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Anyone who has studied classical functional analysis has been exposed to the sequence spaces ($l^p$ spaces) and the Lebesgue spaces ($L^p$ spaces). It's known that there is a reversal of inclusion in these spaces when the underlying manifold of Lebesgue spaces has finite measure. Understanding that sequences are indeed functions defined on the set of natural numbers, the prior observation seems counter-intuitive, when coupling the concept of convergence of real series with the integral test. I would appreciate it if someone could throw some light on this issue, or if I should be a bit more elaborate.

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    $\begingroup$ Might be useful: math.stackexchange.com/q/66029/27978 $\endgroup$
    – copper.hat
    Sep 12, 2018 at 15:44
  • $\begingroup$ Analogously with $\ell^p\subset\ell^q$ if $p<q$, we have $L^p\cap L^\infty\subset L^q\cap L^\infty$ if $p<q$. $\endgroup$
    – user254433
    Sep 13, 2018 at 7:32

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We had the exact same question in our functional analysis tutorial. My intuition (and how I remember the inclusions) goes like this.

In $l^p$ spaces, a sequence $x$ can only be in any one of the $l^p$ spaces if $|x(n)|\geq 1$ for only finitely many $n$. If I increase the $p$-value, i.e. go from $l^p$ to $l^q$ with $q > p$, then those $|x(n)|^q$ become bigger compared to $|x(n)|^p$, where $|x(n)|>1$. But they don't matter at all for the convergence of $$\sum |x(n)|^q$$ because they are only finitely many. Those $|x(n)|$, where $|x(n)| < 1$ become smaller. Thus, if $x\in l^p$, then $x\in l^q$.

On the other hand, if I have $f\in L^P(\Omega)$ with $\mu(\Omega)<\infty$. Those values of $f$, where $|f(t)|\leq 1$ don't matter at all, because they only add up to at most $\mu(\Omega)\cdot 1$. So if you want to deside if $f\in L^q(\Omega)$, i.e. if $$\int_\Omega |f|^q\, d\mu$$ is finite, then only those function values of $f$ are important, where $|f(t)|> 1$. They become bigger with bigger $q$ and smaller with smaller $q$. Therefore the inclusion is reverse: $x\in L^p(\Omega)$ implies $x \in L^q(\Omega)$ if $q<p$.

Summary: For the convergence of the sum of a sequence in $l^p$-spaces those values of the sequence are important which are smaller than 1.They become smaller with bigger p-values. For the convergence of an integral in $L^p$ space where the domain has finite measure, those values of a function are important which are bigger than 1. They increase with bigger p-values.

The reverseal of inclusions becomes less counter intuitive, when you realize that the finite measure of the domain in $L^p$ spaces is essential. For domains with infinite measure, you can actually use an example from the $l^p$ spaces, to show that no inclusion holds:

Consider $f(t) = 1/n^{1/q}$ for $n-1< t \leq n$, where $f: R^+ \rightarrow R$. Then $$\int_{R^+}f^q\,d\mu = \sum \frac{1}{n} = \infty$$. But for all $p>q$: $f \in L^p(R^+)$.

$$\int_{R^+}f^p\,d\mu = \sum \frac{1}{n^{p/q}} < \infty$$

Thus, $L^q(R^+) \subset L^p(R^+)$ is wrong for $q<p$, which was true before. Notice, that this is basically the same example you would use to show, that $l^p \neq l^q$. It depends on function values which are smaller then 1 (!) and works only because you can add them up to infinity, if the domain has measure infinity. If you interprete sequences as a function space from natural numbers to $R$, then you will have to give the natural numbers the discrete measure (is that the write term?), i.e. $\mu(n)=1$ for all $n$, and your domain woud have infinite measure, which is the reason why the normal $L^p$ inclusions wouldnt work.

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Understanding that sequences are indeed functions defined on the set of natural numbers, the prior observation seems counter-intuitive, when coupling the concept of convergence of real series with the integral test.

In the case of $l^p$ the measure space ($\Bbb N$) has infinite measure and the previos theorem does not apply.

EDIT: consider and compare the following cases (do the calculations): $$f(x) = \frac1{x^\alpha},\qquad x\in(0,a),\qquad \alpha,a>0\hbox{ fixed.}$$ $$g(x) = \frac1{x^\alpha},\qquad x\in(b,+\infty),\qquad \alpha,b>0\hbox{ fixed.}$$ $$\sum_{n=1}^\infty\frac1{n^\alpha}.$$ Where is the problem in $\int_0^a f$? When $x\to 0$ ($f$ big in a small set). Where is the problem in $\int_b^\infty g$? When $x\to+\infty$ ($f$ small in a big set). Take $p\ge 1$.

What happens with $\int_0^a f^p$? That $f^p\ge f$ in the "bad set" and $\int_0^a f$ divergent $\implies \int_0^a f^p$ divergent.

What happens with $\int_b^\infty g^p$? That $g^p\le g$ in the "bad set" and $\int_b^\infty g$ can be divergent but $\int_b^\infty g^p$ convergent.

Finally, think: the series is like the first integral or the second integral?

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  • $\begingroup$ Thanks for your response, Mr. Martin-Blas. You've spoken correctly, but I still feel there's a basis for comparison between the two classes of spaces. Perhaps, this could be seen from the boundedness of all sequences in the sequence spaces and the fact that all functions on $\mathbb{N}$ can be taken as analogously continuous when applying the integral test? $\endgroup$
    – Uche Opara
    Sep 12, 2018 at 22:22
  • $\begingroup$ @UcheOpara, see the edit. $\endgroup$ Sep 13, 2018 at 7:10

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