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A Sobolev space is a vector space of functions equipped with a norm that is a combination of Lp-norms of the function itself and its derivatives up to a given order.
32
votes
Compact embeddings of Sobolev spaces: a counterexample showing the Rellich-Kondrachov theore...
Yes, there is a standard way. Take any nonzero $u\in W^{1,p}(U)$, with support in a ball, say w.l.o.g $\operatorname{supp}(u)\subset B(0,r)\subset U$, and consider
$$ u_\epsilon(x):= u\big(\frac{x}{\ …
30
votes
Accepted
For which maps $S^1\to S^1$ is the winding number defined?
A class of maps including both continuous and $H^{1/2}$, where an extension is available, is Vanishing Mean Oscillation, VMO. This has been treated by several authors starting I think with Haïm Brezis …
19
votes
Accepted
Is the $W^{1, \infty}$ limit of differentiable functions also differentiable?
This is true, $f$ is actually everywhere differentiable. It is the "Limit under the Sign of Derivative" Theorem; it also holds for sequences of maps between Banach spaces (and you may even allow count …
13
votes
Arzelà-Ascoli theorem and Hölder spaces
For completeness, let's mention a simpler and more general statement: For $\Omega\subset\mathbb{R}^n$ a bounded open set, $k\in\mathbb{N}$ and $0<\beta<\alpha\le1$ there is a compact embedding
$$ C^{ …
8
votes
Accepted
Convolution with semigroup: does this belong to the Sobolev space $W^{1,1}$?
In other words, if $A$ is the infinitesimal generator of $T$, the mild solution of the abstract inhomogeneous Cauchy problem
$$\begin{cases}\dot v =A v +f\\v(0)=0\end{cases}$$
needs not to be $W^{1,1 …
7
votes
Showing integrability of a locally integrable function on a bounded domain under some additi...
Let $(g_k)_{k\ge0}$ be a sequence of smooth functions such that $g_k(x)=1$ if $\text{dist}(x,\partial\Omega)\ge 2^{-k}$,
$g_k (x)=0$ if $\text{dist}(x,\partial\Omega)\le 2^{-k-1}$ and $0\le g_k\ …
5
votes
Subspaces of a Sobolev space
Yes, all functions in $X_a$ are still symmetric wrto the line generated by $a$. And all traces of these functions on the affine hyperplanes orthogonal to $a$ are radially symmetric in dimension $N-1$. …
5
votes
Accepted
Blowup of Sobolev norms in approximating a non-absolutely continuous function
If $\|u_n\|_{1,p}$ does not diverge to $+\infty$, some subsequence $u_{n_j}$ converges weakly-$W^{1,p}(I)$ to some $g\in W^{1,p}(I)$, and still in $L^\infty(I)$ to $f$. So e.g. $u_{n_j}$ converges …
4
votes
Accepted
When is a Nemytskii map between Sobolev spaces compact?
I think in your setting $F$ is not compact, unless $f$ is a constant map. Indeed, if $f$ is not a constant map, it coincides on some non-trivial interval $[a,b]$ with a smooth diffeomorphism $g:\mat …
4
votes
Accepted
Can we stay invertible while approximating linear maps in Sobolev spaces?
I will recycle the answer I've been writing into some words of explanation on Alex Gavrilov's example, which is more simple and elegant. We can focus on the first column $p_n$ of an approximating sequ …
4
votes
Accepted
Compact radial Sobolev embedding $H^1_{rad}\hookrightarrow L^p$
It's Strauss embedding theorem for radially symmetric functions, proven here:
W. A. Strauss, Existence of solitary waves in higher dimensions, Commun. Math.
Phys. 55 (1977), 149-162.
4
votes
Smooth Sobolev extension from $W^{1,p}(U)$ to $W^{1,p} (\mathbb{R}^n) $
Extension is a well studied topic in the theory of Sobolev spaces, included in any treatise on the topic. Have a look e.g. to Adams' Sobolev spaces, Chapter 4 (Interpolation and Extension theorems). F …
3
votes
Accepted
A Sobolev-type inequality with weights
To complete your computation, let's treat the case of a function supported in interval $(0,1)$. Indeed, for $ f\in C^\infty_c(0,1)$ there is an inequality
$$ \int_0^1 r^{-3}f(r)^6 dr\le C\left(\int_ …
3
votes
Dense set in Sobolev space ${H^1}\left( {0,1} \right)$
Consider, for $m\ge2$, the function $\varphi_m(x):=x(1-x)^m$ . So
$$\varphi_m(0)=0\qquad \varphi_m(1)=0$$ $$ \varphi_m'(0)=1\qquad \varphi_m'(1)=0\ .$$ It is also easy to see that $\| \varphi_m'\ …
3
votes
On the compact embedding of Sobolev space
If $i:X\to Y$ is compact, so is $i^*:Y^*\to X^*$; moreover since here $i$ is dense, $i^*$ is injective.