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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.
1
vote
Accepted
Characterization of the dual of intersection of Banach spaces
In general, in order that the intersection be defined, we should assume that $(U,\|\cdot\|_U)$ and $(V,\|\cdot\|_V)$ are continuously embedded into an ambient Hausdorff TVS $E$. Then, $(U\cap V,\|\cdo …
3
votes
Is the $W^{1, \infty}$ limit of differentiable functions also differentiable?
Taken by a sort of generalisation frenzy I produced the following; I tried to make it as readable as I could. Recall that the upper and lower Dini derivatives are respectively :
$$D^*f(x):=\limsup_{y\ …
19
votes
Accepted
Is the $W^{1, \infty}$ limit of differentiable functions also differentiable?
This is true, $f$ is actually everywhere differentiable. It is the "Limit under the Sign of Derivative" Theorem; it also holds for sequences of maps between Banach spaces (and you may even allow count …
5
votes
Accepted
Blowup of Sobolev norms in approximating a non-absolutely continuous function
If $\|u_n\|_{1,p}$ does not diverge to $+\infty$, some subsequence $u_{n_j}$ converges weakly-$W^{1,p}(I)$ to some $g\in W^{1,p}(I)$, and still in $L^\infty(I)$ to $f$. So e.g. $u_{n_j}$ converges …
3
votes
On the compact embedding of Sobolev space
If $i:X\to Y$ is compact, so is $i^*:Y^*\to X^*$; moreover since here $i$ is dense, $i^*$ is injective.
2
votes
Extension Operator for $W^{1,\infty}(U,X)$
Say $I:=[a,b]$. We may extend $u$ simply putting it constant for $t\ge b$ and $t\le a$, that is $Eu(t):=u((t\vee a)\wedge b)$. This defines a norm-$1$ linear extensor $E$.
13
votes
Arzelà-Ascoli theorem and Hölder spaces
For completeness, let's mention a simpler and more general statement: For $\Omega\subset\mathbb{R}^n$ a bounded open set, $k\in\mathbb{N}$ and $0<\beta<\alpha\le1$ there is a compact embedding
$$ C^{ …
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\ …
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'\ …
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 …
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
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 …
1
vote
Property Sobolev space
Actually one can find a larger space of $g$, taking into account the Sobolev inequalities: if $f\in W^{k,p}$ and $g\in W^{k,q}$, then for any order of derivation $0\le i\le k$, one has $D^if\in L^{p_i …
1
vote
Accepted
Maximizing $\|u\|_q\|u\|_p$ over the unitary sphere in the Sobolev space $H_0^1(\Omega)$
Any maximizer $u\in\{ H^1_0(\Omega): \|\nabla u\|_2=1 \}$ of $\|u\|_p$ is a nonconstant, nonnegative function solving $-\Delta u = \lambda u^{p-1}$, with $\lambda=\lambda_p>0$. So if $u$ maximizes bot …
1
vote
Accepted
Convergence of Sobolev functions near the boundary
I think Question 1 has a positive answer. Denote $B=:B_0(1)$ and $A_r:=\{r<\|x\|<1\} $ for $0<r<1$. For functions $f\in W^{1,2}_0(B)$ we have a Poincaré inequality on $A_r$ :
$$\int_{A_r} f^2dx\le \ …