Proof of Hunt's Interpolation I'm new to weak $L^p$ spaces and I'm doing a book exercise. Can someone enlighten me on the proof of the Hunt's interpolation theorem, which goes as follows:
Theorem

Let $\langle \,M, \mu \, \rangle$ and $\langle \,N,\nu \, \rangle$ be $\sigma$-finite measure spaces and $1 \leq p_1 \leq p_0 \leq \infty$, $1 \leq q_1 \leq q_0 \leq \infty$.
$$\frac{1}{p_t} = \frac{t}{p_1} + \frac{1-t}{p_0}$$ and $$\frac{1}{q_t} = \frac{t}{q_1} + \frac{1-t}{q_0} $$ 
Suppose that $T$ is a bounded linear transformation from $L^{p_0}(M,d\mu)$ to $L^{q_0}(N,d\nu)$ and from $L^{p_1}(M,d\mu)$ to $L^{q_1}(N,d\nu)$. Then for any $t \in (0,1)$, $T$ extends to a bounded linear map of $L_{w}^{p_t}(M,d\mu)$ to $L_{w}^{q_t}(N,d\nu)$. Moreover, $\|\,Tf\,\|_{q_t, w} \leq C_t \|\,f\,\|_{p_t,w}$ where $C_t$ depends only on $t$, $p_t$, $q_t$ and the bounds at the end points. 

In Simon and Reed's Methods of Modern Mathematical Physics Vol II, exercise 26 of chapter IX. The authors provide some hints to approach this.
(a) Let $1 \leq p_0 < p_1 < \infty$, $t \in (0,1)$, $\frac{1}{p} = \frac{t}{p_1} + \frac{1-t}{p_0}$. Show that $f \in L_w^p$ if and only if there exists a C so that for $\lambda > 0 $, $f$ can be decomposed as $f = f_{0,\lambda} + f_{1,\lambda}$ with $f_{0,\lambda} \in L^{p_0}$, $f_{1.\lambda} \in L^{p_1}$ and 
$$\|\,f_{0,\lambda}\,\|_{p_0} \leq C |\,\lambda\,|^{1-p/p_0}, \,\, \|\,f_{1,\lambda}\,\|_{p_1} \leq C |\,\lambda\,|^{1-p/p_1} $$
(b) Prove that $\|\,f\,\|_{p,w} = C$ where C is the smallest constant which can be used in part (a)
(c) Use (a) and (b) to prove Hunt's Interpolation theorem.

So I was able to prove the forward direction of (a) and I got
$$\|\,f_{0,\lambda}\,\|_{p_0} \leq (\frac{p_0}{p-p_0})^{1/p_0} \, \|\,f\,\|_{p,w}^{p/p_0} \, |\,\lambda\,|^{1-p/p_0}$$
$$\|\,f_{a,\lambda}\,\|_{p_a} \leq (\frac{p_a}{p_1-p})^{1/p_1} \, \|\,f\,\|_{p,w}^{p/p_1} \, |\,\lambda\,|^{1-p/p_1} $$
But I don't know how to deal with the reverse direction. As 
$$\|\,f\,\|_{p,w} = \sup_{s > 0} s \, \mu(|\,f\,| > s)$$
and I can split $\mu(|\,f\,| > s)$ into $\mu(|\,f_{0,\lambda}\,| > s/2) + \mu(|\,f_{1,\lambda}\,| > s/2)$, but I don't know how to deal with the $s$ in the front. And how should I approach (b)?
Thanks in advance!
 A: Suppose $f\in L_{w}^{p}$. Fix $\lambda>0$ and define $f_{0,\lambda}:=f\chi_{\left|f\right|\geq\lambda}$ and $f_{1,\lambda}:=f-f_{0,\lambda}$. I claim that $f_{0,\lambda}\in L_{w}^{p_{0}}$ and $f_{1,\lambda}\in L_{w}^{p_{1}}$. Indeed,
\begin{align*}
\left\|f_{0,\lambda}\right\|_{L^{p_{0}}}^{p_{0}}&=\int_{0}^{\infty}p_{0}t^{p_{0}-1}\mu(\left|f_{0,\lambda}\right|>t)dt\\
&=\left(\int_{0}^{\lambda}+\int_{\lambda}^{\infty}\right)p_{0}t^{p_{0}-1}\mu(\left|f_{0,\lambda}\right|>t)dt\\
&=\int_{0}^{\lambda}p_{0}t^{p_{0}-1}\mu(\left|f\right|\geq\lambda)dt+\int_{\lambda}^{\infty}p_{0}t^{p_{0}-1}\mu(\left|f\right|>t)dt\\
&\leq\int_{0}^{\lambda}p_{0}t^{p_{0}-1}\left\|f\right\|_{p,w}^{p}\lambda^{-p}dt+\int_{\lambda}^{\infty}p_{0}t^{p_{0}-1}\left\|f\right\|_{p,w}^{p}t^{-p}dt\\
&\leq\left\|f\right\|_{p,w}^{p}\lambda^{p_{0}-p}+\dfrac{p_{0}}{p-p_{0}}\left\|f\right\|_{p,w}^{p}\lambda^{p_{0}-p}\\
&=\left(1+\dfrac{p_{0}}{p-p_{0}}\right)\left\|f\right\|_{p,w}^{p}\lambda^{p_{0}-p}
\end{align*}
where we use that $p_{0}-1\geq 0$, so the first integral converges, and $\lambda\geq t$ and $p_{0}-p<0$. For the second estimate, observe that $\mu(\left|f_{1,\lambda}\right|>t)=0$ if $t\geq\lambda$, since $f_{1,\lambda}=f\chi_{\left|f\right|<\lambda}$. Also, $\mu(\left|f_{1,\lambda}\right|>t)=\mu(\lambda>\left|f\right|>t)\leq\mu(\left|f\right|>t)$, for $t<\lambda$. So,
\begin{align*}
\left\|f_{1,\lambda}\right\|_{L^{p_{1}}}^{p_{1}}&=\int_{0}^{\infty}p_{1}t^{p_{1}-1}\mu(\left|f_{1,\lambda}\right|>t)dt\\
&=\int_{0}^{\lambda}p_{1}t^{p_{1}-1}\mu(\left|f_{1,\lambda}\right|>t)dt\\
&\leq\int_{0}^{\lambda}p_{1}t^{p_{1}-1}\mu(\left|f\right|>t)dt\\
&\leq\int_{0}^{\lambda}p_{1}t^{p_{1}-1}\left\|f\right\|_{w}^{p}t^{-p}dt\\
&\leq\dfrac{p_{1}}{p_{1}-p}\left\|f\right\|_{w}^{p}\lambda^{p_{1}-p},
\end{align*}
where we use that $p_{1}-p>0$ (i.e. the integral converges). Taking $p_{0}^{th}$ and $p_{1}^{th}$ roots, respectively, we obtain
$$\left\|f_{0,\lambda}\right\|_{L^{p_{0}}}\leq\left(\dfrac{p}{p-p_{0}}\right)^{1/p_{0}}\left\|f\right\|_{p,w}^{p/p_{0}}\lambda^{1-p/p_{0}},\quad\left\|f_{1,\lambda}\right\|_{L^{p_{1}}}\leq\left(\dfrac{p_{1}}{p_{1}-p}\right)^{1/p_{1}}\left\|f\right\|_{p,w}^{p/p_{1}}\lambda^{1-p/p_{1}}$$
Conversely, suppose there exists a $C>0$ such for every $\lambda>0$, there exists such a decomposition $f=f_{0,\lambda}+f_{1,\lambda}$. We then have that
\begin{align*}
\mu(\left|f\right|>\lambda)&\leq\mu(\left|f_{0,\lambda}\right|>\lambda/2)+\mu(\left|f_{1,\lambda}\right|>\lambda/2)\\
&\leq(\lambda/2)^{-p_{0}}\left\|f_{0,1}\right\|_{p_{0}}^{p_{0}}+(\lambda/2)^{-p_{1}}\left\|f_{1,\lambda}\right\|_{p_{1}}^{p_{1}}\\
&\leq C^{p_{0}}(\lambda/2)^{-p_{0}}\lambda^{p_{0}-p}+C^{p_{1}}(\lambda/2)^{p_{1}}\lambda^{p_{1}-p}\\
&=\left((2C)^{p_{0}}+(2C)^{p_{1}}\right)\lambda^{-p}
\end{align*}
which shows that $\left\|f\right\|_{p,w}^{p}\leq (2C)^{p_{0}}+(2C)^{p_{1}}$.
I don't see why (b) should be true and in fact, I don't think it is true (see this question). In any case, you don't need (b) to be true to prove this special case of Hunt's theorem (he actually proved it for more general Lorentz spaces).
To prove the theorem, we need to exploit this estimates on the decomposition $f=f_{0,\lambda}+f_{i,\lambda}$, for $f\in L_{w}^{p}$. Denote the operator norm of $T:L^{p_{i}}\rightarrow L^{q_{i}}$ by $B_{i}$. Let $f\in L_{w}^{p}$ and $\left\|f\right\|_{p,w}=1$. For any $\lambda>0$, write $\lambda=s^{\gamma}$, where $\gamma$ is an exponent to be determined. Observe that by hypothesis,
\begin{align*}
\left\|Tf_{i,\lambda}\right\|_{L^{q_{i}}}&\leq B_{i}\left\|f_{i,\lambda}\right\|_{L^{p_{i}}}\lesssim_{p_{i},p} B_{i}\left\|f\right\|_{p,w}^{p/p_{0}}\lambda^{1-p/p_{i}}=B_{i}s^{\gamma(1-p/p_{i})}
\end{align*}
for $i=0,1$. We want to choose $\gamma>0$ so that $\gamma(1-p/p_{i})=(1-q/q_{i})$. You can check from the definition of $p$ and $q$ that
$$\gamma=\dfrac{1-q/q_{0}}{1-p/p_{0}}=\dfrac{1-q/q_{1}}{1-p/p_{1}}$$
It will then follow from the our argument above that
\begin{align*}
\mu(\left|Tf\right|>s)&\leq\mu(\left|Tf_{0,\lambda}\right|>s/2)+\mu(\left|Tf_{1,\lambda}\right|>s/2)\\
&\leq\left\|Tf_{0,\lambda}\right\|_{L^{q_{0}}}^{q_{0}}(s/2)^{-q_{0}}+\left\|Tf_{1,\lambda}\right\|_{L^{q_{1}}}^{q_{1}}(s/2)^{q_{1}}\\
&\lesssim_{p_{i},q_{i}}2^{q_{0}}B_{0}^{q_{0}}\left\|f\right\|_{p,w}^{q_{0}p/p_{0}}s^{-q_{0}}s^{q_{0}-q}+2^{q_{1}}B_{1}^{q_{1}}\left\|f\right\|_{p,w}^{q_{1}p/p_{1}}s^{-q_{1}}s^{q_{1}-q}\\
&=\left(2^{q_{0}}B_{0}^{q_{0}}+2^{q_{1}}B_{1}^{q_{1}}\right)s^{-q},
\end{align*}
which implies that
$$\left\|Tf\right\|_{q,w}\lesssim_{p,p_{i},q_{i}}\left(2^{q_{0}}B_{0}^{q_{0}}+2^{q_{1}}B_{1}^{q_{1}}\right)^{1/q},\qquad\forall \left\|f\right\|_{p,w}=1$$
Taking the supremum over all $f\in L_{w}^{p}$ with $\left\|f\right\|_{p,w}=1$ completes the proof.
