Tag Info

3

No. The function $$f(t)=\sum_{n=2}^\infty \frac{e^{in t}}{n \log n}$$ is in $H^{1/2}$ but its real part is unbounded near zero. Informally, it's because the series diverges when $t=0$. But a more careful proof is called for. One approach: take small $t>0$, then choose $N=\lfloor 1/t\rfloor$ so that $\cos nt\ge \cos 1>0$ for $2\le n\le N$. The ...

2

The short answer is no. It is possible that $$\|f\|_{H^s}=\infty$$ for all $s>2$ and $$\|f\|_{H^2}<\infty.$$ Take for example the function $f$, with $$\hat f(\xi)=\frac{1}{(1+\xi^2)^{3/2}(1+\log(\xi^2+1))}.$$ Note. If $\|f\|_{H^{s_0}}<\infty$, for some $s_0>2$, then $$\lim_{s\searrow 2}\|f\|_{H^s}=\|f\|_{H^2}.$$

2

I use Sobolev injection Which requires some assumptions on the set $\Omega$. Certainly, being connected is one of them, but also the boundary should not be too irregular: cusps are a problem, as are other features that obstruct passage to the boundary. The usual assumption in the Sobolev embedding theorem is that the domain is of Lipschitz class. (My ...

2

$\newcommand{\R}{\mathbb{R}}$Yes. The approximating argument that you use in the usual case of $C^\infty(0, T;\mathbb{R})\subset W^1(0, T;\mathbb{R})$ (that is, convolution against a mollifying family) works verbatim here. For the details you may look in Evans's book on PDEs (chapter on Sobolev spaces, look in the last paragraph "Spaces involving time"), or ...

1

No. Here is a counterexample, inspired by the answer to this question: If $V \subset H$ compact, is $L^2(0,T;V) \subset L^2(0,T;H)$ compact too? Take $v\in H^1$, $T=\pi$, $$\phi_n(t) = \sin(n t)v.$$ Since $\sin(n \circ)$ converges weakly but not strongly to zero in $L^2(0,T)$, it follows that $\phi_n$ converges weakly to zero in $L^2(0,T;H^1)$. If ...

1

Your third comment is on the right track, but you miss $n$ in the exponent: you can reduce the left hand side of the inequality to $$\|(e^{-2\pi i n\theta }-1)\widehat{f}(n)\|^2_{l^2}$$ To be more explicit, the desired inequality can be stated as $$\sum_{n\in\mathbb Z} |e^{-2\pi i n\theta }-1|^2 |\hat f(n)|^2 \le (2\pi)^2 |\theta|^{2s} \sum_{n\in\mathbb ... 1 Take \phi\in\mathcal{C}^\infty_c(\Omega_1\cup\Omega_2) and let K=\mathrm{supp}\phi be its support; let U_i=\Omega_i\cap K, i=1,2 and take \delta the Lebesgue number of the covering \{U_1, U_2\}. Now let B_1,\ldots, B_m balls of radius r such that 2r<\delta and$$\bigcup B_j\supseteq K\;.$$Let also \{\chi_j\} be a partition of unity ... 1 Apparently your problem is verifying that if a distribution v ( really your D^{\alpha}u) is of function type when restricted to \Omega_1 and \Omega_2 then so is on \Omega_1\cup \Omega_2 --the "sheaf property" of the distributions. It is enough to show that any compactly supported function \phi on \Omega_1\cup \Omega_2 is a sum of two ... 1 The following outlines the necessary steps, but some details are left out: First choose a subsequence (u_{n_k})_k with \Vert u_{n_k} \Vert_{H^1} \to \liminf_n \Vert u_n \Vert_{H^1} and rename the sequence to (u_n)_n again. You know that every bounded sequence has a weakly convergent subsequence (because H^1 is a Hilbert space). Hence u_{n_k} ... 1 Here's an idea for the case when the bundle is trivial, so K(x,y) is just a function on M \times M: I will use K to denote both the operator and its kernel. Think of the square of the L^2 norm of K(x,y) on M \times M as follows:$$ \int \int |K(x,y)|^2 \, dy \, dx = \int K(K(x, \cdot)) \,dx,$$where by K(K(x, \cdot)) I mean the operator K ... 1 I don't know of a theorem to help you in this case. However your operator is a sum of an inclusion from L^2 to H_0^k in the first coordinate and a kth derivative into L^2 from H^k in the second. The differential part is in fact a bounded operator. Therefore you only need to consider the inclusion. I think you can verify directly that the ... 1 The Weierstrass function is Hölder continuous, but is not absolutely continuous, and is not of bounded variation either. (Every function of bounded variation is differentiable at almost every point, but the Weierstrass function is nowhere differentiable.) Conversely, f(x) = 1/\log x is absolutely continuous on [0,1/2] but is not Hölder ... 1 If u vanishes on a set of measure \alpha, then$$ \int_\Omega u^2\le C(\alpha) \int_\Omega (u-u_\Omega)^2 $$because u_\Omega cannot be much larger than \int u^2. Indeed,$$ \int_\Omega u^2\ge \int_\Omega (u-u_\Omega)^2 \ge \int_{\{u=0\}} (u-u_\Omega)^2 =\alpha (u_\Omega)^2 $$hence$$ \int_\Omega u^2 \le 2 (u_\Omega)^2+ 2 \int_\Omega ...

1

Okay, I think this works. Since the function $a^x$ is convex as a function of $x$ we have, putting $a=(1+|\xi|^2)$ that $$(1+|\xi|^2)^r\leq (1-\theta)(1+|\xi|^2)^t +\theta (1+|\xi|^2)^s, \qquad r=(1-\theta)t+\theta s.$$ Multiplying both sides by $|\hat{f}|^2$ and integrating over $\mathbb{R}$ we get that the function $h(t)=\| f\|_{H^t}^2$ is convex on ...

1

$H^{1/2}_{per}[-\pi,\pi]$ contains discontinuous (even unbounded) functions - see the answer of Care Bear. On the other hand, Sobolev inbedding provides that $$H^{1/2+\varepsilon}_{per}[-\pi,\pi]\subset C_{per}[-\pi,\pi],$$ for every $\varepsilon>0$, and in fact the imbedding is compact.

1

If you are talking about Evans & Gariepy, "Measure Theory and Fine Properties of Functions", then either you have a different edition than I have, or your page-number is of, because my version of this book only has 256 pages (before the bibliography). Anyway, the result which I think you are looking for is the following: Let $(T_n)_n$ be a sequence ...

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