Lp Inclusions for Counting Measure and Sequence Spaces

Counting Measure Turns Functions into Sequences

On a countable set with counting measure, integrating a nonnegative function amounts to summing its values. Consequently, the \(L^p\) spaces on \(\mathbb N=\{1,2,\ldots\}\) are precisely the familiar sequence spaces. This identification also makes an inclusion phenomenon transparent: if \(1\le p<q\), an \(\ell^p\) sequence must be in \(\ell^q\), but the reverse inclusion fails.

Throughout, \(\mathbb N\) has its power-set sigma-algebra and counting measure \(\mu\), defined by \(\mu(E)=\#E\), with \(\mu(E)=+\infty\) when \(E\) is infinite. This measure space is complete and sigma-finite, since it is a countable union of finite-measure singletons, but it is not finite. Every real- or complex-valued function on \(\mathbb N\) is measurable, and the only null set is the empty set. Thus equality \(\mu\)-almost everywhere is the same as equality at every index.

The Isometric Identification

Proof. Fix \(1\le r<\infty\). The functions \(s_N=\sum_{n=1}^N |f(n)|^r\chi_{\{n\}}\) are nonnegative measurable simple functions, increase pointwise, and converge to \(|f|^r\). The Monotone Convergence Theorem applies on this measure space and gives \(\int |f|^r\,d\mu=\lim_{N\to\infty}\int s_N\,d\mu=\lim_{N\to\infty}\sum_{n=1}^N |f(n)|^r=\sum_{n=1}^\infty |f(n)|^r\), allowing \(+\infty\). Therefore \(f\in L^r(\mu)\) exactly when its value sequence belongs to \(\ell^r\), and the displayed equality gives equality of the norms. Conversely, every sequence \(a\) defines a measurable function \(f(n)=a_n\), so the map is onto. It is linear because addition and scalar multiplication are pointwise. For \(L^\infty\), every nonempty set has measure at least one, so no exceptional null set can be discarded in computing an essential supremum. Hence \(\|f\|_\infty=\sup_n|f(n)|\), which is finite exactly when \((f(n))\) is bounded. The same pointwise construction proves surjectivity in this case. \(\square\)

Inclusion When the Exponent Increases

Proof. First let \(p<q<\infty\) and \(a\in\ell^p\). If \(\|a\|_p=0\), then \(\sum_n|a_n|^p=0\); nonnegativity of the terms forces every \(a_n=0\), and the conclusion follows. Otherwise, each term satisfies \(|a_n|^p\le\sum_k|a_k|^p=\|a\|_p^p\), so \(|a_n|\le\|a\|_p\). It follows, for every \(n\), that \(|a_n|^q=|a_n|^{q-p}|a_n|^p\le\|a\|_p^{q-p}|a_n|^p\). Summing these nonnegative inequalities gives \(\sum_n|a_n|^q\le\|a\|_p^{q-p}\sum_n|a_n|^p=\|a\|_p^q<\infty\). Thus \(a\in\ell^q\), and taking the \(q\)-th root gives \(\|a\|_q\le\|a\|_p\). For \(q=\infty\), the bound \(|a_n|^p\le\|a\|_p^p\) gives \(|a_n|\le\|a\|_p\) for every \(n\), so taking the supremum proves the claim. \(\square\)

This argument uses the fact that each individual coordinate is controlled by the whole \(\ell^p\) norm. For finite \(q\), that coordinate bound lets us extract the factor \(|a_n|^{q-p}\) while summing the remaining \(p\)-th powers. No finiteness assumption on the measure space is being made; in fact, counting measure on \(\mathbb N\) is infinite.

Why the Inclusion Is Strict

To show that \(\ell^p\subseteq\ell^q\) is strict, it is enough to find one sequence in \(\ell^q\) that is not in \(\ell^p\). The power sequence \(a_n=n^{-1/p}\) does this for every finite \(p\) and every \(q>p\), including \(q=\infty\). We give the needed series test explicitly.

Proof. Group the terms in blocks \(2^k\le n<2^{k+1}\), for integers \(k\ge0\). If \(s>1\), the block has \(2^k\) terms, each at most \(2^{-ks}\). Its sum is at most \(2^{k(1-s)}\). The sum of these bounds over \(k\ge0\) is a convergent geometric series, with ratio \(2^{1-s}<1\), so the original series converges. If \(0<s\le1\), each term in the block is at least \(2^{-(k+1)s}\), so the block sum is at least \(2^k2^{-(k+1)s}=2^{-s}2^{k(1-s)}\ge2^{-s}\). Infinitely many blocks each contribute at least the fixed positive number \(2^{-s}\), so the series diverges. These two cases prove the claim. \(\square\)

Proof. Set \(a_n=n^{-1/p}\). Then \(\sum_n|a_n|^p=\sum_n n^{-1}\), which diverges by the lemma, so \(a\notin\ell^p\). If \(q<\infty\), then \(\sum_n|a_n|^q=\sum_n n^{-q/p}\). Since \(q/p>1\), the lemma shows that this series converges, so \(a\in\ell^q\). If \(q=\infty\), then \(0<a_n\le1\) for all \(n\), so \(a\in\ell^\infty\). Thus in either case there is an element of the larger space that is not in \(\ell^p\). \(\square\)

A Common Mistake: Reversing the Inclusion

A frequent error is to carry over the finite-measure embedding in the wrong setting. On a finite measure space, when \(p<q\), the standard inclusion goes from \(L^q\) into \(L^p\), with a bound involving the total measure. Counting measure on \(\mathbb N\) is not finite, so that result cannot be applied here. For this space the inclusion goes in the other direction: \(\ell^p\subseteq\ell^q\) when \(p<q\).

Wrong calculation: “Since \(1<2\), \(\ell^2\subseteq\ell^1\), so \((1/n)\) must be summable.” This incorrectly uses a finite-measure conclusion on a space of infinite counting measure. The correction is to test the two sums: \(\sum_n|1/n|^2=\sum_n n^{-2}<\infty\), while \(\sum_n|1/n|=\sum_n n^{-1}=+\infty\). Thus \((1/n)\in\ell^2\) but \((1/n)\notin\ell^1\), consistent with the strict inclusion \(\ell^1\subsetneq\ell^2\).

The counting-measure identification is useful because it translates norm questions into series questions without changing their values. The inclusion estimate gives a general implication for all sequences in \(\ell^p\), while the power examples show why it cannot be reversed, even though the underlying set \(\mathbb N\) is countable.