[proofplan] We prove that $\Phi: \ell^q \to (\ell^p)^*$ is an isometric isomorphism in three stages. First, we show $\Phi$ is well-defined and isometric: the [Hölder Inequality](/theorems/516) gives $\|\Phi(a)\| \le \|a\|_{\ell^q}$, and an explicit maximiser achieves equality. Second, we show $\Phi$ is surjective by representing an arbitrary $f \in (\ell^p)^*$ via its action on the standard basis vectors and verifying the resulting sequence lies in $\ell^q$. The cases $1 < p < \infty$ and $p = 1$ require different maximising sequences and different truncation arguments. [/proofplan] [step:Show $\Phi$ is well-defined and bounded with $\|\Phi(a)\|_{(\ell^p)^*} \le \|a\|_{\ell^q}$] Fix $a = (a_k)_{k=1}^\infty \in \ell^q$. For any $x = (x_k)_{k=1}^\infty \in \ell^p$, we must show that the series $\sum_{k=1}^\infty a_k x_k$ converges absolutely and that $|f_a(x)| \le \|a\|_{\ell^q} \|x\|_{\ell^p}$. **Case $1 < p < \infty$** (so $1 < q < \infty$). The counting measure $\mu$ on $\mathbb{N}$ is $\sigma$-finite, and the sequences $a \in \ell^q$ and $x \in \ell^p$ are measurable functions on $(\mathbb{N}, \mathcal{P}(\mathbb{N}), \mu)$. By the [Hölder Inequality](/theorems/516) with conjugate exponents $p, q$ satisfying $1/p + 1/q = 1$, applied to the functions $|x|$ and $|a|$ on this measure space: \begin{align*} \sum_{k=1}^\infty |a_k x_k| = \int_{\mathbb{N}} |a| \cdot |x| \, d\mu \le \|x\|_{\ell^p} \|a\|_{\ell^q}. \end{align*} **Case $p = 1$** (so $q = \infty$). For any $x \in \ell^1$: \begin{align*} \sum_{k=1}^\infty |a_k x_k| \le \|a\|_{\ell^\infty} \sum_{k=1}^\infty |x_k| = \|a\|_{\ell^\infty} \|x\|_{\ell^1}. \end{align*} In both cases, $f_a$ is a well-defined bounded linear functional on $\ell^p$ with $\|f_a\|_{(\ell^p)^*} \le \|a\|_{\ell^q}$. [guided] We need to verify that $\Phi(a) = f_a$ actually maps $\ell^p$ into $\mathbb{R}$ (i.e., the defining series converges) and that $f_a$ is bounded with the correct norm bound. **Case $1 < p < \infty$.** The Hölder inequality is the right tool here. We work on the measure space $(\mathbb{N}, \mathcal{P}(\mathbb{N}), \mu)$ where $\mu$ is the counting measure (assigning measure $1$ to each singleton). This measure space is $\sigma$-finite since $\mathbb{N} = \bigcup_{n=1}^\infty \{1, \ldots, n\}$. The [Hölder Inequality](/theorems/516) requires $p, q \in (1, \infty)$ with $1/p + 1/q = 1$ and measurable functions — both conditions are met. The inequality gives \begin{align*} \sum_{k=1}^\infty |a_k x_k| = \int_{\mathbb{N}} |a| \cdot |x| \, d\mu \le \left(\int_{\mathbb{N}} |x|^p \, d\mu\right)^{1/p} \left(\int_{\mathbb{N}} |a|^q \, d\mu\right)^{1/q} = \|x\|_{\ell^p} \|a\|_{\ell^q}. \end{align*} In particular, the series converges absolutely, so $f_a(x) = \sum_{k=1}^\infty a_k x_k$ is well-defined. Taking the supremum over $\|x\|_{\ell^p} \le 1$: $\|f_a\|_{(\ell^p)^*} \le \|a\|_{\ell^q}$. **Case $p = 1$.** Here $q = \infty$, and Hölder's inequality for $p = 1$ becomes the bound $|a_k x_k| \le \|a\|_{\ell^\infty} |x_k|$. Summing: $\sum_{k=1}^\infty |a_k x_k| \le \|a\|_{\ell^\infty} \|x\|_{\ell^1}$, giving $\|f_a\|_{(\ell^1)^*} \le \|a\|_{\ell^\infty}$. [/guided] [/step] [step:Construct a maximiser to prove $\|\Phi(a)\|_{(\ell^p)^*} = \|a\|_{\ell^q}$] We show the reverse inequality $\|\Phi(a)\|_{(\ell^p)^*} \ge \|a\|_{\ell^q}$ by exhibiting an explicit element $x \in \ell^p$ with $\|x\|_{\ell^p} = 1$ at which $|f_a(x)| = \|a\|_{\ell^q}$. **Case $1 < p < \infty$** (so $1 < q < \infty$). If $a = 0$, both sides are zero. Otherwise, define \begin{align*} x_k := \frac{|a_k|^{q-1} \operatorname{sgn}(a_k)}{\|a\|_{\ell^q}^{q/p}} \quad \text{for each } k \in \mathbb{N}, \end{align*} where $\operatorname{sgn}(a_k) = a_k/|a_k|$ when $a_k \neq 0$ and $\operatorname{sgn}(0) = 0$. We verify $x \in \ell^p$: \begin{align*} \sum_{k=1}^\infty |x_k|^p = \frac{1}{\|a\|_{\ell^q}^q} \sum_{k=1}^\infty |a_k|^{(q-1)p} = \frac{1}{\|a\|_{\ell^q}^q} \sum_{k=1}^\infty |a_k|^q = 1, \end{align*} using $(q-1)p = q$ (which follows from $1/p + 1/q = 1$, i.e., $q(p-1) = p$, i.e., $(q-1)p = q$). So $\|x\|_{\ell^p} = 1$. Now compute: \begin{align*} f_a(x) = \sum_{k=1}^\infty a_k x_k = \frac{1}{\|a\|_{\ell^q}^{q/p}} \sum_{k=1}^\infty |a_k|^q = \frac{\|a\|_{\ell^q}^q}{\|a\|_{\ell^q}^{q/p}} = \|a\|_{\ell^q}^{q - q/p} = \|a\|_{\ell^q}, \end{align*} since $q - q/p = q(1 - 1/p) = q/q = 1$. Hence $\|f_a\|_{(\ell^p)^*} \ge |f_a(x)| = \|a\|_{\ell^q}$. **Case $p = 1$** (so $q = \infty$). If $a = 0$, both sides vanish. Otherwise, for each $N \in \mathbb{N}$, choose $k_N$ with $|a_{k_N}| > \|a\|_{\ell^\infty} - 1/N$, and set $x = \operatorname{sgn}(a_{k_N}) e_{k_N}$ (the $k_N$-th standard basis vector, with appropriate sign). Then $\|x\|_{\ell^1} = 1$ and $f_a(x) = |a_{k_N}| > \|a\|_{\ell^\infty} - 1/N$. Sending $N \to \infty$: $\|f_a\|_{(\ell^1)^*} \ge \|a\|_{\ell^\infty}$. Combining with the previous step: $\|\Phi(a)\|_{(\ell^p)^*} = \|a\|_{\ell^q}$, so $\Phi$ is an isometry (and in particular injective). [guided] The upper bound $\|f_a\| \le \|a\|_{\ell^q}$ was established by Hölder. For the reverse inequality, we need to find a unit vector in $\ell^p$ that "extracts" the full $\ell^q$-norm from $a$. This is the standard duality maximiser construction. **Case $1 < p < \infty$.** The idea is to choose $x_k$ so that $a_k x_k \ge 0$ (to avoid cancellation) and $|x_k|$ is proportional to $|a_k|^{q-1}$ (the exponent that makes Hölder's inequality an equality). Specifically, Hölder's inequality becomes an equality when $|x|^p$ and $|a|^q$ are proportional, which requires $|x_k|^p = c |a_k|^q$ for some constant $c > 0$, i.e., $|x_k| = c^{1/p} |a_k|^{q/p} = c^{1/p} |a_k|^{q-1}$ (since $q/p = q - 1$ follows from $1/p = 1 - 1/q = (q-1)/q$). We choose the constant so that $\|x\|_{\ell^p} = 1$. Define $x_k = |a_k|^{q-1} \operatorname{sgn}(a_k) / \|a\|_{\ell^q}^{q/p}$. The exponent $(q-1)p = q$ follows from the conjugacy relation: $1/p + 1/q = 1$ gives $p(q-1) = p \cdot q(1 - 1/q) \cdot 1/(1-1/q) \cdot (1-1/q)$. More directly, $p = q/(q-1)$, so $(q-1)p = q$. Therefore \begin{align*} \|x\|_{\ell^p}^p = \frac{\sum_{k=1}^\infty |a_k|^{(q-1)p}}{\|a\|_{\ell^q}^q} = \frac{\sum_{k=1}^\infty |a_k|^q}{\|a\|_{\ell^q}^q} = 1. \end{align*} The pairing gives $f_a(x) = \sum_{k=1}^\infty a_k x_k$. Since $a_k x_k = |a_k|^q / \|a\|_{\ell^q}^{q/p} \ge 0$ for each $k$: \begin{align*} f_a(x) = \frac{1}{\|a\|_{\ell^q}^{q/p}} \sum_{k=1}^\infty |a_k|^q = \frac{\|a\|_{\ell^q}^q}{\|a\|_{\ell^q}^{q/p}} = \|a\|_{\ell^q}^{q(1 - 1/p)} = \|a\|_{\ell^q}^{q/q} = \|a\|_{\ell^q}. \end{align*} **Case $p = 1$.** The essential supremum norm cannot be attained in general — we approximate. For each $N$, pick an index $k_N$ with $|a_{k_N}| > \|a\|_{\ell^\infty} - 1/N$ and test against $x = \operatorname{sgn}(a_{k_N}) e_{k_N} \in \ell^1$. Then $\|x\|_{\ell^1} = 1$ and $|f_a(x)| = |a_{k_N}| > \|a\|_{\ell^\infty} - 1/N$. Since $N$ is arbitrary, $\|f_a\| \ge \|a\|_{\ell^\infty}$. [/guided] [/step] [step:Prove surjectivity: represent every $f \in (\ell^p)^*$ as $f = \Phi(a)$ for some $a \in \ell^q$] Let $f \in (\ell^p)^*$. For each $k \in \mathbb{N}$, let $e_k = (0, \ldots, 0, 1, 0, \ldots)$ denote the $k$-th standard basis vector (with $1$ in position $k$), and define $a_k := f(e_k)$. We claim $a := (a_k)_{k=1}^\infty \in \ell^q$ and $f = f_a$. **Case $1 < p < \infty$** (so $1 < q < \infty$). For each $N \in \mathbb{N}$, define the finite sequence \begin{align*} x_k^{(N)} := \begin{cases} |a_k|^{q-1} \operatorname{sgn}(a_k) & \text{if } 1 \le k \le N, \\ 0 & \text{if } k > N. \end{cases} \end{align*} Then $x^{(N)} \in \ell^p$ (it has finite support) with $\|x^{(N)}\|_{\ell^p}^p = \sum_{k=1}^N |a_k|^{(q-1)p} = \sum_{k=1}^N |a_k|^q$ (using $(q-1)p = q$). Evaluate $f$: \begin{align*} f(x^{(N)}) = \sum_{k=1}^N a_k |a_k|^{q-1} \operatorname{sgn}(a_k) = \sum_{k=1}^N |a_k|^q. \end{align*} Since $f$ is bounded, $|f(x^{(N)})| \le \|f\|_{(\ell^p)^*} \|x^{(N)}\|_{\ell^p}$: \begin{align*} \sum_{k=1}^N |a_k|^q \le \|f\|_{(\ell^p)^*} \left(\sum_{k=1}^N |a_k|^q\right)^{1/p}. \end{align*} If $S_N := \sum_{k=1}^N |a_k|^q > 0$, dividing both sides by $S_N^{1/p}$ gives $S_N^{1 - 1/p} = S_N^{1/q} \le \|f\|_{(\ell^p)^*}$, i.e., $S_N \le \|f\|_{(\ell^p)^*}^q$. This bound is uniform in $N$, so $\sum_{k=1}^\infty |a_k|^q \le \|f\|_{(\ell^p)^*}^q < \infty$, proving $a \in \ell^q$. **Case $p = 1$** (so $q = \infty$). For each $k$, $|a_k| = |f(e_k)| \le \|f\|_{(\ell^1)^*} \|e_k\|_{\ell^1} = \|f\|_{(\ell^1)^*}$. Hence $\|a\|_{\ell^\infty} \le \|f\|_{(\ell^1)^*} < \infty$, so $a \in \ell^\infty$. [guided] The candidate for the representing sequence is forced: if $f = f_a$, then $a_k = f_a(e_k) = a_k$. So we define $a_k := f(e_k)$ and must verify two things: (1) $a \in \ell^q$, and (2) $f = f_a$. **Showing $a \in \ell^q$ when $1 < p < \infty$.** The strategy is to test $f$ against cleverly chosen finite sequences that "extract" the $\ell^q$-norm of $a$. We use the same maximiser idea as in the isometry step, but truncated to $N$ terms. Define $x^{(N)}$ with $x_k^{(N)} = |a_k|^{q-1}\operatorname{sgn}(a_k)$ for $k \le N$ and $0$ for $k > N$. Since $x^{(N)}$ has finite support, it lies in $\ell^p$. By linearity and the definition $a_k = f(e_k)$: \begin{align*} f(x^{(N)}) = f\left(\sum_{k=1}^N |a_k|^{q-1}\operatorname{sgn}(a_k) e_k\right) = \sum_{k=1}^N |a_k|^{q-1}\operatorname{sgn}(a_k) \cdot f(e_k) = \sum_{k=1}^N |a_k|^q. \end{align*} The boundedness of $f$ gives $\sum_{k=1}^N |a_k|^q \le \|f\| \cdot \|x^{(N)}\|_{\ell^p}$. Using $\|x^{(N)}\|_{\ell^p} = (\sum_{k=1}^N |a_k|^q)^{1/p}$ and writing $S_N = \sum_{k=1}^N |a_k|^q$: \begin{align*} S_N \le \|f\| \cdot S_N^{1/p} \implies S_N^{1/q} \le \|f\|. \end{align*} The bound $S_N \le \|f\|^q$ holds for all $N$ (when $S_N = 0$, the inequality is immediate), so by the Monotone Convergence Theorem for sums, $\sum_{k=1}^\infty |a_k|^q \le \|f\|^q < \infty$. **Showing $a \in \ell^\infty$ when $p = 1$.** This is simpler: $|a_k| = |f(e_k)| \le \|f\| \cdot \|e_k\|_{\ell^1} = \|f\|$ for every $k$, giving $\|a\|_{\ell^\infty} \le \|f\|$. [/guided] [/step] [step:Verify $f = f_a$ on all of $\ell^p$] Define $g := f - f_a \in (\ell^p)^*$. Then $g(e_k) = f(e_k) - a_k = 0$ for every $k \in \mathbb{N}$. By linearity, $g$ vanishes on every finitely supported sequence. Since the finitely supported sequences are [dense in](/page/Dense%20Subset) $\ell^p$ (for $1 \le p < \infty$, any $x \in \ell^p$ satisfies $\|x - x^{(N)}\|_{\ell^p}^p = \sum_{k=N+1}^\infty |x_k|^p \to 0$ as $N \to \infty$, where $x^{(N)} = (x_1, \ldots, x_N, 0, 0, \ldots)$), and $g$ is continuous (being a bounded linear functional), we conclude $g \equiv 0$. Hence $f = f_a = \Phi(a)$. Since $a \in \ell^q$ (from the previous step) and $\Phi(a) = f$, the map $\Phi$ is surjective. Combined with the isometry property, $\Phi: \ell^q \to (\ell^p)^*$ is an isometric isomorphism. [guided] It remains to show that $f$ and $f_a$ agree on all of $\ell^p$, not just on the basis vectors. The difference $g = f - f_a$ is a bounded linear functional that vanishes on each $e_k$. By linearity, $g$ vanishes on $\operatorname{span}\{e_k : k \in \mathbb{N}\}$, the space of finitely supported sequences. The key fact is that finitely supported sequences are dense in $\ell^p$ for $1 \le p < \infty$: given any $x = (x_k)_{k=1}^\infty \in \ell^p$, define the truncation $x^{(N)} = (x_1, \ldots, x_N, 0, 0, \ldots)$. Then \begin{align*} \|x - x^{(N)}\|_{\ell^p}^p = \sum_{k=N+1}^\infty |x_k|^p, \end{align*} which is the tail of a convergent series (since $x \in \ell^p$) and therefore tends to $0$ as $N \to \infty$. Now, for any $x \in \ell^p$, the sequence $x^{(N)} \to x$ in $\ell^p$, and $g(x^{(N)}) = 0$ for every $N$. By [continuity](/page/Continuity) of $g$: \begin{align*} g(x) = \lim_{N \to \infty} g(x^{(N)}) = 0. \end{align*} Hence $f = f_a$ on all of $\ell^p$. This is a density argument that is specific to $1 \le p < \infty$: it fails for $p = \infty$, where the finitely supported sequences are not dense in $\ell^\infty$. This is consistent with the fact that $(\ell^\infty)^*$ is strictly larger than $\ell^1$ — the dual of $\ell^\infty$ contains singular functionals (related to Banach limits and finitely additive measures) that cannot be represented by sequences. [/guided] [/step]