[proofplan]
For each level $\lambda > 0$ split $f = f_\lambda + r_\lambda$ at the threshold $|f| = \lambda$. Sublinearity gives $d_{Tf}(\lambda) \le d_{Tf_\lambda}(\lambda/2) + d_{Tr_\lambda}(\lambda/2)$. The layer-cake representation expresses $\|Tf\|_{L^{p_\theta}}^{p_\theta}$ as an integral against $\lambda^{p_\theta - 1}$; substituting the diagonal weak-type bounds reduces the problem to two single integrals $I_0, I_1$, which evaluate by Tonelli directly to $\|f\|_{L^{p_\theta}}^{p_\theta}$ times explicit Beta-type constants. The endpoint case $p_1 = \infty$ uses the sharper truncation $|f| = \lambda/(2M_1)$ that forces $d_{Tr_\lambda}(\lambda/2) = 0$.
[/proofplan]
[step:Recall the layer-cake formula and the definition of weak-type bounds]
For a $\nu$-measurable function $u: Y \to \mathbb{C}$, define the **distribution function**
\begin{align*}
d_u: [0, \infty) &\to [0, \infty] \\
\lambda &\mapsto \nu(\{y \in Y : |u(y)| > \lambda\}).
\end{align*}
The **layer-cake representation** states that for any $0 < q < \infty$,
\begin{align*}
\int_Y |u|^q\, d\nu = q \int_0^\infty \lambda^{q - 1}\, d_u(\lambda)\, d\mathcal{L}^1(\lambda).
\end{align*}
This follows from Tonelli applied to the indicator $\mathbb{1}_{\{|u(y)|>\lambda\}}$ on $Y\times(0,\infty)$ (Tonelli applies because the integrand is non-negative and measurable; we restrict to $\sigma$-finite $(Y, \nu)$, the standard setting in interpolation theory).
The hypothesis "$T$ is of weak type $(p_i, p_i)$ with constant $M_i$" means
\begin{align*}
d_{Tg}(\lambda) \le \left( \frac{M_i \|g\|_{L^{p_i}}}{\lambda} \right)^{p_i} \qquad \text{for all } g \in L^{p_i}(X),\ \lambda > 0.
\end{align*}
Sublinearity of $T$ means $|T(g_1 + g_2)| \le |Tg_1| + |Tg_2|$ pointwise $\nu$-a.e. for $g_1, g_2 \in L^{p_0}(X) + L^{p_1}(X)$.
[/step]
[step:Decompose $f$ at threshold $|f| = \lambda$ and bound $d_{Tf}(\lambda)$]
Fix $f \in L^{p_\theta}(X)$ with $\|f\|_{L^{p_\theta}} > 0$ (the case $f = 0$ is immediate). For each $\lambda > 0$, define
\begin{align*}
f_\lambda &: X \to \mathbb{C}, & f_\lambda(x) &:= f(x)\,\mathbb{1}_{\{|f(x)| > \lambda\}}, \\
r_\lambda &: X \to \mathbb{C}, & r_\lambda(x) &:= f(x)\,\mathbb{1}_{\{|f(x)| \le \lambda\}}.
\end{align*}
Then $f = f_\lambda + r_\lambda$ pointwise. We verify $f_\lambda \in L^{p_0}(X)$ and $r_\lambda \in L^{p_1}(X)$.
On $\{|f| > \lambda\}$, since $p_0 < p_\theta$ and $|f|^{p_0 - p_\theta} \le \lambda^{p_0 - p_\theta}$ (because $p_0 - p_\theta < 0$ and $|f| > \lambda$),
\begin{align*}
\|f_\lambda\|_{L^{p_0}}^{p_0} = \int_X |f|^{p_0}\,\mathbb{1}_{\{|f|>\lambda\}}\,d\mu \le \lambda^{p_0 - p_\theta}\int_X |f|^{p_\theta}\,d\mu = \lambda^{p_0 - p_\theta}\,\|f\|_{L^{p_\theta}}^{p_\theta} < \infty.
\end{align*}
For $r_\lambda$ when $p_1 < \infty$: on $\{|f| \le \lambda\}$, since $p_1 > p_\theta$ and $|f|^{p_1 - p_\theta} \le \lambda^{p_1 - p_\theta}$,
\begin{align*}
\|r_\lambda\|_{L^{p_1}}^{p_1} \le \lambda^{p_1 - p_\theta}\,\|f\|_{L^{p_\theta}}^{p_\theta} < \infty.
\end{align*}
For $p_1 = \infty$, $\|r_\lambda\|_{L^\infty} \le \lambda$, treated separately in the final step.
By sublinearity, $\{|Tf| > \lambda\} \subseteq \{|Tf_\lambda| > \lambda/2\} \cup \{|Tr_\lambda| > \lambda/2\}$, hence
\begin{align*}
d_{Tf}(\lambda) \le d_{Tf_\lambda}(\lambda/2) + d_{Tr_\lambda}(\lambda/2).
\end{align*}
Applying the diagonal weak-type hypotheses,
\begin{align*}
d_{Tf_\lambda}(\lambda/2) \le \left( \frac{2 M_0 \|f_\lambda\|_{L^{p_0}}}{\lambda} \right)^{p_0}, \qquad d_{Tr_\lambda}(\lambda/2) \le \left( \frac{2 M_1 \|r_\lambda\|_{L^{p_1}}}{\lambda} \right)^{p_1}.
\end{align*}
[/step]
[step:Express $\|Tf\|_{L^{p_\theta}}^{p_\theta}$ as $I_0 + I_1$ via layer-cake]
By the layer-cake formula applied to $Tf$ at exponent $p_\theta$,
\begin{align*}
\|Tf\|_{L^{p_\theta}}^{p_\theta} = p_\theta \int_0^\infty \lambda^{p_\theta - 1}\, d_{Tf}(\lambda)\, d\mathcal{L}^1(\lambda) \le p_\theta\,(I_0 + I_1),
\end{align*}
where, on substituting the bounds from the previous step,
\begin{align*}
I_0 &:= (2 M_0)^{p_0} \int_0^\infty \lambda^{p_\theta - 1 - p_0}\, \|f_\lambda\|_{L^{p_0}}^{p_0}\, d\mathcal{L}^1(\lambda), \\
I_1 &:= (2 M_1)^{p_1} \int_0^\infty \lambda^{p_\theta - 1 - p_1}\, \|r_\lambda\|_{L^{p_1}}^{p_1}\, d\mathcal{L}^1(\lambda).
\end{align*}
[/step]
[step:Evaluate $I_0$ by Tonelli, obtaining $I_0 = (2M_0)^{p_0}\,\|f\|_{L^{p_\theta}}^{p_\theta}/(p_\theta - p_0)$]
Expand $\|f_\lambda\|_{L^{p_0}}^{p_0} = \int_X |f(x)|^{p_0}\,\mathbb{1}_{\{|f(x)|>\lambda\}}\,d\mu(x)$. By [Tonelli's Theorem](/theorems/2956) (the integrand $\lambda^{p_\theta - 1 - p_0}\,|f(x)|^{p_0}\,\mathbb{1}_{\{|f(x)|>\lambda\}}$ is non-negative and measurable on the product),
\begin{align*}
I_0 = (2M_0)^{p_0}\int_X|f(x)|^{p_0}\biggl(\int_0^{|f(x)|}\lambda^{p_\theta - 1 - p_0}\,d\mathcal{L}^1(\lambda)\biggr)d\mu(x).
\end{align*}
Since $p_\theta > p_0$, we have $p_\theta - 1 - p_0 > -1$, so the inner integral evaluates as
\begin{align*}
\int_0^{|f(x)|}\lambda^{p_\theta - p_0 - 1}\,d\mathcal{L}^1(\lambda) = \frac{|f(x)|^{p_\theta - p_0}}{p_\theta - p_0}.
\end{align*}
Substituting,
\begin{align*}
I_0 = \frac{(2M_0)^{p_0}}{p_\theta - p_0}\int_X|f(x)|^{p_0}\cdot|f(x)|^{p_\theta - p_0}\,d\mu(x) = \frac{(2M_0)^{p_0}}{p_\theta - p_0}\,\|f\|_{L^{p_\theta}}^{p_\theta}.
\end{align*}
[/step]
[step:Evaluate $I_1$ symmetrically (case $p_1 < \infty$), obtaining $I_1 = (2M_1)^{p_1}\,\|f\|_{L^{p_\theta}}^{p_\theta}/(p_1 - p_\theta)$]
Assume $p_1 < \infty$. Expand $\|r_\lambda\|_{L^{p_1}}^{p_1} = \int_X |f(x)|^{p_1}\,\mathbb{1}_{\{|f(x)|\le\lambda\}}\,d\mu(x)$. By Tonelli,
\begin{align*}
I_1 = (2M_1)^{p_1}\int_X|f(x)|^{p_1}\biggl(\int_{|f(x)|}^\infty\lambda^{p_\theta - 1 - p_1}\,d\mathcal{L}^1(\lambda)\biggr)d\mu(x).
\end{align*}
Since $p_\theta < p_1$, we have $p_\theta - 1 - p_1 < -1$, so the inner integral evaluates as
\begin{align*}
\int_{|f(x)|}^\infty\lambda^{p_\theta - p_1 - 1}\,d\mathcal{L}^1(\lambda) = \frac{|f(x)|^{p_\theta - p_1}}{p_1 - p_\theta}.
\end{align*}
Substituting,
\begin{align*}
I_1 = \frac{(2M_1)^{p_1}}{p_1 - p_\theta}\int_X|f(x)|^{p_1}\cdot|f(x)|^{p_\theta - p_1}\,d\mu(x) = \frac{(2M_1)^{p_1}}{p_1 - p_\theta}\,\|f\|_{L^{p_\theta}}^{p_\theta}.
\end{align*}
[/step]
[step:Introduce the threshold scale $c$, optimise, and identify $M_0^{1-\theta} M_1^\theta$]
Repeat the argument from Step 2 with the truncation threshold $c\lambda$ instead of $\lambda$, where $c > 0$ is a free parameter. The bounds in Steps 4--5 become
\begin{align*}
I_0 = \frac{(2M_0)^{p_0}}{p_\theta - p_0}\,c^{p_0 - p_\theta}\,\|f\|_{L^{p_\theta}}^{p_\theta}, \qquad I_1 = \frac{(2M_1)^{p_1}}{p_1 - p_\theta}\,c^{p_1 - p_\theta}\,\|f\|_{L^{p_\theta}}^{p_\theta}.
\end{align*}
(The factor $c^{p_0 - p_\theta}$ in $I_0$ comes from the inner integral $\int_0^{|f|/c}\lambda^{p_\theta - p_0 - 1}\,d\mathcal{L}^1(\lambda) = c^{p_0 - p_\theta}\,|f|^{p_\theta - p_0}/(p_\theta - p_0)$ after the substitution $|f(x)| > c\lambda \iff \lambda < |f(x)|/c$; symmetrically for $I_1$.) Combining,
\begin{align*}
\|Tf\|_{L^{p_\theta}}^{p_\theta}\le p_\theta\,\|f\|_{L^{p_\theta}}^{p_\theta}\,\biggl[\frac{(2M_0)^{p_0}}{(p_\theta - p_0)\,c^{p_\theta - p_0}} + \frac{(2M_1)^{p_1}\,c^{p_1 - p_\theta}}{p_1 - p_\theta}\biggr].
\end{align*}
Differentiating the bracket in $c$ and setting the derivative to zero,
\begin{align*}
c^{p_1 - p_0} = \frac{(2M_0)^{p_0}}{(2M_1)^{p_1}}.
\end{align*}
At this $c = c_{\mathrm{opt}}$, the two summands of the bracket are proportional, and a direct substitution yields a sum equal to a finite constant times $(2M_0)^{p_0(p_1 - p_\theta)/(p_1 - p_0)}\,(2M_1)^{p_1(p_\theta - p_0)/(p_1 - p_0)}$.
**Identification of the exponents.** Set $\beta := (p_\theta - p_0)/(p_1 - p_0) \in (0,1)$. We claim $p_1\beta = \theta\,p_\theta$ and $p_0(1 - \beta) = (1 - \theta)\,p_\theta$. From the convex-combination definition of $p_\theta$,
\begin{align*}
\theta = \frac{1/p_0 - 1/p_\theta}{1/p_0 - 1/p_1} = \frac{p_1(p_\theta - p_0)}{p_\theta(p_1 - p_0)},
\end{align*}
the second equality by multiplying numerator and denominator by $p_0 p_1 p_\theta$. Hence $\theta\,p_\theta = p_1\,(p_\theta - p_0)/(p_1 - p_0) = p_1\beta$. Symmetrically, $(1-\theta)\,p_\theta = p_0\,(p_1 - p_\theta)/(p_1 - p_0) = p_0(1 - \beta)$.
Therefore the minimum of the bracket equals $C^*\,(2M_0)^{(1-\theta)p_\theta}\,(2M_1)^{\theta p_\theta}$ for a constant $C^* = C^*(\theta, p_0, p_1)$ collecting the geometric prefactors $1/(p_\theta - p_0)$, $1/(p_1 - p_\theta)$ and the optimisation constants. Substituting and taking the $p_\theta$-th root,
\begin{align*}
\|Tf\|_{L^{p_\theta}}\le C(\theta, p_0, p_1)\,M_0^{1-\theta}\,M_1^\theta\,\|f\|_{L^{p_\theta}},
\end{align*}
where $C := (2 p_\theta C^*)^{1/p_\theta}\cdot 2$ depends only on $\theta, p_0, p_1$. This is the strong-type bound.
[/step]
[step:Handle the case $p_1 = \infty$ via the sharp truncation $\lambda/(2M_1)$]
When $p_1 = \infty$, the weak-$(\infty, \infty)$ condition reads $\|Tg\|_{L^\infty}\le M_1\,\|g\|_{L^\infty}$ for $g\in L^\infty$. Choose the truncation threshold $c := 1/(2M_1)$ in Step 2 (so $r_\lambda = f\,\mathbb{1}_{\{|f|\le\lambda/(2M_1)\}}$). Then $\|r_\lambda\|_{L^\infty}\le\lambda/(2M_1)$, and
\begin{align*}
\|Tr_\lambda\|_{L^\infty}\le M_1\cdot\frac{\lambda}{2M_1} = \frac{\lambda}{2}.
\end{align*}
Hence $|Tr_\lambda|\le\lambda/2$ pointwise $\nu$-a.e., so $d_{Tr_\lambda}(\lambda/2) = 0$, and $I_1 = 0$. The bound on $\|Tf\|_{L^{p_\theta}}^{p_\theta}$ reduces to $p_\theta\,I_0$ evaluated at $c = 1/(2M_1)$:
\begin{align*}
I_0 = \frac{(2M_0)^{p_0}}{p_\theta - p_0}\,(2M_1)^{p_\theta - p_0}\,\|f\|_{L^{p_\theta}}^{p_\theta},
\end{align*}
giving $\|Tf\|_{L^{p_\theta}}^{p_\theta}\le p_\theta\,(2M_0)^{p_0}(2M_1)^{p_\theta - p_0}\,\|f\|_{L^{p_\theta}}^{p_\theta}/(p_\theta - p_0)$. In the diagonal case $p_1 = \infty$, $\theta = 1 - p_0/p_\theta$, hence $p_0 = (1-\theta)p_\theta$ and $p_\theta - p_0 = \theta p_\theta$. So
\begin{align*}
\|Tf\|_{L^{p_\theta}}^{p_\theta}\le\frac{p_\theta}{p_\theta - p_0}\,(2M_0)^{(1-\theta)p_\theta}\,(2M_1)^{\theta p_\theta}\,\|f\|_{L^{p_\theta}}^{p_\theta},
\end{align*}
and taking the $p_\theta$-th root yields the strong-type bound $\|Tf\|_{L^{p_\theta}}\le C\,M_0^{1-\theta}\,M_1^\theta\,\|f\|_{L^{p_\theta}}$ with $C = 2\,(p_\theta/(p_\theta - p_0))^{1/p_\theta}$ depending only on $\theta, p_0$.
[/step]
[step:Extend the bound from simple functions of finite-measure support to all $f \in L^{p_\theta}$]
The estimate has been established (via the Tonelli evaluations in Steps 4--5 and the optimisation in Step 6, together with the $p_1 = \infty$ case in Step 7) for $f$ such that $\|f\|_{L^{p_\theta}} < \infty$. We restricted attention in Steps 4--5 to the regime where the layer-cake integrals are absolutely convergent; the bounds on $\|f_\lambda\|_{L^{p_0}}$ and $\|r_\lambda\|_{L^{p_1}}$ in Step 2 confirm this.
To verify that the abstract sublinear extension to $L^{p_\theta}$ is well-defined, take a sequence of simple functions $g_n$ of finite-measure support with $g_n\to f$ in $L^{p_\theta}$ — such a sequence exists because simple functions of finite-measure support are dense in $L^{p_\theta}$ when $1\le p_\theta < \infty$. Each $g_n$ lies in $L^{p_0}\cap L^{p_1}$ (it is bounded with finite-measure support), so the strong-type bound applies:
\begin{align*}
\|T g_n\|_{L^{p_\theta}}\le C\,M_0^{1-\theta}\,M_1^\theta\,\|g_n\|_{L^{p_\theta}}.
\end{align*}
By sublinearity, $|T g_n - T g_m|\le|T(g_n - g_m)|$ pointwise $\nu$-a.e., so
\begin{align*}
\|T g_n - T g_m\|_{L^{p_\theta}}\le\|T(g_n - g_m)\|_{L^{p_\theta}}\le C\,M_0^{1-\theta}\,M_1^\theta\,\|g_n - g_m\|_{L^{p_\theta}}\to 0.
\end{align*}
Hence $(T g_n)$ is Cauchy in $L^{p_\theta}$, with limit $Tf\in L^{p_\theta}$. Passing the inequality to the limit,
\begin{align*}
\|Tf\|_{L^{p_\theta}}\le C\,M_0^{1-\theta}\,M_1^\theta\,\|f\|_{L^{p_\theta}},
\end{align*}
which is the strong-type bound asserted by the theorem. This completes the proof.
[/step]
