In this lecture notes by Terry Tao: https://www.math.ucla.edu/~tao/247a.1.06f/notes5.pdf
the first proposition 1.2 says: if $1\leq p<q<\infty$, let $A>0$, let $T:L^p(X)\rightarrow L^p(Y)$ be an arbitrary map (not necessarily linear) and let $r$ be such that the conjugate exponent $r'$ is $q/p$. Then the following are equivalent:
- for any sequence $(f_n)$ in $L^p(X)$, we have
$$|| (\sum_n|Tf_n|^p)^{1/p}||_{L_q(Y)}\leq A||(\sum_n|f_n|^p)^{1/p}||_{L^q(X)}$$
2)for every nonegative $w_2\in L^2(Y)$ there exists a nonnegative $w_1\in L^r(X)$ with $||w_1||_{L^r(X)}\leq ||w_2||_{L^r(Y)}$ such that
$$\int_Y |Tf(y)|^pw_2(y)d\mu_Y(y)\leq A^p \int_X|f(x)|^pw_1(x)\,d\mu_X(x)$$ for all $f\in L^P(X)$.
My questions are:
in both parts do we mean $f_n\in L^q(X)$ and $f\in L^q(X)$ and $T$ maps $L^q(X)$ to $L^q(Y)$? This way the right hand sides of both equations are finite.
in the proof (too long to type), the fourth to last line says "the latter implies that $\nu(F)\geq \lambda(F)$, but how? We have $\nu(F)>1$ and $\lambda(F)>1$ this doesn't imply anything.
in the proof, on page 3 we defined $B$, how is that algebraically open? $F>0$ messes up algebraic openness.
in the proof of 1 impleis 2, it seems that the exact same proof goes through with the much weaker condition $$||Tf||_{L^q(Y)}\leq A||f||_{L^q(X)}$$ mentioned in the notes?