[proofplan] The proof uses the **complex method of interpolation**. By duality, the operator norm $\|T\|_{p_\theta \to q_\theta}$ is determined by the [supremum](/page/Supremum%20and%20Infimum) of $\left|\int (Tf) \cdot g \, d\mu_2\right|$ over unit-norm simple functions $f \in L^{p_\theta}$ and $g \in L^{q_\theta'}$. We embed this bilinear pairing into a holomorphic function $F(z)$ on the strip $S = \{z \in \mathbb{C} : 0 < \operatorname{Re}(z) < 1\}$ by constructing analytic families $f_z$ and $g_z$ that interpolate the exponents. On the boundary lines $\operatorname{Re}(z) = 0$ and $\operatorname{Re}(z) = 1$, the endpoint boundedness hypotheses yield $|F(z)| \le M_0$ and $|F(z)| \le M_1$ respectively. The Hadamard Three-Lines Lemma — a Phragmen-Lindelof-type result for the strip — then gives $|F(\theta)| \le M_0^{1-\theta} M_1^{\theta}$, which is the desired bound. [/proofplan]

[step:Reduce to bounding a bilinear pairing of unit-norm simple functions]We treat the case $q_\theta < \infty$; the case $q_\theta = \infty$ requires only notational changes (replacing the duality pairing by the essential [supremum](/page/Supremum%20and%20Infimum)). Since $1 \le q_\theta < \infty$, its conjugate exponent $q_\theta' := q_\theta/(q_\theta - 1) \in (1, \infty]$ satisfies $1/q_\theta + 1/q_\theta' = 1$. By the duality characterization of the $L^{q_\theta}$ norm, \begin{align*} \|Tf\|_{L^{q_\theta}(E_2)} = \sup \left\{ \left| \int_{E_2} (Tf) \cdot g \, d\mu_2 \right| : g \in L^{q_\theta'}(E_2), \; \|g\|_{L^{q_\theta'}} \le 1 \right\}. \end{align*} Therefore \begin{align*} \|T\|_{p_\theta \to q_\theta} = \sup \left\{ \left| \int_{E_2} (Tf) \cdot g \, d\mu_2 \right| : f \in L^{p_\theta}(E_1), \; g \in L^{q_\theta'}(E_2), \; \|f\|_{L^{p_\theta}} \le 1, \; \|g\|_{L^{q_\theta'}} \le 1 \right\}. \end{align*} By the [Density of Simple Functions in $L^p$](/theorems/893), simple functions are [dense](/page/Dense%20Subset) in $L^{p_\theta}(E_1)$ and in $L^{q_\theta'}(E_2)$ (using $p_\theta < \infty$ and $q_\theta' < \infty$ when $q_\theta > 1$; the case $q_\theta = 1$ requires $g \in L^\infty$ which we handle by truncation). It therefore suffices to prove \begin{align*} \left| \int_{E_2} (Tf) \cdot g \, d\mu_2 \right| \le M_0^{1-\theta} M_1^{\theta} \end{align*} for all simple functions $f: E_1 \to \mathbb{C}$ and $g: E_2 \to \mathbb{C}$ with $\|f\|_{L^{p_\theta}} = 1$ and $\|g\|_{L^{q_\theta'}} = 1$.[/step]

[guided]The first challenge is that $T$ is assumed bounded on $L^{p_0}$ and $L^{p_1}$, but we need to prove boundedness on the intermediate space $L^{p_\theta}$. A direct approach — trying to split an arbitrary $L^{p_\theta}$ function into $L^{p_0}$ and $L^{p_1}$ pieces — leads to uncontrolled errors. Instead, we reformulate the problem using duality. The duality characterization of the $L^q$ norm states that for $1 \le q < \infty$, \begin{align*} \|h\|_{L^q(E_2)} = \sup \left\{ \left| \int_{E_2} h \cdot g \, d\mu_2 \right| : g \in L^{q'}(E_2), \; \|g\|_{L^{q'}} \le 1 \right\}, \end{align*} where $q' = q/(q-1)$ is the conjugate exponent. This identity follows from the [Holder Inequality](/theorems/516) (which gives $\le$) and the explicit extremizer $g = |h|^{q-2} \overline{h} / \|h\|_{L^q}^{q/q'}$ (which gives $\ge$). Applying this with $h = Tf$ and $q = q_\theta$, we reduce the operator norm bound $\|Tf\|_{L^{q_\theta}} \le M_0^{1-\theta}M_1^\theta \|f\|_{L^{p_\theta}}$ to the scalar inequality \begin{align*} \left| \int_{E_2} (Tf) \cdot g \, d\mu_2 \right| \le M_0^{1-\theta} M_1^{\theta} \end{align*} for all $f$ and $g$ with unit norms in $L^{p_\theta}$ and $L^{q_\theta'}$ respectively. This is the bilinear form that we will embed into a holomorphic function. Why restrict to simple functions? Because the analytic family $f_z$ we construct in the next step involves raising $|f|$ to a complex power, which requires $f$ to take only finitely many values to avoid branch-cut issues. By the [Density of Simple Functions in $L^p$](/theorems/893), the simple functions are dense in $L^{p_\theta}(E_1)$ for $p_\theta < \infty$ (which holds since $p_\theta$ is a convex combination of finite quantities $1/p_0$ and $1/p_1$ with $p_0, p_1 \ge 1$). Since both sides of the desired inequality are [continuous](/page/Continuity) in $f$ (in $L^{p_\theta}$) and in $g$ (in $L^{q_\theta'}$), the bound for simple functions extends to all of $L^{p_\theta}$ by density.[/guided]

[step:Construct the analytic family $F(z) = \int_{E_2} (Tf_z) \cdot g_z \, d\mu_2$ by deforming the exponents into the complex strip]Fix simple functions $f = \sum_{j=1}^J a_j \mathbb{1}_{A_j}$ and $g = \sum_{k=1}^K b_k \mathbb{1}_{B_k}$, where $A_j \subset E_1$ and $B_k \subset E_2$ are pairwise disjoint measurable sets of finite measure, with $a_j, b_k \in \mathbb{C} \setminus \{0\}$. Normalize so that $\|f\|_{L^{p_\theta}} = 1$ and $\|g\|_{L^{q_\theta'}} = 1$. Define the exponent functions \begin{align*} \alpha(z) &:= \frac{p_\theta(1-z)}{p_0} + \frac{p_\theta z}{p_1}, \\ \beta(z) &:= \frac{q_\theta'(1-z)}{q_0'} + \frac{q_\theta' z}{q_1'}, \end{align*} where $q_0' = q_0/(q_0-1)$ and $q_1' = q_1/(q_1-1)$ are the conjugates of $q_0$ and $q_1$. These are affine functions of $z$ satisfying $\alpha(\theta) = 1$ and $\beta(\theta) = 1$. Define the analytic families \begin{align*} f_z: E_1 &\to \mathbb{C}, \qquad f_z(x) := |f(x)|^{\alpha(z)} \frac{f(x)}{|f(x)|} = |f(x)|^{\alpha(z)} e^{i\arg f(x)}, \\ g_z: E_2 &\to \mathbb{C}, \qquad g_z(y) := |g(y)|^{\beta(z)} \frac{g(y)}{|g(y)|} = |g(y)|^{\beta(z)} e^{i\arg g(y)}, \end{align*} where we set $f_z(x) = 0$ when $f(x) = 0$ and similarly for $g_z$. Since $f$ and $g$ are simple, the expressions $|f(x)|^{\alpha(z)} = \exp(\alpha(z) \log|f(x)|)$ are well-defined and entire in $z$ on each set $A_j$ (where $|f| = |a_j| > 0$). At $z = \theta$, the normalization gives $f_\theta = f$ and $g_\theta = g$. Now define the holomorphic function on the open strip $S := \{z \in \mathbb{C} : 0 < \operatorname{Re}(z) < 1\}$: \begin{align*} F: \overline{S} &\to \mathbb{C}, \\ z &\mapsto \int_{E_2} (Tf_z)(y) \cdot g_z(y) \, d\mu_2(y). \end{align*} Since $f$ and $g$ are simple, this integral is a finite sum \begin{align*} F(z) = \sum_{j=1}^J \sum_{k=1}^K |a_j|^{\alpha(z)} |b_k|^{\beta(z)} e^{i(\arg a_j + \arg b_k)} \int_{B_k} (T\mathbb{1}_{A_j})(y) \, d\mu_2(y), \end{align*} which is a finite linear combination of exponentials $\exp(c_1 z + c_2)$ with $c_1, c_2 \in \mathbb{C}$. Therefore $F$ is entire (in particular, holomorphic on $S$ and [continuous](/page/Continuity) on $\overline{S}$). For boundedness: writing $z = s + it$, each factor $|a_j|^{\alpha(z)}$ satisfies $\big||a_j|^{\alpha(z)}\big| = |a_j|^{\alpha(s)}$ (since the imaginary part of $\alpha(z)$ contributes only a phase), and $\alpha(s)$ is bounded for $s \in [0,1]$. With finitely many terms, $F$ is bounded on $\overline{S}$.[/step]

[guided]The heart of the complex interpolation method is the construction of the function $F(z)$. The idea is to "move" the exponents continuously from the endpoint pair $(p_0, q_0)$ at $\operatorname{Re}(z) = 0$ to the endpoint pair $(p_1, q_1)$ at $\operatorname{Re}(z) = 1$, passing through the target pair $(p_\theta, q_\theta)$ at $z = \theta$. **Why deform the [test functions](/page/Test%20Function) rather than the operator?** The operator $T$ is fixed — it acts the same way regardless of which $L^p$ space we consider. What changes between the endpoints is the *norm* used to measure inputs and outputs. By modifying $f$ and $g$ through the power $|f|^{\alpha(z)}$, we effectively redistribute the "mass" of $f$ so that: - When $\operatorname{Re}(z) = 0$: the function $f_z$ has its mass distributed as an $L^{p_0}$ function, i.e., $\|f_z\|_{L^{p_0}} = \|f\|_{L^{p_\theta}}^{p_\theta/p_0} = 1$ (by the normalization $\|f\|_{L^{p_\theta}} = 1$ and the definition of $\alpha$). - When $\operatorname{Re}(z) = 1$: similarly $\|f_z\|_{L^{p_1}} = 1$. - When $z = \theta$: $\alpha(\theta) = 1$, so $f_\theta = f$ and $g_\theta = g$. Let us verify the exponent design. We need $\alpha(\theta) = 1$. Substituting $z = \theta$: \begin{align*} \alpha(\theta) = \frac{p_\theta(1-\theta)}{p_0} + \frac{p_\theta \theta}{p_1} = p_\theta \left(\frac{1-\theta}{p_0} + \frac{\theta}{p_1}\right) = p_\theta \cdot \frac{1}{p_\theta} = 1, \end{align*} using the defining relation $1/p_\theta = (1-\theta)/p_0 + \theta/p_1$. Similarly $\beta(\theta) = 1$. **Analyticity of $F$.** On each piece $A_j$, the function $|a_j|^{\alpha(z)}$ equals $\exp(\alpha(z) \log|a_j|)$. Since $\alpha(z)$ is affine in $z$, this is an entire function of $z$. The linearity of $T$ and of integration then gives $F(z)$ as a finite sum of entire functions (one for each pair $(j,k)$), so $F$ is entire — in particular, holomorphic on $S$ and continuous on $\overline{S}$. **Boundedness on the strip.** Each term in the sum has the form $C_{jk} \cdot |a_j|^{\alpha(z)} \cdot |b_k|^{\beta(z)}$ where $C_{jk} \in \mathbb{C}$ is fixed. Writing $z = s + it$ with $s = \operatorname{Re}(z) \in [0,1]$: \begin{align*} |a_j|^{\alpha(z)} = |a_j|^{\operatorname{Re}(\alpha(z))} = |a_j|^{\alpha(s)}, \end{align*} since $\alpha$ is affine and real-valued on $\mathbb{R}$. To see this precisely, write $\alpha(z) = \alpha(s) + it \cdot p_\theta(1/p_1 - 1/p_0)$, so $\operatorname{Re}(\alpha(s+it)) = \alpha(s)$. Therefore \begin{align*} |a_j|^{\alpha(z)} = |a_j|^{\alpha(s)} \cdot \exp\!\left(it \cdot p_\theta\left(\frac{1}{p_1} - \frac{1}{p_0}\right) \log|a_j|\right). \end{align*} The second factor has modulus $1$, so $\big||a_j|^{\alpha(z)}\big| = |a_j|^{\alpha(s)}$. Since $s \in [0,1]$ and $\alpha$ is affine, $\alpha(s)$ ranges between $\alpha(0) = p_\theta/p_0$ and $\alpha(1) = p_\theta/p_1$, both finite positive numbers. Each $|a_j|$ is a positive constant, so $|a_j|^{\alpha(s)}$ is bounded on $[0,1]$. The same applies to $|b_k|^{\beta(z)}$. Therefore $F$ is bounded on $\overline{S}$.[/guided]

[step:State and prove the Hadamard Three-Lines Lemma] [claim:Hadamard Three-Lines Lemma] Let $S := \{z \in \mathbb{C} : 0 < \operatorname{Re}(z) < 1\}$ be the open vertical strip. Suppose $F: \overline{S} \to \mathbb{C}$ is [continuous](/page/Continuity) on $\overline{S}$, holomorphic on $S$, and bounded on $\overline{S}$. Define \begin{align*} M_0 := \sup_{t \in \mathbb{R}} |F(it)|, \qquad M_1 := \sup_{t \in \mathbb{R}} |F(1 + it)|. \end{align*} Then for every $\theta \in (0,1)$, \begin{align*} \sup_{t \in \mathbb{R}} |F(\theta + it)| \le M_0^{1-\theta} \, M_1^{\theta}. \end{align*} [/claim] [proof] If $M_0 = 0$ or $M_1 = 0$, then $F \equiv 0$ on the corresponding boundary line, and the [Maximum Modulus Principle](/theorems/491) applied to bounded subregions gives $F \equiv 0$ on $\overline{S}$, so the inequality holds. Assume $M_0, M_1 > 0$. Define the auxiliary function \begin{align*} G: \overline{S} &\to \mathbb{C}, \\ z &\mapsto M_0^{z-1} \, M_1^{-z} \, F(z), \end{align*} where $M_0^{z-1} := \exp((z-1)\log M_0)$ and $M_1^{-z} := \exp(-z \log M_1)$. The function $G$ is holomorphic on $S$, continuous and bounded on $\overline{S}$ (since $|M_0^{z-1}| = M_0^{\operatorname{Re}(z)-1}$ and $|M_1^{-z}| = M_1^{-\operatorname{Re}(z)}$, both bounded for $\operatorname{Re}(z) \in [0,1]$). On the boundary lines: \begin{align*} |G(it)| &= M_0^{-1} \cdot M_1^{0} \cdot |F(it)| \le M_0^{-1} \cdot M_0 = 1, \\ |G(1+it)| &= M_0^{0} \cdot M_1^{-1} \cdot |F(1+it)| \le M_1^{-1} \cdot M_1 = 1. \end{align*} For $\varepsilon > 0$, define \begin{align*} G_\varepsilon: \overline{S} &\to \mathbb{C}, \\ z &\mapsto e^{\varepsilon z^2} G(z). \end{align*} Then $G_\varepsilon$ is holomorphic on $S$ and continuous on $\overline{S}$. Writing $z = s + it$ with $s \in [0,1]$: \begin{align*} |G_\varepsilon(s + it)| = e^{\varepsilon(s^2 - t^2)} |G(s+it)|. \end{align*} Since $F$ (and hence $G$) is bounded on $\overline{S}$, say $|G(z)| \le C$, the factor $e^{\varepsilon(s^2 - t^2)} \le e^{\varepsilon} \cdot e^{-\varepsilon t^2}$ ensures $|G_\varepsilon(s+it)| \to 0$ as $|t| \to \infty$, uniformly in $s \in [0,1]$. Therefore, for $R > 0$ sufficiently large, $|G_\varepsilon(z)| \le 1$ on $\{z \in \overline{S} : |{\operatorname{Im}(z)}| \ge R\}$. On the rectangle $\Omega_R := \{z \in S : |\operatorname{Im}(z)| < R\}$, the function $G_\varepsilon$ is holomorphic. On the boundary $\partial \Omega_R$: on the vertical sides ($\operatorname{Re}(z) = 0$ or $1$), we have $|G_\varepsilon| \le e^{\varepsilon} \cdot 1 = e^{\varepsilon}$ since $|G| \le 1$ there and $s^2 \le 1$; on the horizontal sides ($\operatorname{Im}(z) = \pm R$), we chose $R$ so that $|G_\varepsilon| \le 1$. By the [Maximum Modulus Principle](/theorems/491), $|G_\varepsilon(z)| \le e^{\varepsilon}$ for all $z \in \overline{\Omega_R}$. Taking $R \to \infty$ and then $\varepsilon \to 0$: \begin{align*} |G(z)| \le 1 \quad \text{for all } z \in \overline{S}. \end{align*} Substituting back: for $z = \theta + it$, \begin{align*} |F(\theta + it)| = M_0^{1-\theta} M_1^{\theta} |G(\theta + it)| \le M_0^{1-\theta} M_1^{\theta}. \end{align*} [/proof] [/step]

[step:Estimate $|F(z)|$ on the boundary lines $\operatorname{Re}(z) = 0$ and $\operatorname{Re}(z) = 1$ using the endpoint bounds]**Boundary $\operatorname{Re}(z) = 0$:** Set $z = it$ for $t \in \mathbb{R}$. We compute $\|f_{it}\|_{L^{p_0}(E_1)}$. Since $f = \sum_j a_j \mathbb{1}_{A_j}$ with the sets $A_j$ pairwise disjoint: \begin{align*} \int_{E_1} |f_{it}|^{p_0} \, d\mu_1 &= \sum_{j=1}^J |a_j|^{p_0 \operatorname{Re}(\alpha(it))} \mu_1(A_j) = \sum_{j=1}^J |a_j|^{p_0 \cdot p_\theta/p_0} \mu_1(A_j) = \sum_{j=1}^J |a_j|^{p_\theta} \mu_1(A_j) = \|f\|_{L^{p_\theta}}^{p_\theta} = 1, \end{align*} using $\operatorname{Re}(\alpha(it)) = \alpha(0) = p_\theta/p_0$ and the normalization $\|f\|_{L^{p_\theta}} = 1$. Therefore $\|f_{it}\|_{L^{p_0}} = 1$. Similarly, $\operatorname{Re}(\beta(it)) = \beta(0) = q_\theta'/q_0'$, and $\|g_{it}\|_{L^{q_0'}} = \|g\|_{L^{q_\theta'}}^{q_\theta'/q_0'} = 1$. Since $T: L^{p_0}(E_1) \to L^{q_0}(E_2)$ with $\|T\|_{p_0 \to q_0} \le M_0$, and $\|f_{it}\|_{L^{p_0}} = 1$, we have $\|Tf_{it}\|_{L^{q_0}} \le M_0$. Applying the [Holder Inequality](/theorems/516) with conjugate exponents $q_0$ and $q_0'$ to the measurable functions $Tf_{it}: E_2 \to \mathbb{C}$ and $g_{it}: E_2 \to \mathbb{C}$ on the $\sigma$-finite measure space $(E_2, \mu_2)$: \begin{align*} |F(it)| = \left| \int_{E_2} (Tf_{it}) \cdot g_{it} \, d\mu_2 \right| \le \|Tf_{it}\|_{L^{q_0}} \, \|g_{it}\|_{L^{q_0'}} \le M_0 \cdot 1 = M_0. \end{align*} **Boundary $\operatorname{Re}(z) = 1$:** Set $z = 1 + it$. The same computation with $\alpha(1) = p_\theta/p_1$ gives $\|f_{1+it}\|_{L^{p_1}} = 1$, and $\beta(1) = q_\theta'/q_1'$ gives $\|g_{1+it}\|_{L^{q_1'}} = 1$. The endpoint bound $\|T\|_{p_1 \to q_1} \le M_1$ and [Holder's Inequality](/theorems/516) yield \begin{align*} |F(1+it)| \le \|Tf_{1+it}\|_{L^{q_1}} \, \|g_{1+it}\|_{L^{q_1'}} \le M_1 \cdot 1 = M_1. \end{align*}[/step]

[guided]This is the step where the endpoint boundedness hypotheses are consumed. The key computation is the norm of $f_z$ on the boundary lines. **Why does $\|f_{it}\|_{L^{p_0}} = 1$?** We defined $f_z(x) = |f(x)|^{\alpha(z)} e^{i\arg f(x)}$, so $|f_z(x)| = |f(x)|^{\operatorname{Re}(\alpha(z))}$. The exponent function $\alpha(z) = p_\theta(1-z)/p_0 + p_\theta z/p_1$ evaluated at $z = it$ gives \begin{align*} \alpha(it) = \frac{p_\theta}{p_0}(1 - it) + \frac{p_\theta}{p_1}(it) = \frac{p_\theta}{p_0} + it \cdot p_\theta\!\left(\frac{1}{p_1} - \frac{1}{p_0}\right). \end{align*} The real part is $\operatorname{Re}(\alpha(it)) = p_\theta/p_0$. Therefore \begin{align*} \|f_{it}\|_{L^{p_0}}^{p_0} = \int_{E_1} |f(x)|^{p_0 \cdot p_\theta/p_0} \, d\mu_1(x) = \int_{E_1} |f(x)|^{p_\theta} \, d\mu_1(x) = \|f\|_{L^{p_\theta}}^{p_\theta} = 1. \end{align*} The exponents are designed so that the $p_\theta$-norm of $f$ reappears as the $p_0$-norm of $f_{it}$: raising $|f|$ to the power $p_\theta/p_0$ and then taking the $p_0$-norm produces $\left(\int |f|^{p_\theta}\right)^{1/p_0}$, which is $1^{1/p_0} = 1$. The same mechanism works for $g_{it}$: we need $\|g_{it}\|_{L^{q_0'}} = 1$. We have $\operatorname{Re}(\beta(it)) = q_\theta'/q_0'$, so \begin{align*} \|g_{it}\|_{L^{q_0'}}^{q_0'} = \int_{E_2} |g(y)|^{q_0' \cdot q_\theta'/q_0'} \, d\mu_2(y) = \int_{E_2} |g(y)|^{q_\theta'} \, d\mu_2(y) = \|g\|_{L^{q_\theta'}}^{q_\theta'} = 1. \end{align*} With these norm identities in hand, we apply the endpoint bounds. Since $f_{it} \in L^{p_0}(E_1)$ with $\|f_{it}\|_{L^{p_0}} = 1$, the hypothesis $\|T\|_{p_0 \to q_0} \le M_0$ gives $\|Tf_{it}\|_{L^{q_0}} \le M_0$. To bound $|F(it)|$, we pair $Tf_{it} \in L^{q_0}(E_2)$ with $g_{it} \in L^{q_0'}(E_2)$ using the [Holder Inequality](/theorems/516). The hypotheses of Holder's inequality require conjugate exponents $q_0$ and $q_0'$ (verified: $1/q_0 + 1/q_0' = 1$ by definition) and integrability ($Tf_{it} \in L^{q_0}$ and $g_{it} \in L^{q_0'}$, both verified). The conclusion: \begin{align*} |F(it)| \le \|Tf_{it}\|_{L^{q_0}} \cdot \|g_{it}\|_{L^{q_0'}} \le M_0 \cdot 1 = M_0. \end{align*} The boundary $\operatorname{Re}(z) = 1$ is handled identically, replacing $(p_0, q_0, M_0)$ by $(p_1, q_1, M_1)$.[/guided]

[step:Apply the Three-Lines Lemma at $z = \theta$ to obtain $\|T\|_{p_\theta \to q_\theta} \le M_0^{1-\theta} M_1^\theta$]We verify the hypotheses of the Hadamard Three-Lines Lemma for $F$: - **[Continuity](/page/Continuity) on $\overline{S}$:** $F$ is a finite sum of entire functions (as computed in the construction step), hence continuous on $\overline{S}$. - **Holomorphicity on $S$:** Same reasoning — $F$ is entire. - **Boundedness on $\overline{S}$:** For $z = s + it \in \overline{S}$, each term in the sum defining $F(z)$ has the form $C_{jk} \cdot |a_j|^{\alpha(s)} \cdot |b_k|^{\beta(s)} \cdot e^{i(\cdots)t}$, where the exponential factor has modulus $1$. Since $\alpha(s)$ and $\beta(s)$ are bounded for $s \in [0,1]$, and there are finitely many terms, $F$ is bounded on $\overline{S}$. - **Boundary estimates:** $\sup_t |F(it)| \le M_0$ and $\sup_t |F(1+it)| \le M_1$, as established in the previous step. The Three-Lines Lemma gives \begin{align*} |F(\theta)| \le M_0^{1-\theta} M_1^\theta. \end{align*} Since $f_\theta = f$ and $g_\theta = g$ (because $\alpha(\theta) = 1$ and $\beta(\theta) = 1$), we have \begin{align*} F(\theta) = \int_{E_2} (Tf)(y) \cdot g(y) \, d\mu_2(y). \end{align*} Therefore \begin{align*} \left| \int_{E_2} (Tf) \cdot g \, d\mu_2 \right| \le M_0^{1-\theta} M_1^\theta \end{align*} for all simple functions $f, g$ with $\|f\|_{L^{p_\theta}} = 1$ and $\|g\|_{L^{q_\theta'}} = 1$.[/step]

[guided]This is where the entire construction pays off. We have: 1. A holomorphic function $F$ on the strip $S = \{0 < \operatorname{Re}(z) < 1\}$, continuous and bounded on $\overline{S}$. 2. Boundary estimates $|F(it)| \le M_0$ and $|F(1+it)| \le M_1$ for all $t \in \mathbb{R}$. These are precisely the hypotheses of the Hadamard Three-Lines Lemma (stated and proved as a claim above). The lemma asserts that for $\theta \in (0,1)$: \begin{align*} |F(\theta + it)| \le M_0^{1-\theta} M_1^\theta \quad \text{for all } t \in \mathbb{R}. \end{align*} In particular, taking $t = 0$: \begin{align*} |F(\theta)| \le M_0^{1-\theta} M_1^\theta. \end{align*} The crucial observation is that the value $z = \theta$ is exactly the point where the deformed functions $f_z$ and $g_z$ recover the originals. Since $\alpha(\theta) = 1$, we have $f_\theta(x) = |f(x)|^1 \cdot e^{i\arg f(x)} = f(x)$. Similarly $g_\theta = g$. Therefore $F(\theta) = \int_{E_2} (Tf) \cdot g \, d\mu_2$, which is precisely the bilinear pairing whose bound we needed. The geometric content of the argument is this: the $L^p$ "landscape" is parameterized by the strip $\{0 \le \operatorname{Re}(z) \le 1\}$, with the left edge encoding $(p_0, q_0)$ and the right edge encoding $(p_1, q_1)$. The bound at the interior point $z = \theta$ is controlled by the boundary bounds through the maximum principle — the same principle that governs harmonic functions and, more generally, holomorphic functions. This is why the result is often called "complex interpolation": the complex-analytic structure of the strip is the engine that drives the estimate.[/guided]

[step:Extend from simple functions to all of $L^{p_\theta}(E_1)$ by density and conclude] We have established that for all simple functions $f: E_1 \to \mathbb{C}$ and $g: E_2 \to \mathbb{C}$ with $\|f\|_{L^{p_\theta}} = 1$ and $\|g\|_{L^{q_\theta'}} = 1$, \begin{align*} \left| \int_{E_2} (Tf) \cdot g \, d\mu_2 \right| \le M_0^{1-\theta} M_1^\theta. \end{align*} Taking the [supremum](/page/Supremum%20and%20Infimum) over all such $g$ and applying the duality characterization of the $L^{q_\theta}$ norm: \begin{align*} \|Tf\|_{L^{q_\theta}(E_2)} \le M_0^{1-\theta} M_1^\theta \|f\|_{L^{p_\theta}(E_1)} \end{align*} for all simple $f \in L^{p_\theta}(E_1)$. By the [Density of Simple Functions in $L^p$](/theorems/893), the simple functions in $L^{p_\theta}(E_1)$ are [dense](/page/Dense%20Subset) in $L^{p_\theta}(E_1)$ (since $p_\theta < \infty$, the hypotheses of Theorem 893 are satisfied on the $\sigma$-finite measure space $(E_1, \mu_1)$). For a general $f \in L^{p_\theta}(E_1)$, choose simple functions $f_m \to f$ in $L^{p_\theta}$. Since $f_m \in L^{p_0}(E_1) \cap L^{p_1}(E_1)$ (simple functions on a $\sigma$-finite space belong to every $L^p$), $Tf_m$ is well-defined, and the bound $\|Tf_m\|_{L^{q_\theta}} \le M_0^{1-\theta}M_1^\theta \|f_m\|_{L^{p_\theta}}$ shows that $(Tf_m)$ is Cauchy in $L^{q_\theta}(E_2)$. By completeness of $L^{q_\theta}(E_2)$, the sequence $Tf_m$ converges in $L^{q_\theta}$ to some $h \in L^{q_\theta}(E_2)$. We define $Tf := h$, and the bound \begin{align*} \|Tf\|_{L^{q_\theta}(E_2)} = \lim_{m \to \infty} \|Tf_m\|_{L^{q_\theta}(E_2)} \le M_0^{1-\theta} M_1^\theta \lim_{m \to \infty} \|f_m\|_{L^{p_\theta}(E_1)} = M_0^{1-\theta} M_1^\theta \|f\|_{L^{p_\theta}(E_1)} \end{align*} holds by [continuity](/page/Continuity) of the norm. This establishes $\|T\|_{p_\theta \to q_\theta} \le M_0^{1-\theta} M_1^\theta$. [/step]