# Tag Info

0

The analogue to the top proposition is true, and it is irrelevant whether or not $X$ is a finite or infinite-dimensional Banach space. The only thing that matters is that you can construct a Borel measure $\mu$ on open $U \subset X$ and that the function $f$ in question is $\mu$-measurable. This is because: The chain rule is valid for Frechet derivatives. ...

0

It is true that $\|\psi_{n}\star f-f\|_{L^2}\rightarrow 0$ and $n\rightarrow \infty$ for any $f \in L^2$. If you start with the wedge $f$ as described, then $\|\psi_{n}\star f - f\|\rightarrow 0$ and $$\|(\psi_{n}\star f)'-f'\|=\|\psi_{n}'\star f-f'\|=\|\psi_{n}\star f'-f'\|\rightarrow 0.$$ This is because, even though $f$ has only a piecewise ...

0

The reason is that both sides of the inequality are continuous with respect to the $H^1$ norm. When you have two continuous function $f,g$ on some space, and $f\le g$ holds on a dense subset, then it follows that $f\le g$ everywhere. The continuity of $u\mapsto \int_{\mathbb{R}^3} |\nabla u(x)|^2 dx$ is immediate from the definition of $H^1$ norm. Let ...

1

It is well known that for functionals of the form $J(u) = \int_{\Omega}f(x,\nabla u)\,dx$ a sufficient condition for sequential lower semicontinuity is that $\xi \mapsto f(x,\xi)$ is convex. Proofs of this result are a little bit involved, for a reference look at Direct Methods in the Calculus of Variation by B. Dacorogna or Modern Methods in the Calculus ...

0

We have, by definition of $H^{t_\sigma}(\def\R{\mathbf R}\R^d)$, that $$\def\norm#1{\left\|#1\right\|}\def\F{\mathscr F} \norm{u}_{H^{t_\sigma}} = \norm{ (1 + \def\abs#1{\left|#1\right|}\abs\cdot^2)^{t_\sigma/2}\F u}_{L^2}$$ We have for $\frac 1\alpha + \frac 1\beta = \frac 12$ by Hölder \begin{align*} \norm{u}_{H^{t_\sigma}} &= \norm{ (1 + ...

1

Let $M=\|u\|_{\infty}$. Since $G'$ is continuous, there exists a constant $K$ such that $|G'(s)|\leq K$ for all $s\in [-M,M]$. Thus $|G'\circ u|\leq K$; it follows that $|G'\circ u|^p$ is bounded by $K^p$. The desired conclusion follows from Hölder's inequality as you explained with a little correction (you have to change parenthesis by norms): ...

1

For $\varphi \in C^{\infty}_c$ one defines $\langle\partial_{x_1}f,\varphi\rangle = -\int_{\Omega}f\partial_{x_1}\varphi$. Extending this definition to $H^1_0(\Omega)$ we obtain a linear functional on $H^1_0(\Omega)$. Continuity, i.e. boundedness, follows from Holder's inequality (together with Poincare's inequality, depending on the norm you define on ...

0

Q1/ Yes. If you know the basic fact that $H^2(0,1)$ is dense in $H^1(0,1)$, then just take $v\in V$, then there is $u_n$ a sequence of $H^2$ converging to $v$. Then $\tilde{u}_n:=u_n-u_n(0)$ belongs to $H^2(0,1)\cap V$ and as $u_n(0)\to_{n\to\infty} v(0)=0$ (I guess you know that $H^1$ convergence implies uniform convergence, in dimension 1), you have ...

1

Even though Matt has solved the original problem properly, I feel obligated to give a proper solution to the fixed version. This is just for completeness: Step 1: By scaling, we may suppose that $t=1$. This is a calculation just like the one done by Matt above. Step 2: By translating $f$ if necessary, we may restrict ourselves to the evaluation of $|T_1 ... 2 So we're in dimension$n=3$according to the problem statement. I agree with you that we can get the inequality without the$1/t$factor basically by Cauchy-Schwarz. But am I being silly or does the stated inequality not hold by scaling? Suppose the inequality were true. Let$f_{t}=f(\cdot/t)$. Then ... 0 Is might be of interest that the$d$-fold tensor product of$W^k_2(\mathbb{R})$is not only dense in$W^k_2(\mathbb{R}^d)$(for any k and d) but in the space$W^k_{2,\text{mix}}(\mathbb{R}^d):=\{ f : \partial_{\alpha}f\in L^2 \quad \forall |\alpha|_{\infty}\leq k\}$(equipped with the obvious norm). Note that this space has a stronger norm than ... 1 To answer your edit: once you have found the correct domain of definition for the Laplacian, you have that$(I-t\Delta)^{-1} = \sum \limits _{k=0} ^\infty t^k \Delta^k$. Since$\int (\Delta u) v = \int u (\Delta v)$(on test functions, at least), you may use induction and prove that$\int (\Delta^k u) v = \int u (\Delta^k v)$, so$\int (\sum \limits _{k=0} ...

0

In the sense of (1), it can be defined for $u_i,v\in L^1_{\text{loc}}(\Omega)$. More generally, it can be defined for arbitrary distributions. There are $u \in L^2(\Omega; \mathbb{R}^d)$, such that there is no $v \in L^1_{\text{loc}}(\Omega)$ with $v = \operatorname{div} u$. But $u$ has always a distributional divergence. If each component of $u$ is weakly ...

1

The answer is yes if $u$ has zero-boundary values and $\Omega$ is sufficiently regular. This follows from $H^2$-regularity of \begin{align}-\Delta u &= f \text{ in } \Omega \\ u &= 0 \text{ on }\partial\Omega\end{align} with $f = -\Delta u \in L^2(\Omega)$. Otherwise it may not hold.

0

Here is a suggestion for a proof. Some details are missing, but I am confident that these can be filled. Let us assume that there is a functional $J : H_0^1(\Omega) \to \mathbb{R}$, such that $J$ is Fréchet-differentiable at $u \in H_0^1(\Omega)$ with $$J'(u)\,v = \int_\Omega \nabla u \cdot \nabla v - f(u,\nabla u)\,v\,\mathrm{d}x$$ for all $v \in ... 0 I think it's implicitly assumed that$k\ge 2$; otherwise the gradient need not be in$L^2$. To begin with, all the three limits of$\{\phi_j\}$are the same. Indeed, let$\phi$be its weak limit in$W^{1,k}$. Since the embedding into$L^2$is compact (and compact operators map weakly convergent sequences to strongly convergent ones), we have$\phi_j\to ...

0

Take $v_\epsilon \equiv 1$. Then, your assumption is satisfied for all $a,b > 0$. But $$\frac1{(2t(\epsilon))^2} \int_{-t(\epsilon)}^{t(\epsilon)}1 \mathrm{d}s = \frac1{2t(\epsilon)} \to \infty$$ as $\epsilon \to 0$. After the question has completely changed, still a similar construction is possible: $$v_\epsilon(t) = \max\{1, n_\epsilon \, |t|\}$$ with ...

0

Hint: the derivative of $(I+t\Delta)^{-1}$ w.r.t. $t$ is $-(I+t\Delta)^{-1}\Delta(I+t\Delta)^{-1}$.

0

You probably know that the Fourier transform takes derivatives to polynomials, ie $F (\partial_i u) = \xi_i \cdot Fu$. So, if all $m$-th order derivatives are in $L^2$ you get that $(1 + |\xi|^2)^{m/2} Fu$ is in $L^2$. This provides an easy way to define Sobolev spaces for $\mathbb{R}^d$. One gets a natural generalization to arbitrary $m \in \mathbb{R}$. ...

1

Its the distributional divergence, that is for $v \in L^2(\Omega)$ we have $v = \operatorname{div}u$ iff $$\int_\Omega v \, \varphi \, \mathrm{d}x = -\int_\Omega u \cdot \nabla \varphi \, \mathrm{d}x \quad\forall \varphi \in C_0^\infty(\Omega).$$ To be compared with https://en.wikipedia.org/wiki/Integration_by_parts#Higher_dimensions.

1

Rewrite the inequality that you want to prove as $$\|u\|_{p}^{p/2}\|(-\Delta)^{\sigma/4}|u|^{\frac{p+m-1}{2}}\|_{2}\geq C\|u\|_{\frac{N(2p+m-1)}{2N-\sigma}}^{\frac{2p+m-1}{2}}$$ Set $v:=|u|^{\frac{p+m-1}{2}}$. The LHS above becomes $$\|v\|_{q}^{\alpha}\|(-\Delta)^{\sigma/4}v\|_{2}$$ with $q=\frac{2p}{p+m-1}$ and $\alpha=\frac{p}{p+m-1}$. Apply the NGN ...

0

There are a few related concepts here: 1) A bump function is a term for a smooth function with compact support. The set of all bump functions forms a vector space. If these functions are on $\mathbb{R}^n$, then it is often denoted $C_c^\infty(\mathbb{R}^n)$. In distribution theory, this is what is most commonly referred to when one refers to test functions, ...

4

A mollifier is a function $f$ that you convolve with another function $g$ to get a function which is "close" to $g$ but "nicer". For instance $f$ might be a general $L^1$ function and $g*f$ might be a smooth, compactly supported approximation to $f$. Really a mollifier is not one function but a sequence, or even sometimes a one-parameter continuous family. ...

1

The assumption is implicitely used here: \begin{align*} \int_{\mathbb{R}^n}\int_{\Omega'} \tilde{u}(x-y) J_{\varepsilon}(y) D^{\alpha}\phi(x){\rm d}x {\rm d}y &= \int_{B_\varepsilon(0)}\int_{\Omega'} \tilde{u}(x-y) J_{\varepsilon}(y) D^{\alpha}\phi(x){\rm d}x {\rm d}y \\ &= (-1)^{\lvert \alpha \rvert} \int_{B_\varepsilon(0)}\int_{\Omega'} ...

1

No, you can't. Consider $\Omega$ being a unit circle on $\mathbb{R}^2$. Then: $$\int\limits_{\Omega} \sum_{|\alpha|=2} | D^{\alpha}u | = \int\limits_{x^2 + y^2 < 1} \left( \left| \frac{\partial^2 u}{\partial x^2} \right| + \left| \frac{\partial^2 u}{\partial x \partial y} \right| + \left| \frac{\partial^2 u}{\partial y^2} \right| ... 0 It will also depend on the dimension of \Omega. Roughly speaking, the embedding theorem works the same for fraction ordered Sobolev space. Please check Theorem 6.14 from this book, and some related theorems from there as well. 0 Since v(0) = v(a) = 0, you have$$v(x) = \int_0^x v'(x) \, \mathrm{d} x = -\int_x^a v'(x) \, \mathrm{d} x.$$Hence,$$|v(x)| \le \int_0^x |v'(x)| \, \mathrm{d} x$$and$$|v(x)| \le \int_x^a |v'(x)| \, \mathrm{d} x.$$Thus,$$|v(x)| \le \min\{\int_0^x |v'(x)| \, \mathrm{d} x, \int_x^a |v'(x)| \, \mathrm{d} x\} \le \frac12 \int_0^a |v'(x)|\,\mathrm{d}x.$$0 if p\in[1,6), we have \nabla u_n\to \nabla u weakly in L^2 implies u_n\to u strongly in L^p (this is sobolev compact embedding). Hence, you have l.s.c. of F. if u_n\to u strongly in H^1, it implies that u_n\to u strongly in L^6. (Note that if u_n\to u weakly in H^1, you only have u_n\to u strongly in L^p for p<6, like in ... 0 For f\in H^1(\Omega), you already know that the weak derivatives f_\alpha := \partial^\alpha f exist on all of \Omega. Thus, you only need to show f_\alpha \in L^p(\Omega). For this, it suffices to show (why?) f\alpha \in L^p(\Omega_1) and f_\alpha \in L^p(\Omega_2), which follows (apparently) from Sobolev embedding. 1 I have found the answer in Kobayashi & Nomizu, volume 1, page 124. If K is a tensor of type (r,s), then one may construct a new tensor \nabla K of type (r, s+1), defined by$$(\nabla K) (X_1, \dots, X_s, Y) = (\nabla _Y K) (X_1, \dots, X_s)$$and thus define inductively \nabla ^k K as \nabla (\nabla ^{k-1} K). Choosing now K to be f, a ... 1 At the beginning of this section, Evans makes the assumptions$$ a^{ij},b^i,c\in L^\infty(U), f\in L^2(U). $$The following inequalities are crucial: if w\in H^1(U), then$$ \|w\|_L^2(U) \leq \|w\|_{H^1(U)}, \|w_{x_i}\|_{L^2(U)} \leq \|w\|_{H^1(U)}. $$This is pretty much immediate from the definition of the Sobolev norm. You may need to tack on ... 2 Remember that f \in W^{1,1} (\gamma_1) means that f, f' \in L^1 (\gamma_1) where f' is understood in the weak sense. Remember also that f' in the Sobolev sense in the same as f' in the distributional sense, therefore let us compute the latter one. Finally, we shall use that \gamma_1 = \dfrac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2}. Note, ... 2 The functions U, U_x and U_y are locally bounded, except possibly at (x,y)=(0,0). If B_1 is the unit ball centred at (0,0), We have$$ \int_{B_1} \lvert u\rvert = \int_0^{2\pi}\int_0^1 r\lvert u(r\cos\vartheta,r\sin\vartheta)\rvert\,dr\,d\vartheta= \int_0^{2\pi}\int_0^1 r\lvert U(r)\rvert\,dr\,d\vartheta=2\pi\int_0^1 r\lvert U(r)\rvert\,dr.  ...

1

As stated, existence is trivial: $u\equiv 0$ is a solution. When a nontrivial solution exists, it is not unique since one can multiply it by a constant. But this happens only for some specific $\lambda$: those that are in the spectrum of the Dirichlet Laplacian. You can't infer this kind of structure from Lax-Milgram. Rather, sine Fourier series should be ...

1

You can't just choose the function to be identically zero outside $U$ in general, because this can cause the weak derivative to fail to be $L^p$ because of "bad regularity" at the boundary. For instance when $n=1$, all $W^{k,p}$ functions are actually continuous, so extending $f(x)=1$ on $(0,1)$ to be identically zero elsewhere certainly does not give a ...

1

It's false. For a counterexample consider $n = k = p = 1$, $U = (0,1)$ and $f \equiv 1$. It is possible to extend Sobolev functions under certain regularity assumptions on the boundary of the set $U$, but it is delicate. Any book on Sobolev spaces addresses the issue.

Top 50 recent answers are included