Let \(1\le p<q<\infty\). Determine which statements are true and which are false:
\(L^p(\mathbb R)\subset L^q(\mathbb R)\);
\(L^q(\mathbb R)\subset L^p(\mathbb R)\);
\(L^p([0,1])\subset L^q([0,1])\);
\(L^q([0,1])\subset L^p([0,1])\);
\(\ell^p(\mathbb Z)\subset\ell^q(\mathbb Z)\);
\(\ell^q(\mathbb Z)\subset\ell^p(\mathbb Z)\).
Give counterexamples for the false statements. Also determine all \(s\ge1\) for which \[ L^p(\mathbb R)\cap L^q(\mathbb R)\subset L^s(\mathbb R). \]
Solution
The answers are \[ \begin{array}{c|cccccc} &\text{(i)}&\text{(ii)}&\text{(iii)}&\text{(iv)}&\text{(v)}&\text{(vi)}\\ \hline &\text{false}&\text{false}&\text{false}&\text{true}&\text{true}&\text{false}. \end{array} \] Moreover, \[ L^p(\mathbb R)\cap L^q(\mathbb R)\subset L^s(\mathbb R) \quad\Longleftrightarrow\quad p\le s\le q. \]
1. Counterexamples for (i), (ii), and (iii).
Choose \(a\) with \[ \frac1q\le a<\frac1p. \] Then \[ f(x)=x^{-a}\mathbf1_{(0,1)}(x) \] belongs to \(L^p\) but not \(L^q\), because \[ \int_0^1x^{-ar}\,dx<\infty \quad\Longleftrightarrow\quad ar<1. \] This disproves both (i) and (iii).
For (ii), choose \(b\) with \[ \frac1q<b\le\frac1p \] and set \[ g(x)=x^{-b}\mathbf1_{[1,\infty)}(x). \] Then \(g\in L^q(\mathbb R)\) but \(g\notin L^p(\mathbb R)\), because \[ \int_1^\infty x^{-br}\,dx<\infty \quad\Longleftrightarrow\quad br>1. \]
2. Statements (iv) and (v) are true, while (vi) is false.
Since \([0,1]\) has finite measure, Hölder’s inequality gives \[ \|f\|_p\le \|f\|_q \] for \(f\in L^q([0,1])\), proving (iv).
For sequences, if \(a=(a_n)\in\ell^p\), then \(a_n\to0\), so all but finitely many terms satisfy \(|a_n|\le1\). Hence for all sufficiently large \(n\), \[ |a_n|^q\le|a_n|^p, \] which proves \(a\in\ell^q\) and therefore (v).
For (vi), choose \(c\) with \[ \frac1q<c\le\frac1p \] and set \(a_n=n^{-c}\) for \(n\ge1\) and \(a_n=0\) for \(n\le0\). Then \(a\in\ell^q\) but \(a\notin\ell^p\).
3. \(L^p(\mathbb R)\cap L^q(\mathbb R)\subset L^s(\mathbb R)\) exactly when \(p\le s\le q\).
If \(p\le s\le q\), choose \(\theta\in[0,1]\) such that \[ \frac1s=\frac\theta p+\frac{1-\theta}{q}. \] For \(f\in L^p\cap L^q\), Hölder interpolation gives \[ \|f\|_s\le \|f\|_p^\theta\|f\|_q^{1-\theta}, \] so \(f\in L^s\).
Now suppose \(1\le s<p\). Choose \(b\) with \[ \frac1p<b\le\frac1s. \] Then \[ f(x)=x^{-b}\mathbf1_{[1,\infty)}(x) \] belongs to both \(L^p\) and \(L^q\) but not to \(L^s\).
Finally suppose \(s>q\). Choose \(a\) with \[ \frac1s\le a<\frac1q. \] Then \[ g(x)=x^{-a}\mathbf1_{(0,1)}(x) \] belongs to both \(L^p\) and \(L^q\) but not to \(L^s\).
Thus the inclusion holds exactly for \[ \boxed{p\le s\le q}. \]