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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.

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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 …
Pietro Majer's user avatar
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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 …
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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 $N-1$. …
Pietro Majer's user avatar
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2 votes
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weak convergence in Sobolev spaces and pointwise convergence?

Just this: for all $r > 0$ the restriction of $u_m$ to the ball $B_r$ is compact in $L^1({B_r})$, hence it has a subsequence converging a.e. there. By a standard diagonal argument, there is a single s …
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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'\ …
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5 votes
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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 …
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1 vote
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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 \ …
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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\ …
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3 votes
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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_ …
Pietro Majer's user avatar
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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.
Pietro Majer's user avatar
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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}{\ …
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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\ …
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19 votes
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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 …
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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 …
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1 vote
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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 …
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