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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
1
vote
1
answer
75
views
Name for a uniform local boundedness property of a function
I am working with a function $f : \mathbb{R}^N \to \mathbb{R}$ having the property that for every $R > 0$, there exists $M > 0$ such that if $x, y \in \mathbb{R}^N$ and $\vert x - y \vert \le R$, then …
1
vote
About a completion of a Sobolev space
I think that the trace is defined by the completion. Indeed for every $u \in H^1 (\Omega)$, you have $\Vert\operatorname{tr} u\Vert_{L^2 (\Omega)} \le \Vert u \Vert_{\varepsilon}$. Since $H^1 (\mathca …
2
votes
Do extracted weak $H^{1,2}$-limits and $C^0$-limits coincide?
Both convergences imply the weak convergence in $L^2 (I)$, which has a unique limit.
5
votes
A question on optimal Sobolev inequality.
Here is a variational argument to prove that the maximizers do not change sign.
If $f \in H^1 (\mathbb{R}^N)$ be a maximizer, $u$ can be written as
$$
f = f_+ - f_-,
$$
with $f_+ \ne 0$ and $f_- \ …
2
votes
Simultaneous Orthogonal basis for $L^2(\mathbb{R}^n)$ and $H^1(\mathbb{R}^n)$
I your aim is to apply the Galerkin method, you do not need simultaneous orthonormal basis.
An inspection of Evans’ proof shows that you need a sequence of linear maps $(P_n)_{n \in \mathbb{N}}$ such …
3
votes
Accepted
Why $\|u\|_{\tau}\leq C[u]_{W^{s,p}}^a\|u\|_{L^q}^{1-a}$ not correct for $p=1$?
The issue with $p = 1$ is that some definitions of fractional Sobolev spaces that were equivalent when $p > 1$ (by the Gagliardo seminorm that you gave, by interpolation between functional spaces, by …
6
votes
Accepted
Are Sobolev spaces on non-compact manifolds separable?
Yes they are.
Step 1 There exists measurable sections $e_1, e_2, \dotsc, e_m$, where $m = \dim M$, of $TM$ (measurable functions mapping a point $x$ to a vector of its tangent plane $T_xM$) such that …
2
votes
0
answers
137
views
Points are removable for weakly differentiable functions
If $\Omega \subseteq \mathbb{R}^N$ is an open set and $N \ge 2$, then any point $a \in \Omega$ is removable for weakly differentiable maps: for each function $u \in W^{1, 1} (\Omega \setminus \{a\})$, …
2
votes
Friedrichs/Poincare inequality on $S_n \times (0,\infty)$?
By Fubini's theorem and by taking functions that do not depend on the $\mathbb{S}^n$ variable, your estimate is equivalent with the inequality
$$
\int_0^\infty \vert u \vert^2 \le C \int_0^\infty \v …
2
votes
Accepted
Does Trudinger inequality implies this critical Sobolev embedding?
This is not the case. Functions can have a high level of integrability and still not have bounded mean oscillation. For example, let $u : B_1 \to \mathbb{R}$ be defined for $x \in B_1 \setminus \{0\} …
1
vote
Accepted
$u_n$ bounded in $L^\infty(0,T;H) \cap L^2(0,T;V)$ implies $u_n \to u$ strongly in $L^2(0,T;...
As it is stated, this property does not hold: indeed consider the sequence of functinos $(u_n)_{n \in \mathbb{N}}$ defined for $t \in [0, T]$ by
$$
u_n (t) = \sin (2n\pi t) v,
$$
where $v \in V$ is …
2
votes
Accepted
Sobolev's lemma on manifolds
This follows from its counterpart in the Euclidean space by local charts.
If you want to have an estimate on the derivative $D^r f$, then you should impose some bounds on derivatives of the curvatur …
2
votes
Sobolev trace map: is the fractional seminorm bounded by just the gradient?
This should follow from the nonhomogeneous trace inequalty
$$
\vert T u \vert_{H^{1/2} (\partial M)} \le C \lVert \nabla u \rVert_{L^2} + \lVert u \rVert_{L^2},
$$
and and from the classical Poincar …
2
votes
Isocapacity inequalities in the theory of Sobolev Spaces
For $p > n$, assume that the Hölder continuous representative vanishes at some point $a \in Q^d$.
Sinco $p > n$, by the Sobolev--Morrey embedding, for every $x \in Q^d$
$$
\vert u (x) \vert= \vert u …
5
votes
Accepted
Lieb: Stability of matter, problem with variational method
You are minimizing under the constraint that $\rho \ge 0$. Hence your variation $\rho_m + t \eta$ might not be admissible. (Say if $\eta \ne 0$ and $\vert t\vert$ is large enough.)
The trick is to co …