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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.

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    $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Commented Oct 28, 2015 at 3:37
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    $\begingroup$ 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 :-) $\endgroup$
    – Nate Eldredge
    Commented Oct 28, 2015 at 3:51
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    $\begingroup$ Why necessary bounded? Finite measure is enough. $\endgroup$
    – Fedor Petrov
    Commented Oct 28, 2015 at 6:44
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    $\begingroup$ 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 $f$? $\endgroup$
    – Pietro Majer
    Commented Oct 28, 2015 at 7:02
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    $\begingroup$ 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 $||f||_{L^2(\mathbb R^n)}$ ? $\endgroup$
    – Jean Duchon
    Commented Oct 29, 2015 at 13:48

2 Answers 2

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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 $1$.

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

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

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