5
$\begingroup$

It is well-known that if $\Omega\in \mathbb{R}^n$ is a bounded domain, then we have the embedding $$ L^4({\Omega})\subset L^2({\Omega}) $$ since $||f||_{L^2(\Omega)}\leq C(\Omega) ||f||_{L^4(\Omega)}$ by Hölder's inequality. However, the above inequality no longer holds when $\Omega$ is unbounded.

Now, for general domain $\Omega$, let us consider the set $\mathcal{F}=\{f:\Omega\to\mathbb{R}:||f||_{L^2(\Omega)}\leq C||f||_{L^4(\Omega)}< \infty\}$, and $C$ is an universal constant independent of $f$. I am wondering if there exists a function space $X$ such that $X\subset\mathcal{F}$, where $X$ contains functions with non-compact support. It is okay to assume $f$ is smooth.

$\endgroup$
20
  • 2
    $\begingroup$ Trivially, $X=0$, or $X = C^\infty_c(U)$ for some bounded set $U \subset \Omega$. Maybe you should be more specific about what properties you want $X$ to have. $\endgroup$ Oct 28, 2015 at 3:37
  • 14
    $\begingroup$ Let $X$ be the one-dimensional space spanned by $f(x) = e^{-|x|^2}$, or your other favorite non-compactly-supported function $f \in L^2(\Omega) \cap L^4(\Omega)$? (I pride myself on the ability to produce annoying trivial examples that fit requirements but are obviously not what people want... so keep trying :-) $\endgroup$ Oct 28, 2015 at 3:51
  • 2
    $\begingroup$ Why necessary bounded? Finite measure is enough. $\endgroup$ Oct 28, 2015 at 6:44
  • 1
    $\begingroup$ The definition of the class $\mathcal F $ is not completely clear to me: (1) is $\|f\|_{L^4(\Omega)}$ assumed to be finite? (2) Does $C$ depend on $f$? $\endgroup$ Oct 28, 2015 at 7:02
  • 2
    $\begingroup$ If $X\subset\mathcal F$ cannot be a vector space, what could it be that make sense? A ball in a normed space? The underlying question may be: somehow I got an estimate $||f||_{L^4(\mathbb R^n}\le M$, what else should I know to be able to bound $||f||_{L^2(\mathbb R^n)}$ ? $\endgroup$ Oct 29, 2015 at 13:48

2 Answers 2

1
$\begingroup$

Making such functions is easy. For example take $f_n = 1_{[0,1]} + \frac{1}{n} 1_{[1,n]}$. Then all $L^p$ norms ($p>1+\delta$) are comparable and close to $1$.

$\endgroup$
1
  • $\begingroup$ the question does not ask for functions, but for function spaces. $\endgroup$ Oct 24, 2017 at 10:48
0
$\begingroup$

If two Banach spaces are continuously embedded as subspaces of a topological vector space, then their vector sum and their intersection have natural structures of Banach spaces (in the first case as a quotient, in the second case as a closed subspace of their product). Unless I am misunderstanding your question, you can take the intersection of $L^2$ and $L^4$ (which are subspaces of the space of equivalence classes of meaurable functions) to get a Banach space with the required property.

$\endgroup$
1
  • 1
    $\begingroup$ $L^2\cap L^4$ does not have the required property: take $f_n=1_{[0,n]}$, you have $||f_n||_2=n^{1/2}$ while $||f_n||_4=n^{1/4}$... $\endgroup$ Nov 3, 2015 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.