Tag Info

Hot answers tagged

5

In the definition of $Z$, you probably want $|v|>N$ instead of $v>N$. Also, in item 1, the definition of $B$, you have the Sobolev norm of $x$, so it's better to use norm notation for that. Let's also not use subscripts in superscripts... say, $p<q$ and the embedding is $H^q\to H^p$. The $H^q$ norm is given by $$\|f\|_{H^q}^2 = ...


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

I think smoothness is the wrong term to focus on; the difference concerns the continuity of $f$. To get first-order classical derivative of $f$, we would need $f\in W^{2,p}$ with $p>n$; compare with item 4 below. The fact that $f$ has a weak derivative makes it locally absolutely continuous on almost every line. If the derivative is also in ...


3

Forming the convolution of the (scaled) bump function with indicator functions you get "qausi-indicator functions" in $\mathscr D$, in particular, there are $\psi_n\in\mathscr D(\mathbb R)$ which are positive and equl to $1$ on $[-n,n]$. It is then easy to see that the elements of $E^{loc}$ are those distributions which, on every compact set, have the same ...


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

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

Yes. $x \rho_t(x)$ is smooth and compactly supported, and any such function $f$ is in $H^s$ for every $s$. The easiest way to see this may be this: Let $k$ be any integer larger than $s$. Verify from the definition that $H^k \subset H^s$. (Note that $(1+|\xi|^2)^s \le (1+|\xi|^2)^k$). Using the fact that the Fourier transform takes differentiation to ...


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

This proof cover all cases: $a,b$ finite or not. In the case that $a,b$ are not finite, $u(a)$ is understood as $\lim_{x\to-\infty}u(x)$. Analogous for $b$. Also, if $a,b$ are not finite, then we weill consider locally things, i.e. $BV_{loc}((a,b))$. I am also assuming that $H_0^1((a,b))$ is the closure of $C_0^1((a,b))$ with respect to the $H^1((a,b))$ ...


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

First, let $a\in\mathbb{R}$ be arbitrary. As you noted, this implies $$ \limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}\leq\lim_{n\rightarrow\infty}\left\Vert f-a\right\Vert _{L^{p}\left(\Omega_{n}\right)}=\left\Vert f-a\right\Vert _{L^{p}\left(\bigcup\Omega_{n}\right)}, $$ and thus $$ \limsup_{n\rightarrow\infty}\left\Vert f\right\Vert ...


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

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

Let $X$ be a Banach space and $u\in L^1(0,T,X)$. We say that $u'\in L^1(0,T,X)$ the weak derivative of $u$ if $$\int_0^T u(t)\phi'(t)dt=-\int_0^T u'(t)\phi(t)dt,\forall\ \phi\in C_0^\infty(0,T).$$ According to this definition, the problem is not that $u'(t)$ does not belong to $H^1$, it does belong to $H^1$. The question is, if it belong to $L^\infty$, ...


1

First, you apply Hölder to get $$ \|u(t)\| \leq \|u(s)\| + \int_0^T\|u'(\tau)\|\,\mathrm{d}\tau \leq \|u(s)\| + T^{1-1/p}\|u'\|_{L^p(0,T;X)} . $$ Then, taking the $p$-th power of both sides, and integrating with respect to $s$, we have $$ T\|u(t)\|^p \leq c\int_0^T\|u(s)\|^p\,\mathrm{d}s + cT\cdot T^{p-1}\|u'\|_{L^p(0,T;X)}^p = c\|u\|_{L^p(0,T;X)}^p + ...


1

Start with an example. So, look at $\Omega = [0,\pi]\subset \mathbb{R}^{1}$, where you have an orthonormal basis of eigenfunctions for $-\Delta=-\frac{d^{2}}{dx^{2}}$ given by $\{ e_{n}(x)=\sqrt{2/\pi}\sin(nx)\}_{n=1}^{\infty}$. To solve the equation in this case, write the solution as $$ u(x,t) = \sum_{n=1}^{\infty}a_{n}(t)e_{n}(x) $$ Then $$ ...


1

Only large frequencies matter for smoothness. For every $M$, the part of Fourier transform with $\{\xi:|\xi|\le M\}$ contributes a real-analytic term to the function. You know that integrability of $|\xi|^\alpha \hat u(\xi)$ implies certain smoothness of $u$. So you want to show that this product is integrable for every $\alpha$. On every ball $\{|\xi|\le ...



Only top voted, non community-wiki answers of a minimum length are eligible