Special case of Riesz-Thorin Interpolation Theorem $L^{p_0} \cap L^{p_1} \to L^1$ Let $1 \le p_0,p_1 \le \infty$ and consider a linear map $T : L^{p_0} \cap L^{p_1} \to L^1$ such that there exist constants $M_0, M_1 > 0$ so that
$$(\forall f \in L^{p_0} \cap L^{p_1}) \quad \|Tf\|_{L^1} \le M_0 \|f\|_{L^{p_0}} \text{ and } \|Tf\|_{L^1} \le M_1 \|f\|_{L^{p_1}}.$$
I need to prove that for $t \in \langle 0,1\rangle$ we have
$$(\forall f \in L^{p_0} \cap L^{p_1}) \quad \|Tf\|_{L^1} \le M_0^{1-t}M_1^t \|f\|_{L^{p_t}}$$
where $p_t$ is defined by $\frac1{p_t} = \frac{t}{p_1} + \frac{1-t}{p_0}$.
This is a special case of Riesz-Thorin Interpolation theorem, outlined e.g. here. The proof of this special case is supposed to be simple and is omitted from most proofs. However, I seem to be missing something here.
My attempt:
By multiplying inequalities $\|Tf\|_{L^1}^{1-t} \le M_0^{1-t} \|f\|_{L^{p_0}}^{1-t}$ and $\|Tf\|_{L^1}^t \le M_1^t \|f\|_{L^{p_1}}^t$ we get
$$\|Tf\|_{L^1} \le M_0^{1-t}M_1^t \|f\|_{L^{p_0}}^{1-t}\|f\|_{L^{p_1}}^{t}.$$
Now there is the simple interpolation inequality
$$\|f\|_{L^{p_t}
}\le \|f\|_{L^{p_0}}^{1-t}\|f\|_{L^{p_1}}^{t}$$
but it is in the wrong direction. I'm sure I need to use it somehow.
 A: I believe its like this...let me first recall bits of that proof. The inequality $\|Tf\|_{q_\theta} \le  M_0^{1-\theta} M_1^\theta  \|f\|_{p_\theta}$ is equivalent to
$$ \sup_{\substack{\|f\|_{p_\theta}=1\\ \|g\|_{q_\theta'}=1} 
    } \int_Y (T f)g \le  M_0^{1-\theta} M_1^\theta  $$
Where the sup is over all functions $f\in L^{p_\theta}(X)$ and $g\in L^{q_\theta}(Y)$.
In the given proof that assumes $q_{0,1}\neq1$ and $p_{0,1}\neq\infty$, this LHS ends up being reduced to a double supremum only over simple functions(which are a dense subclass), at which point its equal to $F(\theta)$, where
$$\frac{1}{p(z)}=\frac{1-z}{p_{0}}+\frac{z}{p_{1}}, \quad \frac{1}{q^{\prime}(z)}=\frac{1-z}{q_{0}^{\prime}}+\frac{z}{q_{1}^{\prime}},\\f_{z}(x)=\sum_{j=1}^{n}\left|a_{j}\right|^{\frac{p_\theta}{p(z)}} \frac{a_{j}}{\left|a_{j}\right|} \chi_{A_{j}}, \quad g_{z}(x)=\sum_{\ell=1}^{m}\left|b_{\ell}\right|^{\frac{q^{\prime}_\theta}{q^{\prime}(z)}} \frac{b_{\ell}}{\left|b_{\ell}\right|} \chi_{B_{\ell}},
\\
F(z)=\int g_{z} T f_{z} = \sum_{j=1}^{n} \sum_{\ell=1}^{m}\left|a_{j}\right|^{\frac{p_\theta}{p(z)}} \frac{a_{j}}{\left|a_{j}\right|}\left|b_{\ell}\right|^{\frac{q^{\prime}_\theta}{q^{\prime}(z)}} \frac{b_{\ell}}{\left|b_{\ell}\right|} \int_{B_{\ell}} T \chi_{A_{j}}$$
Lets also ignore $p_\theta = \infty$ for now (see end of post). If you reflect on what $q'(z)$ is used for, its to interpolate between the $q_0'$ and $q_1'$ norms; but for us, these are all just $\infty$. So we can just replace $q'_\theta/q'(z)$ by $1$, and just use $g$ in place of $g_z$. Another issue is that simple functions with compact support are not dense in $L^\infty$. But now that we don't need to introduce the function $g_z$, we do not need to consider only step functions for $g$, and can allow the sup to range over the entire space $L^\infty$. Therefore: lets consider the following function $F$ instead:
$$ F(z) := \int_Y g Tf_z
=
\sum_{j=1}^{n} \left|a_{j}\right|^{\frac{p_\theta}{p(z)}} \frac{a_{j}}{\left|a_{j}\right|}\int_{Y} g T \chi_{A_{j}}$$
The $p_0=p_1=p_\theta$ ($\leqq \infty$)  endpoint, for any $q_{0,1}$ is recovered by the Lebesgue interpolation you mentioned $\|Tf\|_{q_\theta} \le \|Tf\|^{1-\theta}_{q_0}\|Tf\|_{q_1}^{\theta} \le M_0^{1-\theta} M_1^\theta \|f\|_{\infty}$.
That should cover all cases. The most detail I've ever seen explicitly written about the edge cases is in Tao's blog-notes. He suggests monotone convergence which also works; I basically verified this here.
