- Prove the following inclusions of \(L^p\) spaces for \(m(X) < \infty\): \[ L^\infty(X) &\subset L^2(X) \subset L^1(X) \\ \ell^2(\ZZ) &\subset \ell^1(\ZZ) \subset \ell^\infty(\ZZ) .\]
Erratum: the second line of the display should read \(\ell^1(\ZZ) \subset \ell^2(\ZZ) \subset \ell^\infty(\ZZ)\). On a finite measure space the \(L^p\) spaces decrease as \(p\) increases, and for counting measure on \(\ZZ\) the \(\ell^p\) spaces increase as \(p\) increases.
Solution
1. If \(m(X) < \infty\), then \(L^\infty(X) \subseteq L^2(X) \subseteq L^1(X)\).
1.1. \(\|f\|_1 \le m(X)^{1/2}\|f\|_2\) for \(f \in L^2(X)\).
By the Cauchy–Schwarz inequality, \(\int|f| = \int |f|\cdot 1 \le \|f\|_2\|1\|_2 = \|f\|_2\, m(X)^{1/2}\).
1.2. \(\|f\|_2 \le m(X)^{1/2}\|f\|_\infty\) for \(f \in L^\infty(X)\).
\(\int|f|^2 \le \|f\|_\infty^2\, m(X)\).
1.3. On \([0,1]\) with Lebesgue measure, \(x^{-1/2} \in L^1 \setminus L^2\) and \(\log x \in L^2 \setminus L^\infty\).
\(\int_0^1 x^{-1/2}\,dx = 2\) and \(\int_0^1 x^{-1}\,dx = \infty\). Also \(\int_0^1 (\log x)^2\,dx = 2\) and \(\log x\) is unbounded near \(0\).
1.4. Q.E.D.
Steps 1 and 2 give the inclusions, and step 3 shows that they are strict.
2. For counting measure on \(\ZZ\), \(\ell^1(\ZZ) \subseteq \ell^2(\ZZ) \subseteq \ell^\infty(\ZZ)\), with strict inclusions.
2.1. If \(\sum_n |a_n| < \infty\) then \(\sum_n |a_n|^2 < \infty\).
The terms of a convergent series tend to \(0\), so \(|a_n| \le 1\) for all but finitely many \(n\), and for those \(n\), \(|a_n|^2 \le |a_n|\).
2.2. If \(\sum_n |a_n|^2 < \infty\) then \(|a_n| \le \|a\|_2\) for every \(n\).
\(|a_n|^2 \le \sum_k |a_k|^2\).
2.3. \(a_n = 1/|n|\) (\(n \neq 0\)) is in \(\ell^2 \setminus \ell^1\), and \(a_n = 1/\sqrt{|n|}\) (\(n \neq 0\)) is in \(\ell^\infty \setminus \ell^2\); set \(a_0 = 0\) in both.
\(\sum_{n \geq 1} 1/n\) diverges and \(\sum_{n \geq 1} 1/n^2\) converges, and \(\sup_n |n|^{-1/2} = 1\).
2.4. Q.E.D.
Steps <2>1–<2>3.
3. Q.E.D.
Step 1 proves the first line of the display. Step 2 proves the second line with \(\ell^1\) and \(\ell^2\) exchanged; by step 2 3, the inclusion \(\ell^2(\ZZ) \subset \ell^1(\ZZ)\) printed in the problem is false.