutely continuous ...


3

Your phenomenon cannot be true as we have $W^{1,p}(0,1) \subseteq W^{1,q}(0,1)$ if $1 \leq q < p \leq \infty$. And indeed, for $1 \leq p < \infty$, we have $$ \int_0^1 u'(x)^p \, dx = \frac{1}{2^p} \int_0^1 x^{-\frac{p}{2}} \, dx = \begin{cases} -\frac{1}{2^p \left( \frac{p}{2} - 1 \right)} \left(1 - \lim_{x \to 0} x^{-\frac{p}{2} + 1} \right) & ...


3

The closure of the domain in $L^2$ is simply $L^2$: Obviously it holds $C_0^\infty(0,1)\subset D(A_0)$. The set of smooth function is dense in $L^2(0,1)$, hence its closure is $L^2(0,1)$. This implies that the closure of $D(A_0)$ is $L^2(0,1)$ as well.


1

Let $\tilde V$ denote the closure of $\mathfrak D$ with respect to the $H^1$-norm. We want to show $\mathcal V = \tilde V$. Clearly, $\tilde V$ is a closed subspace of $\mathcal V$. Now let $f\in \mathcal V$ be given such that $f(v)=0$ for all $v\in \tilde V$. This holds in particular for all $v\in \mathfrak D$. Hence it follows that $f$ is a gradient, i.e....


3

This argument works if $U$ is bounded and has Lipschitz boundary. Since $(u_h)$ converges weak-star in $W^{1,\infty}(U)$, it is bounded in $W^{1,\infty}(U)$ and also bounded in $W^{1,n+1}(U)$ since $U$ is bounded. The space $W^{1,n+1}(U)$ is continuously embeddded into the space of H"older continuous functions $C^{0,1/n}(\bar U)$, Morrey embedding theorem. ...


0

Yet another term: I think of these as homogeneity arguments. A simple example (has to be simple, I don't know much about Sobolev spaces): You know that if $f$ is a function on the line and $f'\in L^2$ then $f\in Lip_{1/2}$. Are there any other $Lip_\alpha$ spaces for which this is true? No. Suppose $$\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}\le c||f'||...


1

That's not true. For example, when $n=1$, $\sin x$ is an element in $W^{1,p}_0(0,\pi)$ since it is a $W^{1,p}$-limit of elements in $C^\infty_c((0,\pi))$. However, $\sin x$ does not have compact support $(0,\pi)$. In general, if $\Omega$ has a nice boundary (for example if it's $C^1$), $W^{k,p}_0(\Omega)$ are those in $W^{k,p}(\Omega)$ which vanishes at the ...


2

So first of all, $H^1$ does not really have a restriction map, even to interior points, much less boundary points (except in one dimension, where there is a continuous embedding into $C^0$). The better way of thinking about this is to start out with smooth test functions, then look at the equation that you get and identify the solution space and the test ...


2

It is usual to assume some regularity property for $\Gamma$. For example, in Evan's PDE, if $\Omega$ is bounded and has a $C^1$ boundary, then there is (bounded) trace operator $$T : H^1(\Omega) \to L^2(\Gamma)$$ so that $T u = u|_\Gamma$ if $u \in H^1(\Omega) \cap C(\overline\Omega)$. It is also proved that $$\{ H^1(\Omega) : Tu = 0\} = H^1_0(\Omega),...


1

One needs to assume something about the regularity of the boundary and of the boundary data. Simply saying "satisfies Dirichlet boundary conditions" is not useful without further information on the boundary data. Example Let $\Omega\subset \mathbb{R}^2$ be the unit disk. Thinking of $\mathbb{R}^2$ as complex plane $\mathbb{C}$, let $u(z) = \operatorname{...


1

Yes, your argument is correct. You first prove the inequality on a dense subspace and both sides of the inequality depend continuously (w.r.t. $H^1$) on $v$. Hence, you can pass to the limit via density and obtain the inequality on all of $H_0^1$. This approach also works for many other properties of $H_0^1$.


1

let's take $\Omega = [0,1]$, $\varphi,f \in C^\infty_c((0,1))$. Then $$\langle f'',\varphi \rangle = \int_0^1 f''(x) \varphi (x) dx = f'(1)\varphi (1)-f'(0)\varphi (0) - \int_0^{1} f'(x) \varphi '(x) dx$$ $\varphi,f$ have their support strictly inside $(0,1)$ so $f'(0)\varphi (0) = f'(1)\varphi (1)= 0$ and $$\langle f'',\varphi \rangle =-\langle f',\...


0

For the first question, since $|G'(s)|\le M$ for all $s\in\mathbb{R}$, it follows that $|G(u)-G(u_n)|\le M|u-u_n|$, and hence $\|G\circ u - G\circ u_n\|_{L^p(\Omega)}\le M\|u-u_n\|_{L^p(\Omega)}\rightarrow 0$. For the second question, notice that $$ (G'\circ u)\frac{\partial u}{\partial x_i} - (G'\circ u_n)\frac{\partial u_n}{\partial x_i} = (G'\circ u - G'\...


0

One can define $\Delta$ in $H^1$ because $\Delta$ is in divergence form $$\Delta u = \text{div}\nabla u$$ Thus we define $\Delta : H^1_0(\Omega) \to H^1_0(\Omega)^*$ by $$\Delta f (\phi) := -\int_\Omega \nabla f \cdot \nabla \phi.$$ The right hand side is well defined for $f , \phi \in H^1_0(\Omega)$. If $f$ is in $H^2_0$, then integration by part gives ...


1

The function $$F(a) = \int_{B_4(0)} \frac{\mid\nabla u(x) \mid^p}{\mid x-a \mid^{n-1}} dx$$ is the convolution of $|\nabla u|^p$ (restricted to the ball) with the Riesz potential $I(y) = 1/|y|^{n-1}$. Moreover, since only the values of $x,a$ with $|x-a|\le 8$ appear in the integral, we can just as well convolve with the restriction of the $I$, namely $$J(y) ...


0

The number of components is not the dimension. Here, because there is 1 "constraint" (that the normal component of the tangential derivative is zero), the boundary is indeed a codimension 1 submanifold. You probably don't actually believe the number of components is the dimension, but this particular scenario has confused you. For example, in $\mathbb{R}^n$ ...


1

Since the mapping $$ \iota\colon\mathcal{D}\to H $$ is a continuous linear operator (which is the "generalization" of bounded operators to the setting of locally convex spaces), there is a (continuous) transpose $$ \iota^t \colon H' \to \mathcal{D}' $$ satisfying $$ \langle \varphi, \iota^t(T)\rangle = \langle \iota(\varphi), T\rangle $$ for all $\varphi\in\...


1

Your explanation for the choice of the function space is right. Note that $u \in H_0^2(\Omega)$ if and only if $u = 0$ and $\nabla u = 0$ on $\partial \Omega$. This, of course, is not the right function space for your Neumann problem!


1

Let us start by discussing the following Neumann problem: \begin{align} - \Delta u &= f \quad \textrm{ in } \Omega, \\ \frac{\partial u}{\partial \nu} &= 0 \quad \textrm{ on } \partial \Omega. \end{align} Here, $\Omega \subset \mathbb{R}^n$ is a bounded Lipschitz domain. As Michal has already mentioned, a solution $u$ of this boundary value ...


1

This is not possible. Every $v \in W^{2,4}(\Omega)$ satisfies $\nabla v \in C(\bar\Omega)$. Hence, if $\varphi^\varepsilon \in W^{2,4}(\Omega)$ converges in $W^{1,\infty}(\Omega)$ towards $v$, we get $\nabla v \in C(\bar\Omega)$, since the derivatives converge in $L^\infty(\Omega)$. But $\nabla \varphi \in L^\infty(\Omega)$ can be discontinuous.


0

As has been mentioned in a comment your question is very broad. One possible answer is to use the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$, it consists of the linear functional continuous with respect to $\mathcal{S}(\mathbb{R}^n)$ which is the Schwartz space. In other words it can be shown that the ...


0

The $\varphi$ in question does not necessarily have compact support in $\mathbb{R}\times (0,\infty)$, but it vanishes only in $(x,0)$, which is a boundary point of the (open) domain of definition. So the definition does not apply.


0

This is not true for dimension $n>1$. In the book "Some Applications of Functional Analysis in Mathematical Physics" by Sobolev the following example is constructed. Consider function $\varphi(x,y)=f_1(x)+f_2(y)$, where $f_1$ and $f_2$ are continuous on $\mathbb{R}$ nowhere differentiable functions. Then $\varphi$ doesn't have strong (classical) ...


0

Let's formulate a lemma: Suppose $f\in L^p(\mathbb{R}\times(0,+\infty))$ and $\varphi\in C^\infty(0,+\infty)$ as in the question. Then $$\lim\limits_{\varepsilon\to 0}\|f(x_1,x_2)\varphi(x_2/\varepsilon)-f(x_1,x_2)\|_{L^p}=0$$ Proof of the lemma. Note that $$ \int\limits_{\mathbb{R}\times(0,+\infty)}|f(x_1,x_2)\varphi(x_2/\varepsilon)-f(x_1,x_2)|^pd\mu ...


0

If the weak Laplacian is in $L^2$, then the function is in $H^2$ locally. Indeed, the Poisson equation with $L^2$ data admits an $H^2_{\rm loc}$ solution $w$ (e.g., here. Then the difference $u-w$ satisfies the Laplace equation in the weak sense, and therefore (by Weyl's Lemma) is a classical harmonic function, in particular $C^\infty$. (I think you can also ...


0

I think your second argument is somewhat going into the wrong direction. You want to show that $H'\subset\mathcal{D}'$, so have to show two things: 1) Given $f\in H'$, you have $f\in\mathcal{D}'$, which follows since $\mathcal{D}$ is continuously embedded into $H$ (that you still have to show). 2) If two functionals $f,g\in H'$ coincide on $\mathcal{D}$, ...


0

First notice that the pointwise limit of the integrand is $0$. Indeed, fix $x \in \mathbb{R} \times (0,\infty)$ and let $\epsilon$ be so small that $x_2/\epsilon > 1$. Now recall that since $\phi$ is constant for $s \ge 1$ its derivative is $0$ for $s > 1$. What we are left to prove is that we can pass the limit inside the integral signs. By Lebesgue'...


0

For any $f\in H_s$, $R(f)$ is compactly supported because of the definition of $R(f)$ as $\int_{U} e^{i x\cdot\xi}r(x,\xi)\hat{f}(\xi)d\xi$, where $r$ is the symbol of $R$. $R(f)$ inherits the support of $r(x,\xi)$. One sees that a limit $\lim_{n\rightarrow\infty} R(f_n)$ of such images must be supported similarly. The key here being that not only is each ...


0

You are not allowed to do equip $ \mathcal{D}$ with the restriction $ \|\cdot\|_{\mathcal{D}}$ of the norm induced by $⟨⋅,⋅⟩_H$, because if you change the norm on $ \mathcal{D}$ you change $ \mathcal{D}' $. For example condider $ C^{\infty}_{c}$ with the $ L^2 $ norm then the dual is $ L^2$.


2

Let me handle the case $d = 1$ for simplicity of notation. Since $H$ is a Hilbert space, $F$ is represented by an element $g \in H$. Thus, for all $\phi \in \mathcal{D} \subset H$ with $\operatorname{supp} \phi = K$ we have $$ |F(\phi)| = \left| \left<\phi, g \right>_H \right| \leq ||g||_H ||\phi||_H = ||g||_H \left( \int_{\Omega} \phi \cdot \phi + \...


0

Thanks to user1952009's comment, the first box itself does not imply the later two boxes. One would have the conclusions by looking at the Fourier transform of $\partial^\alpha f$ and applying the Plancherel Theorem.


1

If $N > 1$ and $\Omega = B(0,1)$ then $u(x) = \log \log \left( 1 + \frac{1}{||x||} \right) \in W^{1,N}(\Omega)$, the function is continuous on $B(0,1) \setminus \{ 0 \}$ and unbounded at the origin and so does not equal a.e to any continuous function defined on $B(0,1)$. If $N = 1$ then any Sobolev function has a (locally) absolutely continuous ...


0

I think we are the same, I mainly studied this definition in "Real Analysis" by Folland. The definition can be justified as follows, note that the maps $\omega_s \cdot : \varphi(\xi) \in \mathcal{S}(\mathbb{R}^n) \longmapsto \omega_s(\xi)\varphi(\xi) \in \mathcal{S}(\mathbb{R}^n)$, where $\omega_s(\xi):=(1+|\xi|^2)^{s/2}$ is continuous with respect to ...


2

If you use $(u, v)_{H_0^1} = (\nabla u, \nabla v)_{L^2}$, you get $R = -\Delta$, as you have shown. And there is another scalar product on $H_0^1$, such that $R = I - \Delta$. Which one?


1

For $p<n$, the function $$f(x)=\|x\|^{-\epsilon}$$ works (on the unit open ball, for example), when $\epsilon>0$ is sufficiently small. How small? Well, the gradient has one degree of homogeneity less, so $|\nabla f(x)|=O(\|x\|^{-\epsilon-1})$. And we want this to be in $L^p$, which requires $p(-\epsilon-1)>-n$. Rearrange to get $$\epsilon<\frac{...


1

You can find the general statement and proof in Chapter 6 of the book "Sobolev Spaces" by Robert A. Adams and John. J. F. Fournier.


0

For $x\in\mathbb{R}^d$, define $\displaystyle G(x)=\int_0^\infty t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}dt$. Since $t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}>0$, $0<G(x)\le\infty$ is well-defined. By Tonelli's theorem $\displaystyle||G||_1=\int_{\mathbb{R}^d}dx\int_0^\infty t^{s-1-d/2}e^{-t}e^{-\frac{|x|^2}{4t}}dt=\int_0^\infty t^{s-1-d/2}e^{-t}dt\int_{\...


3

A counterexample for $d=2$: let $\Omega$ be the disk $\{x:\|x\|<\exp(-\exp(\pi))\}$, and $$ f(x) = \sin \log \log \frac{1}{\|x\|} $$ This function is in $W_0^{1,2}(\Omega)\cap L^\infty(\Omega)$ (relevant calculations here) but has a discontinuity at $0$, and moreover cannot be made continuous by redefining it on a set of measure zero. On the other ...


1

A function $u \in W^{1,2}(\Omega)$ might be unbounded for $n \ge 2$. Hence, it is not possible by continuous functions. In the case $p > 2$, I think one can develop an argument along the following lines: There is a function $u \in W^{1,2}(\Omega)$, which has a singularity "which is stronger as the space $W^{1,p}(\Omega)$ allows". Hence, the one-sided ...



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