Show that for \(0 < p < q \le \infty\), \[ \|a\|_{\ell^q} \le \|a\|_{\ell^p} \] for any sequence \(a = (a_j)_{j=1}^\infty \in \mathbb{C}^\mathbb{N}\), where \(\|a\|_{\ell^\infty} := \sup_j |a_j|\) and \(\|a\|_{\ell^r} := \left(\sum_{j=1}^\infty |a_j|^r\right)^{1/r}\) for \(r < \infty\).
Solution
1. Case 1: \(q = \infty\) (where \(0 < p < \infty\)).
1.1. If \(\|a\|_{\ell^p} = \infty\), then the right-hand side of \(\|a\|_{\ell^\infty} \le \|a\|_{\ell^p}\) is \(\infty\), and every extended real number is \(\le \infty\), so the inequality holds.
1.2. Assume \(\|a\|_{\ell^p} < \infty\). For every fixed index \(k \ge 1\):\[|a_k|^p \le \sum_{j=1}^\infty |a_j|^p = \|a\|_{\ell^p}^p.\]
1.3. Taking the \((1/p)\)-th power of both sides (since \(t \mapsto t^{1/p}\) is strictly increasing on \([0, \infty)\)):\[|a_k| \le \|a\|_{\ell^p} \quad \text{for all } k \ge 1.\]
1.4. Taking the supremum over all \(k \ge 1\):\[\|a\|_{\ell^\infty} = \sup_{k \ge 1} |a_k| \le \|a\|_{\ell^p}.\]
2. Case 2: \(q < \infty\) (where \(0 < p < q < \infty\)).
2.1. If \(\|a\|_{\ell^p} = 0\), then \(a_j = 0\) for all \(j\), so \(\|a\|_{\ell^q} = 0 = \|a\|_{\ell^p}\).
2.2. If \(\|a\|_{\ell^p} = \infty\), the inequality holds.
2.3. Assume \(0 < \|a\|_{\ell^p} < \infty\). Define the normalized sequence \(b = (b_j)_{j=1}^\infty\) by \(b_j = \frac{a_j}{\|a\|_{\ell^p}}\).
2.4. The normalized sequence satisfies\[\|b\|_{\ell^p}^p = \sum_{j=1}^\infty |b_j|^p = \sum_{j=1}^\infty \frac{|a_j|^p}{\|a\|_{\ell^p}^p} = 1.\]
2.5. In particular, for every index \(j \ge 1\), \(|b_j|^p \le \sum_{k=1}^\infty |b_k|^p = 1\), which implies \(|b_j| \le 1\).
2.6. Since \(q > p > 0\) and \(|b_j| \le 1\), the exponent comparison gives\[|b_j|^q = |b_j|^{q-p} |b_j|^p \le 1^{q-p} |b_j|^p = |b_j|^p \quad \text{for every } j \ge 1.\]
2.7. Summing over all \(j \ge 1\):\[\|b\|_{\ell^q}^q = \sum_{j=1}^\infty |b_j|^q \le \sum_{j=1}^\infty |b_j|^p = \|b\|_{\ell^p}^p = 1.\]
2.8. Taking the \((1/q)\)-th power gives \(\|b\|_{\ell^q} \le 1^{1/q} = 1\).
2.9. Multiplying both sides by the positive scalar \(\|a\|_{\ell^p}\):\[\|a\|_{\ell^q} = \|a\|_{\ell^p} \|b\|_{\ell^q} \le \|a\|_{\ell^p} \cdot 1 = \|a\|_{\ell^p}.\]
3. Q.E.D.
Steps 1 and 2 cover \(q = \infty\) and \(q < \infty\).