用户: Solution/ 解答: GTM249 Chapter 1 Part II

1.3 Interpolation

1.3.1.	This just requires some tiny changes in the proof of the original theorem.
1.3.2.	Interpolating between $L^1$ and $L^r$ using Marcinkiewicz, we have $$\begin{array}{rl}\Vert T{\Vert }_{L^{\frac{p+1}{2}}\to L^{\frac{p+1}{2}}}& \le 2(\frac{\frac{p+1}{2}}{\frac{p+1}{2}-1}+\frac{\frac{p+1}{2}}{r-\frac{p+1}{2}}{)}^{\frac{1}{p}}{A}_0^{\frac{\frac{2}{p+1}-\frac{1}{r}}{1-\frac{1}{r}}}{A}_1^{\frac{1-\frac{2}{p+1}}{1-\frac{1}{r}}}\\ & =2(\frac{p+1}{p-1}+\frac{p+1}{2r-p-1}{)}^{\frac{1}{p}}{A}_0^{\frac{2r-p-1}{(r-1)(p+1)}}{A}_1^{\frac{r(p-1)}{(p+1)(r-1)}}.\end{array}$$Then using Riesz–Thorin to interpolate between $L^{\frac{p+1}{2}}$ and $L^r$, we have $$\begin{array}{rl}\Vert T{\Vert }_{L^p\to L^p}& \le \left(2\right(\frac{p+1}{p-1}+\frac{p+1}{2r-p-1}{)}^{\frac{1}{p}}{A}_0^{\frac{2r-p-1}{(r-1)(p+1)}}{A}_1^{\frac{r(p-1)}{(p+1)(r-1)}}{)}^{1-\theta }{A}_1^{\theta },\end{array}$$where $\theta =\frac{r(1-p)}{p(p+1-2r)}$ by solving $\frac{1}{p}=\frac{2(1-\theta )}{p+1}+\frac{\theta }{r}$. Thus $$\begin{array}{rl}\Vert T{\Vert }_{L^p\to L^p}& \le 2^{\frac{(p-r)(1+p)}{p(p+1-2r)}}(\frac{p+1}{p-1}+\frac{p+1}{2r-p-1}{)}^{\frac{(r-p)(p+1)}{p^2(2r-p-1)}}{A}_0^{\frac{\frac{1}{p}-\frac{1}{r}}{1-\frac{1}{r}}}{A}_1^{\frac{1-\frac{1}{p}}{1-\frac{1}{r}}}.\end{array}$$Now we come to estimate the constant. Note that $r-p\le \frac{1}{2}(2r-p-1)$, we have then $\frac{r-p}{2r-p-1}\le \frac{1}{2}$. Thus $$\frac{(p-r)(1+p)}{p(p+1-2r)}=\frac{(1+p)}{p}\frac{r-p}{2r-p-1}\le 2^1\cdot \frac{1}{2}=1,$$$$\frac{p+1}{p-1}+\frac{p+1}{2r-p-1}\le \frac{2(p+1)}{p-1}.$$With these estimates, we have $$\begin{array}{rl}\Vert T{\Vert }_{L^p\to L^p}& \le 2(\frac{2(p+1)}{p-1}{)}^{\frac{1}{p}}{A}_0^{\frac{\frac{1}{p}-\frac{1}{r}}{1-\frac{1}{r}}}{A}_1^{\frac{1-\frac{1}{p}}{1-\frac{1}{r}}}\\ & \le 2^3(p-1{)}^{-\frac{1}{p}}{A}_0^{\frac{\frac{1}{p}-\frac{1}{r}}{1-\frac{1}{r}}}{A}_1^{\frac{1-\frac{1}{p}}{1-\frac{1}{r}}}\\ & \le 8(p-1{)}^{-\frac{1}{p}}{A}_0^{\frac{\frac{1}{p}-\frac{1}{r}}{1-\frac{1}{r}}}{A}_1^{\frac{1-\frac{1}{p}}{1-\frac{1}{r}}},\end{array}$$with the use of $(2(1+p){)}^{\frac{1}{p}}\le 4$.
1.3.3.	(a) It is a special case of (b). To be specific, take $p_0=1$ in (b), then the norm becomes $$\begin{array}{rl}{p}^{1+\frac{1}{p}}[\frac{B(2,p-1)}{(p-1{)}^{p-1}}{]}^{\frac{1}{p}}{A}_0^{\frac{1}{p}}{A}_1^{1-\frac{1}{p}}& =p^{1+\frac{1}{p}}[\frac{\Gamma (2)\Gamma (p-1)}{\Gamma (p+1)(p-1{)}^{p-1}}{]}^{\frac{1}{p}}{A}_0^{\frac{1}{p}}{A}_1^{1-\frac{1}{p}}\\ & =p^{1+\frac{1}{p}}[\frac{1}{p(p-1{)}^p}{]}^{\frac{1}{p}}{A}_0^{\frac{1}{p}}{A}_1^{1-\frac{1}{p}}=\frac{p}{p-1}{A}_0^{\frac{1}{p}}{A}_1^{1-\frac{1}{p}}.\end{array}$$ (b) It suffices to study the case of $f\ge 0$ due to the assumption $|T(f)|\le T(|f|)$. For $\alpha >0$, write $f=f_0+f_1$, where $f_0=(f-\frac{\lambda \alpha }{A_1}){\chi }_{f\ge \frac{\lambda \alpha }{A_1}}$, then we have the relation $\{|T(f)|>\alpha \}\subset \{|T(f_0)|>(1-\lambda )\alpha \}$ since $\Vert T(f_1){\Vert }_{L^{\infty }}\le \Vert T{\Vert }_{L^{\infty }\to L^{\infty }}\Vert f_1{\Vert }_{L^{\infty }}\le A_1\frac{\lambda \alpha }{A_1}=\lambda \alpha$. Thus $$\begin{array}{rl}\Vert T(f){\Vert }_{L^p}^p& =p{\int }_0^{\infty }{\alpha }^{p-1}{d}_{T(f)}(\alpha )d\alpha \le p{\int }_0^{\infty }{\alpha }^{p-1}{d}_{T(f_0)}((1-\lambda )\alpha )d\alpha \\ & \le \frac{p{A}_0^{p_0}}{(1-\lambda {)}^{p_0}}{\int }_0^{\infty }{\alpha }^{p-p_0-1}\Vert f_0{\Vert }_{L^{p_0}}^{p_0}d\alpha =\frac{p{A}_0^{p_0}}{(1-\lambda {)}^{p_0}}{\int }_0^{\infty }{\alpha }^{p-p_0-1}{\int }_{f\ge \frac{\lambda \alpha }{A_1}}(f-\frac{\lambda \alpha }{A_1}{)}^{p_0}d\mu d\alpha \\ & =\frac{p{A}_0^{p_0}}{(1-\lambda {)}^{p_0}}{\int }_X{\int }_0^{\frac{A_{1}f(x)}{\lambda }}{\alpha }^{p-p_0-1}(f-\frac{\lambda \alpha }{A_1}{)}^{p_0}d\alpha d\mu \\ & =\frac{p{A}_0^{p_0}{A}_1^{p-p_0}}{(1-\lambda {)}^{p_0}{\lambda }^{p-p_0}}{\int }_X{\int }_0^{f(x)}{\alpha }^{p-p_0-1}(f-\alpha {)}^{p_0}d\alpha d\mu \\ & =\frac{p{A}_0^{p_0}{A}_1^{p-p_0}}{(1-\lambda {)}^{p_0}{\lambda }^{p-p_0}}B(p_0+1,p-p_0){\int }_X|f{|}^{p}d\mu \\ & =\frac{p{A}_0^{p_0}{A}_1^{p-p_0}}{(1-\lambda {)}^{p_0}{\lambda }^{p-p_0}}B(p_0+1,p-p_0)\Vert f{\Vert }_{L^p}^p.\end{array}$$Now take $\lambda =1-\frac{p_0}{p}$ to end this proof. (c) Again, write $f=f_0+f_1$, where $f_0=(f-\frac{\delta \alpha }{A_1}){\chi }_{f\ge \frac{\delta \alpha }{A_1}}$. Thus $$\begin{array}{rl}\Vert T(f){\Vert }_{L^p}^p& =p{\int }_0^{\infty }{\alpha }^{p-1}{d}_{T(f)}(\alpha )d\alpha \\ & \le p{\int }_0^{\infty }{\alpha }^{p-1}{d}_{T(f_0)}((1-\lambda )\alpha )d\alpha +p{\int }_0^{\infty }{\alpha }^{p-1}{d}_{T(f_1)}(\lambda \alpha )d\alpha \\ & =\frac{p{A}_0^{p_0}}{(1-\lambda {)}^{p_0}}{\int }_0^{\infty }{a}^{p-p_0-1}\Vert f_0{\Vert }_{L^{p_0}}^{p_0}d\alpha +\frac{p{A}_1^{p_1}}{{\lambda }^{p_1}}{\int }_0^{\infty }{\alpha }^{p-p_1-1}\Vert f_0{\Vert }_{L^{p_0}}^{p_0}d\alpha \\ & =\frac{p{A}_0^{p_0}}{(1-\lambda {)}^{p_0}}{\int }_0^{\infty }{\alpha }^{p-p_0-1}{\int }_{f\ge \delta \alpha }(f-\delta \alpha {)}^{p_0}d\alpha \\ & +\frac{p{A}_1^{p_1}}{{\lambda }^{p_1}}{\int }_0^{\infty }{\alpha }^{p-p_1-1}({\int }_{f\ge \delta \alpha }{\delta }^{p_1}{\alpha }^{p_1}d\mu +{\int }_{f<\delta \alpha }{f}^{p_1}d\mu )d\alpha \\ & =\frac{p{A}_0^{p_0}}{(1-\lambda {)}^{p_0}}{\int }_X{\int }_0^{\frac{f(x)}{\delta }}{\alpha }^{p-p_0-1}(f-\delta \alpha {)}^{p_0}d\alpha d\mu \\ & +\frac{p{A}_1^{p_1}}{{\lambda }^{p_1}}({\int }_X{\int }_0^{\frac{f(x)}{\delta }}{\alpha }^{p-1}{\delta }^{p_1}d\alpha d\mu +{\int }_X{\int }_{\frac{f(x)}{\delta }}^{\infty }{\alpha }^{p-p_1-1}{f}^{p_1}d\alpha d\mu )\\ & =\frac{p{A}_0^{p_0}}{(1-\lambda {)}^{p_0}}{\int }_X{\int }_0^1(\frac{f}{\delta }\alpha {)}^{p-p_0-1}(f-f\alpha {)}^{p_0}d\left(\frac{f}{\delta }\alpha \right)d\mu \\ & +\frac{p{A}_1^{p_1}}{{\lambda }^{p_1}}({\delta }^{p_1-p}{\int }_X{f}^{p}d\mu +\frac{{\delta }^{p_1-p}}{p_1-p}{\int }_X{f}^{p}d\mu )\\ & =\Vert f{\Vert }_{L^p}^p(\frac{p{A}_0^{p_0}}{(1-\lambda {)}^{p_0}{\delta }^{p-p_0}}B(p-p_0,p_0+1)+\frac{p{A}_1^{p_1}{\delta }^{p_1-p}}{{\lambda }^{p_1}}\frac{p_1-p-1}{p_1-p}).\end{array}$$Now take $\delta$ such that $\frac{A_0^{p_0}}{{\delta }^{p-p_0}}=\frac{A_1^{p_1}}{{\delta }^{p-p_1}}$ and optimize over $\lambda$ to end the estimate.
1.3.7.	Write$$f=\sum _{k=0}^{\infty }f{\chi }_{S_k}$$where $S_k=\{2^k\le |f|<2^{k+1}\}$ when $k>1$ and $S_0=\{|f|<2\}$. Using Hölder’s inequality and the hypotheses on $T$, we obtain that $$\begin{array}{rl}{\int }_Y|T(f{\chi }_{S_k})|d\nu & \le \left({\int }_{Y}d\nu {)}^{\frac{1}{k+1}}\right({\int }_Y|Tf{|}^{\frac{k+1}{k}}{\chi }_{S_k}d\nu {)}^{\frac{k}{k+1}}\\ & \le \nu (Y{)}^{\frac{1}{k+1}}A(\frac{k+1}{k}-1{)}^{-\alpha }({\int }_{S_k}|f{|}^{\frac{k+1}{k}}d\mu {)}^{\frac{k}{k+1}}\\ & \le 2^{k+1}\nu (Y{)}^{\frac{1}{k+1}}A{k}^{\alpha }\mu (S_k{)}^{\frac{k}{k+1}}.\end{array}$$If $\mu (S_k)\le \frac{1}{3^{k+1}}$, then $$\begin{array}{r}{2}^{k+1}\nu (Y{)}^{\frac{1}{k+1}}A{k}^{\alpha }\mu (S_k{)}^{\frac{k}{k+1}}\le 2\nu (Y{)}^{\frac{1}{k+1}}A{k}^{\alpha }(\frac{2}{3}{)}^k.\end{array}$$If $\mu (S_k)\ge \frac{1}{3^{k+1}}$, then $$\begin{array}{rl}{\int }_Y|T(f{\chi }_{S_k})|d\nu & \le \nu (Y{)}^{\frac{1}{k+1}}A{k}^{\alpha }\mu (S_k{)}^{-\frac{1}{k+1}}{\int }_{S_k}|f|d\mu \\ & \le 3\nu (Y{)}^{\frac{1}{k+1}}A{\int }_{S_k}|f|k^{\alpha }d\mu \\ & \le 3\nu (Y{)}^{\frac{1}{k+1}}A{\int }_{S_k}|f|({\log }_2^{+}|f|{)}^{\alpha }d\mu .\end{array}$$For $S_0$, we have $$\begin{array}{rl}{\int }_Y|T(f{\chi }_{S_0})|d\nu & \le \left({\int }_{Y}d\nu {)}^{\frac{1}{2}}\right({\int }_Y|T(f{\chi }_{S_0}){|}^{2}d\nu {)}^{\frac{1}{2}}\\ & \le \nu (Y{)}^{\frac{1}{2}}A\cdot 2\mu (X{)}^{\frac{1}{2}}.\end{array}$$Add the three parts together and use the countable subadditivity of $T$, we then conclude that $$\begin{array}{rl}{\int }_Y|T(f)|d\nu & \le \sum _{k=0}^{\infty }{\int }_Y|T(f{\chi }_{S_k})|d\nu \\ & \le 6A\max (1,\nu (Y){)}^{\frac{1}{2}}[{\int }_X|f|({\log }_2^{+}|f|{)}^{\alpha }d\mu +\sum _{k=1}^{\infty }{k}^{\alpha }(\frac{2}{3}{)}^k+\mu (X{)}^{\frac{1}{2}}]\\ & \le 6A(1+\nu (Y){)}^{\frac{1}{2}}[{\int }_X|f|({\log }_2^{+}|f|{)}^{\alpha }d\mu +C_{\alpha }+\mu (X{)}^{\frac{1}{2}}].\end{array}$$
1.3.8.	Use the change of variables $s=e^{\pi t}\in (0,\infty )$, and the first integral becomes $$\begin{array}{rl}& \frac{\sin (\pi x)}{2}{\int }_{-\infty }^{+\infty }\frac{1}{\cosh (\pi t)+\cos (\pi x)}dt\\ & =\sin (\pi x){\int }_0^{\infty }\frac{1}{s+s^{-1}+2\cos (\pi x)}\frac{ds}{\pi s}\\ & =\frac{\sin (\pi x)}{\pi }{\int }_0^{\infty }\frac{ds}{(s+\cos (\pi x){)}^2+\sin (\pi x{)}^2}\\ & =\frac{1}{\pi }{\int }_0^{\infty }\frac{d\left(\frac{s}{\sin (\pi x)}\right)}{(\frac{s}{\sin (\pi x)}+\cot (\pi x){)}^2+1}\\ & =\frac{1}{\pi }{\int }_{\cot (\pi x)}^{\infty }\frac{du}{u^2+1}=x.\end{array}$$For the second one, $$\begin{array}{rl}& \frac{\sin (\pi x)}{2}{\int }_{-\infty }^{+\infty }\frac{1}{\cosh (\pi t)-\cos (\pi x)}dt\\ & =\sin (\pi x){\int }_0^{\infty }\frac{1}{s+s^{-1}-2\cos (\pi x)}\frac{ds}{\pi s}\\ & =\frac{\sin (\pi x)}{\pi }{\int }_0^{\infty }\frac{ds}{(s-\cos (\pi x){)}^2+\sin (\pi x{)}^2}\\ & =\frac{1}{\pi }{\int }_0^{\infty }\frac{d\left(\frac{s}{\sin (\pi x)}\right)}{(\frac{s}{\sin (\pi x)}-\cot (\pi x){)}^2+1}\\ & =\frac{1}{\pi }{\int }_{-\cot (\pi x)}^{\infty }\frac{du}{u^2+1}=1-x.\end{array}$$

1.4. Lorentz Spaces

1.4.1.	(a) Since $$\begin{array}{r}{\int }_{A}gd\mu ={\int }_{X}g{\chi }_{A}d\mu ={\int }_0^{\infty }(g{\chi }_A{)}^{\ast }(t)dt={\int }_0^{\mu (A)}(g{\chi }_A{)}^{\ast }(t)dt\le {\int }_0^{\mu (A)}{g}^{\ast }(t)dt.\end{array}$$ (b) Since $$\begin{array}{rl}{\int }_X|fg|d\mu & ={\int }_X{\int }_0^{\infty }{\int }_0^{\infty }{\chi }_{\{|f|\ge s\}}{\chi }_{\{|g|\ge t\}}dtdsd\mu ={\int }_X{\int }_0^{\infty }{\int }_0^{\infty }{\chi }_{\{|f|\ge s\}\cap \{|g|\ge t\}}dtdsd\mu \\ & ={\int }_0^{\infty }{\int }_0^{\infty }\mu (\{|f|\ge s\}\cap \{|g|\ge t\})dtds\\ & \le {\int }_0^{\infty }{\int }_0^{\infty }\min (\mu (\{|f|\ge s\}),\mu (\{|g|\ge t\}))dtds\\ & ={\int }_0^{\infty }{\int }_0^{\infty }\min (\mu (\{f^{\ast }\ge s\}),\mu (\{g^{\ast }\ge t\}))dtds\\ & ={\int }_0^{\infty }{\int }_0^{\infty }{\int }_0^{\infty }{\chi }_{\{f^{\ast }\ge s\}}(u){\chi }_{\{g^{\ast }\ge t\}}(u)dudtds\\ & ={\int }_0^{\infty }{f}^{\ast }(u)g^{\ast }(u)du.\end{array}$$
1.4.2.	We omit the proof of the $s=\infty$ case since $L^{p,\infty }(X)\subset L^{p,q}(X)$ always holds for $0<q<\infty$. Now for $q_1<\infty$ case, write $\gamma d_f(\gamma {)}^{\frac{1}{q_0}}\le C_1$ and $\gamma d_f(\gamma {)}^{\frac{1}{q_1}}\le C_2$. Since $q\in (q_0,q_1)$, we may always find ${\delta }_1\in (0,1)$ and ${\delta }_2>1$, such that $$\frac{1}{q}=\frac{r_i}{q_0}+\frac{{\delta }_i-r_i}{q_1}$$for each $i=0,1$ and some $r_i\in (0,{\delta }_i)$. Thence we may continue our estimate as follows: $$\begin{array}{rl}{\int }_0^{\infty }(d_f(\gamma {)}^{\frac{1}{q}}\gamma {)}^s\frac{d\gamma }{\gamma }& ={\int }_0^1(d_f(\gamma {)}^{\frac{1}{q}}\gamma {)}^s\frac{d\gamma }{\gamma }+{\int }_1^{\infty }(d_f(\gamma {)}^{\frac{1}{q}}\gamma {)}^s\frac{d\gamma }{\gamma }\\ & \le {\int }_0^1(\frac{C_1^{r_1}{C}_2^{{\delta }_1-r_1}}{{\gamma }^{{\delta }_1}}\gamma {)}^s\frac{d\gamma }{\gamma }+{\int }_1^{\gamma }(\frac{C_1^{r_2}{C}_2^{{\delta }_2-r_2}}{{\gamma }^{{\delta }_2}}\gamma {)}^s\frac{d\gamma }{\gamma }=\frac{C_1^{r_1}{C}_2^{{\delta }_1-r_1}}{s(1-{\delta }_1)}+\frac{C_1^{r_2}{C}_2^{{\delta }_2-r_2}}{s({\delta }_2-1)}.\end{array}$$Thus $$\Vert f{\Vert }_{L^{q,s}}\le q^{\frac{1}{s}}(\frac{C_1^{r_1}{C}_2^{{\delta }_1-r_1}}{s(1-{\delta }_1)}+\frac{C_1^{r_2}{C}_2^{{\delta }_2-r_2}}{s({\delta }_2-1)}{)}^{\frac{1}{s}}.$$For the $q_1=\infty$ case, suppose $|f|\le C_2$ a.e., then $d_f(\gamma )=0$ for $\gamma >C_2$. Then $$\begin{array}{rl}{\int }_0^{\infty }(d_f(\gamma {)}^{\frac{1}{q}}\gamma {)}^s\frac{d\gamma }{\gamma }& ={\int }_0^{C_2}(d_f(\gamma {)}^{\frac{1}{q}}\gamma {)}^s\frac{d\gamma }{\gamma }\\ & \le {\int }_0^{C_2}(d_f(\gamma {)}^{\frac{1}{q}}{\gamma }^{\frac{q_0}{q}}{\gamma }^{\frac{q-q_0}{q}}{)}^s\frac{d\gamma }{\gamma }\\ & \le {\int }_0^{C_2}{C}_1^{\frac{s}{q}}\frac{d\gamma }{{\gamma }^{\frac{q(1-s)+q_{0}s}{q}}}=\frac{q}{s(q-q_0)}{C}_1^{\frac{s}{q}}{C}_2^{\frac{s(q-q_0)}{q}}.\end{array}$$Thus $$\Vert f{\Vert }_{L^{q,s}}\le (\frac{q^2}{s(q-q_0)}{C}_1^{\frac{s}{q}}{C}_2^{\frac{s(q-q_0)}{q}}{)}^{\frac{1}{s}}.$$
1.4.3.	(a) For $t<\mu (X)$, we have $$\begin{array}{rl}((f+g{)}^{\ast \ast }(t){)}^r& =\underset{\mu (E)\ge t}{\sup }\frac{1}{\mu (E)}{\int }_E|f+g{|}^{r}d\mu \\ & \le \underset{\mu (E)\ge t}{\sup }\frac{1}{\mu (E)}{\int }_E|f{|}^{r}d\mu +\underset{\mu (E)\ge t}{\sup }\frac{1}{\mu (E)}{\int }_E|g{|}^{r}d\mu =(f^{\ast \ast }(t){)}^r+(g^{\ast \ast }(t){)}^r\end{array}$$since $r\le 1$. Otherwise, $$\begin{array}{rl}((f+g{)}^{\ast \ast }(t){)}^r& =\frac{1}{t}{\int }_X|f+g{|}^r\le \frac{1}{t}{\int }_X|f{|}^r+\frac{1}{t}{\int }_X|g{|}^r\\ & =(f^{\ast \ast }(t){)}^r+(g^{\ast \ast }(t){)}^r.\end{array}$$For the second part, since $\frac{q}{r}>1$, we have $$\begin{array}{rl}{\left|\left|\left|f+g\right|\right|\right|}_{L^{p,q}}^r& =({\int }_0^{\infty }(t^{\frac{1}{p}}(f+g{)}^{\ast \ast }(t){)}^q\frac{dt}{t}{)}^{\frac{r}{q}}=({\int }_0^{\infty }(t^{\frac{r}{p}-\frac{r}{q}}((f+g{)}^{\ast \ast }(t){)}^r{)}^{\frac{q}{r}}\frac{dt}{t}{)}^{\frac{r}{q}}\\ & \le ({\int }_0^{\infty }(t^{\frac{r}{p}-\frac{r}{q}}(((f{)}^{\ast \ast }(t){)}^r+((g{)}^{\ast \ast }(t){)}^r){)}^{\frac{q}{r}}\frac{dt}{t}{)}^{\frac{r}{q}}\\ & \le ({\int }_0^{\infty }(t^{\frac{r}{p}-\frac{r}{q}}((f{)}^{\ast \ast }(t){)}^r{)}^{\frac{q}{r}}\frac{dt}{t}{)}^{\frac{r}{q}}+({\int }_0^{\infty }(t^{\frac{r}{p}-\frac{r}{q}}((g{)}^{\ast \ast }(t){)}^r{)}^{\frac{q}{r}}\frac{dt}{t}{)}^{\frac{r}{q}}\\ & ={\left|\left|\left|f\right|\right|\right|}_{L^{p,q}}^r+{\left|\left|\left|g\right|\right|\right|}_{L^{p,q}}^r.\end{array}$$When $r=1$, we have the property of scalar multiplication, whence ${\left|\left|\left|\cdot \right|\right|\right|}_{L^{p,q}}^r$ is a norm. (b) The first inequality holds since $f^{\ast }(t)\le f^{\ast \ast }(t)$. This is quite obvious when $t\ge \mu (X)$, since $f^{\ast }(t)=0$ in this case. For $t<\mu (X)$, take $s=f^{\ast }(t)$, then $d_f(s)\le t$. Then $\forall \epsilon >0$, we have $d_f(s-\epsilon )>t$ by definition. Set $\tilde{E}:=\{|f|\ge s-\epsilon \}$, we have $$\begin{array}{r}{f}^{\ast \ast }(t)\ge (\frac{1}{\tilde{E}}{\int }_{\tilde{E}}|f{|}^{r}d\mu {)}^{\frac{1}{r}}\ge s-\epsilon .\end{array}$$Since $\epsilon$ is arbitrary, we have $$f^{\ast \ast }(t)\ge s=f^{\ast }(t).$$Thus $$\begin{array}{rl}{\left|\left|\left|f\right|\right|\right|}_{L^{p,q}}& =({\int }_0^{\infty }(t^{\frac{1}{p}}{f}^{\ast \ast }(t){)}^q\frac{dt}{t}{)}^{\frac{1}{q}}\ge ({\int }_0^{\infty }(t^{\frac{1}{p}}{f}^{\ast }(t){)}^q\frac{dt}{t}{)}^{\frac{1}{q}}=\Vert f{\Vert }_{L^{p,q}}.\end{array}$$For the second inequality, we may seek for a proof for $f^{\ast \ast }(t)\le (\frac{p}{p-r}{)}^{\frac{1}{r}}{f}^{\ast }(t)$, but unfortunately this is untrue. (For $t>\mu (X)$, the right side is definitely zero but unnecessarily for the left.) Note that the $f^{\ast \ast }$ defined is controlled by $(\frac{1}{t}{\int }_0^t|f^{\ast }(s){|}^{r}ds{)}^{\frac{1}{r}}$. We just show the case when $0\le t\le \mu (X)$. This is because $$\begin{array}{rl}\underset{\mu (E)\ge t}{\sup }(\frac{1}{\mu (E)}{\int }_{\mu (E)}|f{|}^{r}d\mu {)}^{\frac{1}{r}}& =\underset{\mu (E)\ge t}{\sup }(\frac{1}{\mu (E)}{\int }_{\mu (E)}|f{\chi }_E{|}^{r}d\mu {)}^{\frac{1}{r}}\\ & =\underset{\mu (E)\ge t}{\sup }(\frac{1}{\mu (E)}{\int }_0^{\mu (E)}(f{\chi }_E{)}^{\ast }(s{)}^{r}ds{)}^{\frac{1}{r}}\\ & \le \underset{\mu (E)\ge t}{\sup }(\frac{1}{\mu (E)}{\int }_0^{\mu (E)}{f}^{\ast }(s{)}^{r}ds{)}^{\frac{1}{r}}\stackrel{\mathrm{decreasing}}{\le }(\frac{1}{t}{\int }_0^t{f}^{\ast }(s{)}^{r}ds{)}^{\frac{1}{r}}.\end{array}$$Thus, $$\begin{array}{rl}{\left|\left|\left|f\right|\right|\right|}_{L^{p,q}}& \le \left({\int }_0^{\infty }\right(t^{\frac{1}{p}}(\frac{1}{t}{\int }_0^t(f^{\ast }(t){)}^{r}ds{)}^{\frac{1}{r}}{)}^q\frac{dt}{t}{)}^{\frac{1}{q}}\le \left(\right({\int }_0^{\infty }(\frac{1}{t}{\int }_0^t(f^{\ast }(t){)}^{r}ds{)}^{\frac{q}{r}}{)}^q{t}^{\frac{q}{p}-\frac{q}{r}}\frac{dt}{t}{)}^{\frac{r}{q}}{)}^{\frac{1}{r}}\\ & \stackrel{\text{Hardy}}{\le }\left(\frac{\frac{q}{r}}{\frac{q}{p}-\frac{q}{r}}{)}^{\frac{1}{r}}\right({\int }_0^{\infty }(t(f^{\ast }(t){)}^r{)}^{\frac{q}{r}}{t}^{\frac{q}{p}-\frac{q}{r}}\frac{dt}{t}{)}^{\frac{1}{q}}=(\frac{p}{p-r}{)}^{\frac{1}{r}}\Vert f{\Vert }_{L^{p,q}}.\end{array}$$ (c) When $p,q>1$, we may take $r=1$, then $\Vert \cdot {\Vert }_{L^{p,q}}$ will be equivalent to the norm ${\left|\left|\left|\cdot \right|\right|\right|}_{L^{p,q}}$.
1.4.4.	(a) Write $f={\sum }_{n\in \mathbb{Z}}f{\chi }_{(2^n,2^{n+1}]}$ and $f_n=f{\chi }_{(2^n,2^{n+1}]}$. Since $\gamma d_f(\gamma {)}^{\frac{1}{p}}\le C<\infty$, we may fix $q<\infty$, then $$\begin{array}{r}\Vert f_n{\Vert }_{L^{p,q}}^q=p{\int }_0^{\infty }(\gamma d_{f_n}(\gamma {)}^{\frac{1}{p}}{)}^q\frac{d\gamma }{\gamma }\le p{\int }_{2^n}^{2^{n+1}}{C}^q\frac{d\gamma }{\gamma }=p{C}^q\log 2.\end{array}$$Thus, for any $f_n$, choose finitely combined simple functions ${\varphi }_{n,\epsilon }$ such that $$\Vert f_n-{\varphi }_{n,\epsilon }{\Vert }_{L^{p,q}}\le \frac{\epsilon }{2^{|n|+1}}(\frac{p}{q}{)}^{\frac{1}{q}}.$$Then the estimate follows that (for $p>1$, using the fact that $L^{p,\infty }$ is actually metrizable) $$\begin{array}{rl}\Vert f-\sum _{n\in \mathbb{Z}}{\varphi }_{n,\epsilon }{\Vert }_{L^{p,\infty }}& \le \sum _{n\in \mathbb{Z}}\Vert f_n-{\varphi }_{n,\epsilon }{\Vert }_{L^{p,\infty }}\le \left(\frac{q}{p}{)}^{\frac{1}{q}}\sum _{n\in \mathbb{Z}}\frac{\epsilon }{2^{|n|+1}}\right(\frac{p}{q}{)}^{\frac{1}{q}}<\epsilon .\end{array}$$For $0<p\le 1$, just take the power of $r$ for some suitable $0<r<p$ described in [Ex.1.1.12], and make some necessary changes to get the same estimate. (b) Write $f={\sum }_{i=1}^n{a}_i{\chi }_{A_i}$ and $A={\bigcup }_{i=1}^n{A}_i$. Set $M=\max a_i$, then for $x<(\lambda -M{)}^{-p}$, we have $$\Vert f-h{\Vert }_{L^{p,\infty }}\ge \underset{\lambda >M}{\sup }\frac{\lambda }{\lambda -M}=1.$$
1.4.10.	(a) Take $u_n=f_n{\chi }_{|f_n|\le \frac{\alpha }{2}}$, $v_n=f_n{\chi }_{|f_n|>\frac{\alpha }{2{\lambda }_n}}$ and $w_n=f_n{\chi }_{\frac{\alpha }{2}<|f_n|\le \frac{\alpha }{2{\lambda }_n}}$. Thus by setting $u={\sum }_n{\lambda }_n{u}_n$, $v={\sum }_n{\lambda }_n{v}_n$ and $w={\sum }_n{\lambda }_n{w}_n$, we obtain a decomposition $f=u+v+w$. Clearly $|u|\le \frac{\alpha }{2}$. For $v$, we have $$\begin{array}{rl}\mu (\{v\ne 0\})& \le \sum _n\mu (|f_n|>\frac{\alpha }{2{\lambda }_n})=\sum _n\mu (|{\lambda }_n{f}_n|>\frac{\alpha }{2})\\ & \le \sum _n\frac{2\lambda n}{\alpha }\le \frac{2}{\alpha }.\end{array}$$Plus, for $\{|w|>\frac{\alpha }{2}\}$, we have $$\begin{array}{rl}\frac{\alpha }{2}\mu (\{|w|>\frac{\alpha }{2}\})& \le {\int }_{\{|w|>\frac{\alpha }{2}\}}|w|d\mu \le {\int }_X|w|d\mu \\ & \le \sum _n{\int }_X|f_n|{\chi }_{\frac{\alpha }{2}<|f_n|\le \frac{\alpha }{2{\lambda }_n}}d\mu \\ & \le \sum _n{\lambda }_n({\int }_{\frac{\alpha }{2}}^{\frac{\alpha }{2{\lambda }_n}}{d}_{f_n}(t)dt+{\int }_0^{\frac{\alpha }{2}}{d}_{f_n}(\frac{\alpha }{2}\left)dt\right)\\ & \le \sum _n{\lambda }_n({\int }_{\frac{\alpha }{2}}^{\frac{\alpha }{2{\lambda }_n}}\frac{dt}{t}+{\int }_0^{\frac{\alpha }{2}}\frac{2}{\alpha }dt)\\ & \le \sum _n{\lambda }_n(\log ({\lambda }_n)+1)\le K+1.\end{array}$$Now combined together, we have $$\begin{array}{rl}\mu (\{|f|>\alpha \})& \le \mu (\{|u|>\frac{\alpha }{2}\})+\mu (\{v\ne 0\})+\mu (\{|w|>\frac{\alpha }{2}\})\\ & \le 0+\frac{2}{\alpha }+\frac{2}{\alpha }(K+1)=\frac{2(K+2)}{\alpha },\end{array}$$which is exactly $\Vert f{\Vert }_{L^{1,\infty }}\le 2(K+2)$. (b) Since $$\begin{array}{rl}\Vert Tf{\Vert }_{L^{1,\infty }}& =\Vert \sum _n{\lambda }_n{T}_{n}f{\Vert }_{L^{1,\infty }}\\ & \le 2(K+2)\underset{n}{\sup }\Vert T_{n}f{\Vert }_{L^{1,\infty }}\le 2B(K+2)\Vert f{\Vert }_{L^1}.\end{array}$$ (c) Since $$\begin{array}{rl}\mu (\{\sum _{n=1}^{\infty }{2}^{-\delta n}{f}_n>\alpha \})& \le \sum _{n=1}^{\infty }\mu (\{2^{-\delta n}{f}_n>(2^{\epsilon }-1)2^{-\epsilon n}\alpha \})\\ & \le \sum _{n=1}^{\infty }\frac{2^{-\delta n}}{(2^{\epsilon }-1)2^{-\epsilon n}\alpha }=\sum _{n=1}^{\infty }\frac{2^{(\epsilon -\delta )n}}{(2^{\epsilon }-1)\alpha }\\ & \le \frac{2^{\delta -\epsilon }-1}{(2^{\epsilon }-1)\alpha }.\end{array}$$
1.4.11.	First we show (a) (Fatou’s Lemma in $L^{p,q}$) (b) We know $$\underset{\alpha >0}{\sup }\alpha d_{g_n-g}(\alpha {)}^{\frac{1}{s}}\le (\frac{s}{q}{)}^{\frac{1}{s}}\Vert g_n-g{\Vert }_{L^{q,s}}$$Since $g_n\stackrel{L^{q,s}}{\to }g$, we have by this control that $g_n\to g$ in measure. Now choose a subsequence $g_{n_k}$ such that $g_{n_k}\to g$ a.e.
1.4.13.	(Dyadic Decomposition) Set $c_n=2^{\frac{n}{p}}{f}^{\ast }(2^n)$, $A_n:=\{f^{\ast }(2^{n+1})\le |f|\le f^{\ast }(2^n)\}$ and $f_n=c_n^{-1}f{\chi }_{A_n}$. Obviously, we have $f={\sum }_{n\in \mathbb{Z}}{f}_n$. Since $f^{\ast }$ is decreasing, we have $$\begin{array}{r}|f_n|\le 2^{-\frac{n}{p}}{f}^{\ast }(2^n{)}^{-1}{f}^{\ast }(2^n)=2^{-\frac{n}{p}}.\end{array}$$Clearly, $\mathrm{support}(f_n)\subset A_n$, which has measure $2^n$. For the last matter, set $E_n=[2^n,2^{n+1}]$. Thus $$\begin{array}{rl}\Vert f{\Vert }_{L^{p,q}}^q& ={\int }_0^{\infty }(t^{\frac{1}{p}}{f}^{\ast }(t){)}^{\frac{1}{q}}\frac{dt}{t}=\sum _{n\in \mathbb{Z}}{\int }_{E_n}(t^{\frac{1}{p}}{f}^{\ast }(t){)}^{\frac{1}{q}}\frac{dt}{t}\\ & \ge \sum _{n\in \mathbb{Z}}{2}^{\frac{nq}{p}}{f}^{\ast }(2^{n+1}{)}^q{\int }_{E_n}\frac{dt}{t}=\log 2\sum _{n\in \mathbb{Z}}{2}^{\frac{(n-1)q}{p}}{f}^{\ast }(2^n{)}^q\\ & =2^{-\frac{q}{p}}\log 2\Vert \{c_k{\}}_k{\Vert }_{l^q},\end{array}$$which is enough to show the boundedness of $\Vert \{c_k{\}}_k{\Vert }_{l^q}$. Another direction follows from the estimate on each $E_n$ that $f^{\ast }(t)\le f^{\ast }(2^n)$ and $t\le 2^{n+1}$ and use a shift.
1.4.14.	(a) Take $E^{\prime }=E\cap \{|f|\le A(1-\gamma {)}^{-\frac{1}{p}}\mu (E{)}^{-\frac{1}{p}}\}$, then $$d_f(A(1-\gamma {)}^{-\frac{1}{p}}\mu (E{)}^{-\frac{1}{p}})\le (1-\gamma )\mu (E)$$by the definition of $A$. Thus $$\begin{array}{r}\left|{\int }_{E^{\prime }}fd\mu \right|\le {\int }_{E^{\prime }}|f|d\mu \le \mu (E)A(1-\gamma {)}^{-\frac{1}{p}}\mu (E{)}^{-\frac{1}{p}}=A(1-\gamma {)}^{-\frac{1}{p}}\mu (E{)}^{1-\frac{1}{p}}.\end{array}$$ (b) Since $\{|f|>\alpha \}\subset \{\mathrm{Re}f>\frac{\alpha }{\sqrt{2}}\}\cup \{\mathrm{Re}f<-\frac{\alpha }{\sqrt{2}}\}\cup \{\mathrm{Im}f>\frac{\alpha }{\sqrt{2}}\}\cup \{\mathrm{Im}f<-\frac{\alpha }{\sqrt{2}}\}={\bigcup }_{i=1}^4{F}_i$, set $E_{n,i}=X_n\cap F_i$, where $\{X_n\}$ is an ascending chain of finite measure subspace with union equal to $X$. Then by condition, we have $$\begin{array}{r}\left|{\int }_{E_{n,i}^{\prime }}fd\mu \right|\ge \frac{\alpha }{\sqrt{2}}\gamma \mu (E_{n,i}).\end{array}$$Thus $\alpha \mu (E_n{)}^{\frac{1}{p}}\le B\sqrt{2}{\gamma }^{-1}$. Let $n\to \infty$, then $\alpha \mu (F_i{)}^{\frac{1}{p}}\le B\sqrt{2}{\gamma }^{-1}$. Thus, $$\begin{array}{r}{\alpha }^p\mu (\{|f|>\alpha \})\le \sum _{i=1}^4{\alpha }^p\mu (F_i)\le 4B^p{\sqrt{2}}^p{\gamma }^{-p},\end{array}$$which means $\Vert f{\Vert }_{L^{p,\infty }}\le B{4}^{\frac{1}{p}}\sqrt{2}{\gamma }^{-1}$.