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I want to learn the embedding properties of $L^p$ spaces, but the Wiki description is too sophisticated.

In one sentence, if a function $f \in L^p(\Omega)$, then it must be (or, under what conditions) in $L^q(\Omega)$ for any $1 \le p \le q \le \infty$, i.e., $L^p \subseteq L^q$ for $1 \le p \le q \le \infty$, or the opposite? How to show that?

I think $f \in L^p(\Omega)$ has the norm \begin{align} \|f\|_{L^p} = \left(\int_{\Omega} |f(x)|^p dx\right)^\frac{1}{p} < \infty \end{align} and we have $\|f\|_{L^p} \ge \|f\|_{L^q}$(?) for $1 \le p \le q \le \infty$, thus $\|f\|_{L^q} < \infty$ as well.

BTW, I saw someone denotes $L^p$ as superscript while some other denotes $L_p$ as subscript, which is the formal notation for the Lebesgue integrable function spaces?

Analysis Newbie
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  • See my post here https://math.stackexchange.com/questions/2799738/is-l2-norm-always-smaller-than-l1-norm-even-in-infinite-dimensional-space/2799755#2799755 – Lorenzo Q May 29 '18 at 09:30
  • None of the inclusions are true in general. Look at the positive real line with the Lebesgue measure. The function $1/\sqrt{x}$ on $[0,1]$ and zero everywhere else is in $L^1$ but not in $L^2$. The opposite is true for $1/x$ defined on $[1,\infty)$ and zero everywhere else. – Manan May 29 '18 at 09:50
  • Regarding the question about notation: both are possible, just pick one style and stick to it. – harfe May 29 '18 at 10:47

2 Answers2

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Sometimes one of the embeddings is true, sometimes the other, sometimes none.

An important case is, when $\Omega$ has finite measure. Then it holds that $$\|f\|_{L^q} \leq C \|f\|_{L^p}, \quad\text{and}\quad L^q(\Omega)\subset L^p(\Omega)$$ for $1\leq p \leq q \leq \infty$ and a suitable constant $C$. This can be shown using Hölder's inequality.

Another case is the case of $\ell^p$ (ie. $\Omega$ is countable and we consider the counting measure). Then it is the other way around.

If we consider $\mathbb R^n$ with the $\ell^p$-Norm, then both embeddings are true.

However, if we use $\Omega=\mathbb R$, then no embedding can be found. So there are functions that are in $L^1(\mathbb R)\setminus L^2(\mathbb R)$, and there are functions in $L^1(\mathbb R)\setminus L^2(\mathbb R)$.

harfe
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  • How can we prove the case of $l^p(\mathbb{N})$ for the counting measure that $l^p(\mathbb{N}) \subseteq l^q(\mathbb{N})$ for $p<q$? By using Hölder's inequality as well? – Analysis Newbie Sep 17 '18 at 18:54
  • @AnalysisNewbie i found https://math.stackexchange.com/questions/4094/how-do-you-show-that-l-p-subset-l-q-for-p-leq-q – harfe Sep 19 '18 at 08:56
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If we have a finite measure space $X$, then $\|f\|_p\mu(X)^{-1/p}\leq \|f\|_q \mu(X)^{-1/q}$ if $p\leq q$. One can use Holder's inequality to show this. If the measure space $X$ is such that every set of positive measure has measure at least $m$ then $\|f\|_pm^{-1/p}\geq \|f\|_q m^{-1/q}$ if $p\leq q$. This can be shown by first showing the corresponding result for sequence spaces and then realizing that if one of the sides has to be finite, then there can be at most countably many atoms in the space. As a reference, I can suggest the 1st chapter of the 4th book of Stein and Shakarchi, and also Terence Tao's blog notes on $L^p$ spaces.

Manan
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