用户: Solution/ 解答: GTM249 Chapter 1 Part I

1.1 $L^p$ and Weak $L^p$

1.1.1	(a) For an arbitrary sequence $\{{\alpha }_n\}\downarrow \alpha$, we have a pointwise convergence ${\chi }_{|f|>{\alpha }_n}\downarrow {\chi }_{|f|>\alpha }$. If $d_f(\alpha )$ is unbounded, then the proposition holds automatically. Otherwise, ${\chi }_{|f|>{\alpha }_n}$ are all bounded by ${\chi }_{|f|>\alpha }$, thus $$d_f(\alpha )=\int {\chi }_{|f|>\alpha }d\mu \stackrel{\mathrm{DCT}}{=}\underset{n\to \infty }{\lim }\int {\chi }_{|f|>{\alpha }_n}d\mu =\underset{n\to \infty }{\lim }{d}_f({\alpha }_n).$$ (b) Let $E=\{|f|>\alpha \}$ and $E_n=\{|f_n|>\alpha \}$. Thus by noting that $\mu ({\bigcap }_{n=m}^{\infty }{E}_n)\le \underset{n\to \infty }{\mathrm{liminf}}\mu (E_n)$ and $E\subset {\bigcup }_{m=1}^{\infty }{\bigcap }_{n=m}^{\infty }{E}_n$, $\mu$-a.e., we have $d_f\le \underset{n\to \infty }{\mathrm{liminf}}{d}_{f_n}$ by definition. (c) We have $\underset{n\to \infty }{\mathrm{limsup}}{d}_{f_n}\le d_f$ by $|f_n|\uparrow |f|$. Now the result comes with (b) that $$\underset{n\to \infty }{\mathrm{limsup}}{d}_{f_n}\le d_f\le \underset{n\to \infty }{\mathrm{liminf}}{d}_{f_n}.$$
1.1.2.	(a) Use induction. Suppose this holds for the $n-1$ case, and consider the $n$ case. If $p_n=\infty$, then since $\sum _{k=1}^{n-1}\frac{1}{p_k}=\frac{1}{p}$, we have by pulling out the essential supremum of $|f_n|$ that $$\Vert f_1\cdots f_n{\Vert }_{L^p}\le \Vert f_1\cdots f_{n-1}{\Vert }_{L^p}\Vert f_n{\Vert }_{L^{\infty }}\le \Vert f_1{\Vert }_{L^{p_1}}\cdots \Vert f_{n-1}{\Vert }_{L^{p_{n-1}}}\Vert f_n{\Vert }_{L^{\infty }}.$$Otherwise, if $p_n<\infty$, then automatically $p<\infty$. Set $r=\frac{p_n}{p_n-p}$ and $q=\frac{p_n}{p}$. Then $$\begin{array}{rl}\Vert f_1\cdots f_n{\Vert }_{L^p}& =\Vert |f_1\dots f_n{|}^p{\Vert }_{L^1}^{\frac{1}{p}}\stackrel{\text{Ho¨lder}}{\le }\Vert |f_1\cdots f_{n-1}{|}^p{\Vert }_{L^r}^{\frac{1}{p}}\Vert |f_n{|}^p{\Vert }_{L^q}^{\frac{1}{p}}\\ & =\Vert f_1\cdots f_{n-1}{\Vert }_{L^{pr}}\Vert f_n{\Vert }_{L^{pq}}\\ & \le \Vert f_1{\Vert }_{L^{p_1}}\cdots \Vert f_{n-1}{\Vert }_{L^{p_{n-1}}}\Vert f_n{\Vert }_{L^{p_n}}.\end{array}$$Also, note that the $n=2$ case is contained in the proof. (b) The equality condition is the same as the one of Hölder’s inequality in the $n=2$ case. (c) It suffices to show that $\Vert fg{\Vert }_{L^1}\Vert g^{-1}{\Vert }_{L^{\frac{q}{1-q}}}\ge \Vert f{\Vert }_{L^q}$, which is straightforward by $(a)$.
1.1.3.	(a) If $\Vert f{\Vert }_{L^{\infty }}=\infty$, then define $E_n=\{|f|>n\}$. Since $\mu (E_n)>0$ for each $n$, we have $\Vert f{\Vert }_{L^p}\ge \Vert f{\Vert }_{L^p(E_n)}\ge n\mu (E_n{)}^{\frac{1}{p}}$. Let $p\to \infty$ to get $\underset{p\to \infty }{\mathrm{liminf}}\Vert f{\Vert }_{L^p}\ge n$, then let $n$ vary. If $\Vert f{\Vert }_{L^{\infty }}<\infty$, now $\Vert f{\Vert }_{L^p}\le \Vert f{\Vert }_{L^{\infty }}^{\frac{p-p_0}{p}}\Vert f{\Vert }_{L^{p_0}}^{\frac{p_0}{p}}$. Let $p\to \infty$, we obtain $\underset{p\to \infty }{\mathrm{limsup}}\Vert f{\Vert }_{L^p}\le \Vert f{\Vert }_{L^{\infty }}$. For another, set $E_{\gamma }=\{|f|>\gamma \Vert f{\Vert }_{L^{\infty }}\}$ for arbitrary $\gamma \in (0,1)$. Then $\mu (E_{\gamma })>0$ and thus $\Vert f{\Vert }_{L^{p_0}(E_{\gamma })}>0$, and $$\Vert f{\Vert }_{L^p}\ge (\gamma \Vert f{\Vert }_{L^{\infty }}{)}^{\frac{p-p_0}{p}}\Vert f{\Vert }_{L^{p_0}(E_{\gamma })}^{\frac{p_0}{p}}.$$Again let $p\to \infty$, we get $\underset{p\to \infty }{\mathrm{liminf}}\Vert f{\Vert }_{L^p}\ge \gamma \Vert f{\Vert }_{L^{\infty }}$. Now let $\gamma \to 1$. (b) By Jensen’s inequality, we have $${\int }_X\log |f|d\mu \le \log ({\int }_X|f|d\mu ).$$It yields $$\Vert f{\Vert }_{L^p}=({\int }_X|f{|}^{p}d\mu {)}^{\frac{1}{p}}\ge \left(\exp \right({\int }_{X}p\log |f|d\mu ){)}^{\frac{1}{p}}=\exp ({\int }_X\log |f|d\mu ).$$ (c) Fix a sequence $\{p_n\}\downarrow 0$ bounded by $p_0$ and define $$h_n(x)=\frac{1}{p_0}(|f(x){|}^{p_0}-1)-\frac{1}{p_n}(|f(x){|}^{p_n}-1).$$Using that ${\int }_1^t{s}^{p-1}ds=\frac{1}{p}(t^p-1)\downarrow \log t$ as $p\downarrow 0$, the Lebesgue monotone convergence theorem yields ${\int }_X{h}_{n}d\mu \uparrow {\int }_{X}hd\mu$, hence $${\int }_X\frac{1}{p_n}(|f(x){|}^{p_n}-1)d\mu \downarrow {\int }_X\log |f|d\mu .$$Using part (b) and taking $t={\int }_X|f{|}^{p_n}d\mu$ in the inequality $t\le e^{t-1}$, we get $$\exp ({\int }_X\log |f|d\mu )\le ({\int }_X|f{|}^{p_n}d\mu {)}^{\frac{1}{p_n}}\le \exp ({\int }_X\frac{1}{p_n}(|f{|}^{p_n}-1)d\mu ).$$Thus take $n\to \infty$ and we have $\underset{p\to 0}{\lim }\Vert f{\Vert }_{L^p}=\exp ({\int }_X\log |f|d\mu )$.
1.1.4.	(a) Use Jensen’s inequality of integration to the $f(x)=\sum _{k=1}^{\infty }{a}_k{\chi }_{[k-1,k)}$ and $X=\mathbb{R}-(-\infty ,0]$, since $x^{\theta }$ is convex for $\theta \in (0,1]$. Another way is to assume that $1=({\sum }_{j=1}^{\infty }{a}_j{)}^{\theta }$ by homogeneity, which implies $a_j\le 1$, and subsequently $1={\sum }_{j=1}^{\infty }{a}_j\le {\sum }_{j=1}^{\infty }{a}_j^{\theta }$. (b) This is the same as the preceding question with a change in convexity. Or observe that $({\sum }_{j=1}^{\infty }{a}_j^{\theta }{)}^{1/\theta }\le {\sum }_{j=1}^{\infty }(a_j^{\theta }{)}^{1/\theta }={\sum }_{j=1}^{\infty }{a}_j$. (c) Use the discrete Jensen’s inequality. The problem is the same as $$(\frac{{\sum }_{k=1}^n{a}_k}{n}{)}^{\theta }\le \sum _{k=1}^n{a}_k^{\theta }.$$ (d) The same as (c).
1.1.5.	(a) Since $p\ge 1$, we have $$\begin{array}{rl}\Vert \sum _{j=1}^N{f}_j{\Vert }_{L^p}^p& ={\int }_X|\sum _{j=1}^N{f}_j{|}^{p}d\mu ={\int }_X|\sum _{j=1}^N{f}_j||\sum _{j=1}^N{f}_j{|}^{p-1}d\mu \\ & \le \sum _{i=1}^N{\int }_X|f_i||\sum _{j=1}^N{f}_j{|}^{p-1}d\mu \stackrel{\text{Ho¨lder}}{\le }(\sum _{i=1}^N\Vert f_j{\Vert }_{L^p}\left)\right({\int }_X|\sum _{j=1}^N{f}_j{|}^{p}d\mu {)}^{\frac{p-1}{p}}\\ & =\sum _{i=1}^N\Vert f_j{\Vert }_{L^p}\frac{\Vert {\sum }_{j=1}^N{f}_j{\Vert }_{L^p}^p}{\Vert {\sum }_{j=1}^N{f}_j{\Vert }_{L^p}}.\end{array}$$The rest is by multiplying both sides by $\Vert {\sum }_{j=1}^N{f}_j{\Vert }_{L^p}^{1-p}$. (b) This is just the same with (a). Here the first “$\le$” turns to “$=$” and the Hölder’s inequality turns its direction. (c) This is direct from 1.1.4 (a) pointwise on $X$, and subsequently 1.1.4 (c) where $a_j={\int }_X|f_j{|}^{p}d\mu$. (d) Take $n$ disjoint sets $M_j$ with the same measure $\alpha >0$ and take $f_j={\chi }_{M_j}$. Then $\Vert {\sum }_{j=1}^N{f}_j{\Vert }_{L^p}=N^{\frac{1}{p}}{\alpha }^{\frac{1}{p}}$ and ${\sum }_{j=1}^N\Vert f_j{\Vert }_{L^p}=N{\alpha }^{\frac{1}{p}}$.
1.1.6.	(a) The $p=1$ case is trivial. When $p>1$, let $q$ be the conjugate number of $p$, then set $G(t)={\int }_{X}F(x,t)d\mu (x)$, we have $$\begin{array}{rl}\Vert G{\Vert }_{L^p(\nu )}^p& ={\int }_T({\int }_{X}F(x,t)d\mu (x))G(t{)}^{p-1}d\nu (t)={\int }_X{\int }_{T}F(x,t)G(t{)}^{p-1}d\nu (t)d\mu (x)\\ & \stackrel{\text{Ho¨lder}}{\le }{\int }_X({\int }_{T}F(x,t{)}^{p}d\nu (t){)}^{\frac{1}{p}}\Vert G{\Vert }_{L^p(\nu )}^{p-1}d\mu (x).\end{array}$$Now eliminating $\Vert G{\Vert }_{L^p(\nu )}^{p-1}$ on both sides finishes the proof when $\Vert G{\Vert }_{L^p(\nu )}<\infty$. Otherwise, choose ascending sequences of subsets $X_n\subset X$ and $T_n\subset T$, such that ${\bigcup }_n{X}_n=X$ and ${\bigcup }_n{T}_n=T$, and $\mu (X_n),\nu (T_n)<\infty$ for all $n$. Now take $F_k=\max (F,k)$. Thence $$\left[{\int }_{T_m}\right({\int }_{X_n}{F}_k(x,t)d\mu (x){)}^{p}d\nu (t){]}^{\frac{1}{p}}\le {\int }_{X_n}[{\int }_{T_m}F(x,t{)}^{p}d\nu (t){]}^{\frac{1}{p}}d\mu (x).$$for all $m,n$ and $k\in \mathbb{N}$. By monotony, we obtain the desired result by taking $m,n\to \infty$ and then $k$. (b) (c) Since Hölder’s inequality reverses when $0<p<1$. (d)
1.1.7.	(a) Since $$\begin{array}{rl}\Vert \sum _{j=1}^N{f}_j{\Vert }_{L^{p,\infty }}& =\sup \{\gamma d_{f_1+\cdots +f_N}(\gamma {)}^{\frac{1}{p}}:\gamma >0\}\le \sum _{j=1}^N\sup \{\gamma d_{f_j}(\frac{\gamma }{N}{)}^{\frac{1}{p}}:\gamma >0\}\\ & =N\sum _{j=1}^N\sup \{\gamma d_{f_j}(\gamma {)}^{\frac{1}{p}}:\gamma >0\}=N\sum _{j=1}^N\Vert f_j{\Vert }_{L^{p,\infty }}.\end{array}$$ (b) Since $$\begin{array}{rl}\Vert \sum _{j=1}^N{f}_j{\Vert }_{L^{p,\infty }}& =\sup \{\gamma d_{f_1+\cdots +f_N}(\alpha {)}^{\frac{1}{p}}:\gamma >0\}\le N^{\frac{1}{p}-1}\sum _{j=1}^N\sup \{\gamma d_{f_j}(\frac{\gamma }{N}{)}^{\frac{1}{p}}:\gamma >0\}\\ & =N^{\frac{1}{p}}\sum _{j=1}^N\sup \{\gamma d_{f_j}(\gamma {)}^{\frac{1}{p}}:\gamma >0\}=N^{\frac{1}{p}}\sum _{j=1}^N\Vert f_j{\Vert }_{L^{p,\infty }}.\end{array}$$
1.1.8.	Suppose $\{f_n\}$ is an $L^p$ Cauchy sequence. Let $\{n_i\}$ be the sequence such that $\Vert f_{n_{i+1}}-f_{n_i}{\Vert }_{L^p}\le \frac{1}{2^i}$. Thus the series $f=f_{n_1}+{\sum }_{i=1}^{\infty }(f_{n_{i+1}}-f_{n_i})$ converges in $L^p$. Of course $f_n\stackrel{L^p}{⟶}f$, since for all $\epsilon >0$, take $N$ such that $\Vert f_n-f_{n_k}{\Vert }_{L^p}<\epsilon$. Thus $$\begin{array}{r}{\int }_X|f_n-f{|}^{p}d\mu ={\int }_X\underset{k\to \infty }{\mathrm{liminf}}|f_n-f_{n_k}{|}^{p}d\mu \stackrel{\text{Fatou}}{\le }\underset{k\to \infty }{\mathrm{liminf}}{\int }_X|f_n-f_{n_k}{|}^{p}d\mu =\underset{k\to \infty }{\mathrm{liminf}}\Vert f_n-f_{n_k}{\Vert }_{L^p}^p<{\epsilon }^p.\end{array}$$which implies $\Vert f_n-f{\Vert }_{L^p}<\epsilon$. Thus we have shown the completeness.
1.1.9.	Note that for $\epsilon >0$, $$\{x\in X\mid f_n(x)\to f(x)\}⫅\bigcup _{m=1}^{\infty }\bigcap _{n=m}^{\infty }\{x\in X\mid |f_n(x)-f(x)|<\epsilon \}⫅X,$$in which both sides have measure $\mu (X)$. Since $$\bigcap _{n=m}^{\infty }\{x\in X\mid |f_n(x)-f(x)|<\epsilon \}⫅\bigcap _{n=m+1}^{\infty }\{x\in X\mid |f_n(x)-f(x)|<\epsilon \},$$we have$$\bigcap _{n=N}^{\infty }\{x\in X\mid |f_n(x)-f(x)|<\epsilon \}>\mu (X)-\epsilon$$for some large $N$, thus $$\mu (\{x\in X\mid |f_n(x)-f(x)|<\epsilon \})<\epsilon$$for $n>N$.
1.1.10.	(a) This is due to$$\begin{array}{rl}{d}_{f_{\gamma }}(\alpha )& =\mu (\{|f|>\gamma \}\cap \{|f|>\alpha \})=d_f(\max (\alpha ,\gamma )).\\ d_{f^{\gamma }}(\alpha )& =d_f(\alpha )-d_{f_{\gamma }}(\alpha )=\{\begin{array}{ll}0,& \alpha \ge \gamma \\ d_f(\alpha )-d_f(\gamma ),& \alpha <\gamma \end{array}.\end{array}$$ (b) Using (a) we obtain $$\begin{array}{rl}\Vert f_{\gamma }{\Vert }_{L^p}^p& =p{\int }_0^{\infty }{\alpha }^{p-1}{d}_f(\max (\alpha ,\gamma ))d\alpha =p({\int }_{\gamma }^{\infty }{\alpha }^{p-1}{d}_f(\alpha )d\alpha +{\int }_0^{\gamma }{\alpha }^{p-1}{d}_f(\gamma )d\alpha )=p{\int }_{\gamma }^{\infty }{\alpha }^{p-1}{d}_f(\alpha )d\alpha +{\gamma }^p{d}_f(\gamma ).\\ \Vert f^{\gamma }{\Vert }_{L^p}^p& =p{\int }_0^{\gamma }{\alpha }^{p-1}(d_f(\alpha )-d_f(\gamma ))d\alpha =p{\int }_0^{\gamma }{\alpha }^{p-1}{d}_f(\alpha )d\alpha -{\gamma }^p{d}_f(\gamma ).\\ {\int }_{\gamma <|f|\le \delta }& |f{|}^{p}d\mu =\Vert f^{\delta }{\Vert }_{L^p}^p-\Vert f^{\gamma }{\Vert }_{L^p}^p=p{\int }_{\gamma }^{\delta }{\alpha }^{p-1}{d}_f(\alpha )d\alpha -{\delta }^p{d}_f(\delta )+{\gamma }^p{d}_f(\gamma ).\end{array}$$ (c) For $q>p$, from (b) and $q-p-1>-1$ we get $$\begin{array}{r}\Vert f^{\gamma }{\Vert }_{L^q}^q=q{\int }_0^{\gamma }{\alpha }^{q-1}{d}_f(\alpha )d\alpha -{\gamma }^q{d}_f(\gamma )=q^p{\int }_0^{\gamma }{\alpha }^{q-p-1}\Vert f{\Vert }_{L^{p,\infty }}^{p}d\alpha -{\gamma }^q{d}_f(\gamma )<\infty .\end{array}$$Plus, for $q<p$, from (b) and $q-p-1<-1$ we get $$\begin{array}{r}\Vert f_{\gamma }{\Vert }_{L^q}^q=q{\int }_{\gamma }^{\infty }{\alpha }^{q-1}{d}_f(\alpha )d\alpha +{\gamma }^q{d}_f(\gamma )\le q{\int }_{\gamma }^{\infty }{\alpha }^{q-p-1}\Vert f{\Vert }_{L^{p,\infty }}^{p}d\alpha +{\gamma }^q{d}_f(\gamma )<\infty .\end{array}$$Finally, $L^{p,\infty }\subseteq L^{p_0}+L^{p_1}$ is a direct consequence of $f=f_{\gamma }+f^{\gamma }$.
1.1.11.	(a) Take $A=\mu (E{)}^{-\frac{1}{p}}\Vert f{\Vert }_{L^{p,\infty }}$, then we have $$\begin{array}{rl}{\int }_E|f(x){|}^{q}d\mu (x)& =q{\int }_0^A{\alpha }^{q-1}{d}_f(\alpha )d\alpha +q{\int }_A^{\infty }{\alpha }^{q-1}{d}_f(\alpha )d\alpha \le A^q\mu (E)+q{\int }_A^{\infty }{\alpha }^{q-p-1}\Vert f{\Vert }_{L^{p,\infty }}^{p}d\alpha \\ & =A^q\mu (E)+\frac{q}{p-q}{A}^{q-p}\Vert f{\Vert }_{L^{p,\infty }}^p=\frac{p}{p-q}\mu (E{)}^{1-\frac{q}{p}}\Vert f{\Vert }_{L^{p,\infty }}^q.\end{array}$$ (b) Take $E=X$ in the preceding problem, then this means $L^{p,\infty }(X,\mu )\subset L^q(X,\mu )$.
1.1.12.	(a) By Exercise 1.1.11 (a), we have $$\begin{array}{rl}{\left|\left|\left|f\right|\right|\right|}_{L^{p,\infty }}& =\underset{0<\mu (E)<\infty }{\sup }\mu (E{)}^{-\frac{1}{r}+\frac{1}{p}}({\int }_E|f{|}^{r}d\mu {)}^{\frac{1}{r}}\\ & \le \underset{0<\mu (E)<\infty }{\sup }\mu (E{)}^{-\frac{1}{r}+\frac{1}{p}}(\frac{p}{p-r}{)}^{\frac{1}{r}}\mu (E{)}^{\frac{1}{r}-\frac{1}{p}}\Vert f{\Vert }_{L^{p,\infty }}\le (\frac{p}{p-r}{)}^{\frac{1}{r}}\Vert f{\Vert }_{L^{p,\infty }}.\end{array}$$ (b) Choose an ascending chain of subsets $X_n\subset X$ such that $\mu (X_n)<\infty$ and ${\bigcup }_n{X}_n=X$. Now set $E=\{|f|>\alpha \}\cap X_n$, we have $$\begin{array}{r}{\left|\left|\left|f\right|\right|\right|}_{L^{p,\infty }}\ge \mu (E{)}^{-\frac{1}{r}+\frac{1}{p}}({\int }_E|f{|}^{r}d\mu {)}^{\frac{1}{r}}\ge \mu (E{)}^{-\frac{1}{r}+\frac{1}{p}}({\alpha }^r\mu (E){)}^{\frac{1}{r}}=\alpha d_f(\alpha {)}^{\frac{1}{p}}.\end{array}$$Since $\alpha$ is arbitrary, we have $\Vert f{\Vert }_{L^{p,\infty }}\le {\left|\left|\left|f\right|\right|\right|}_{L^{p,\infty }}$ for finite measure spaces $X_n$, and so for $X$ by definition. (c) It’s easy to verify that $\left|\left|\left|f\right|\right|\right|$ is a norm for $p>1$ by Minkowski’s inequality and taking proper $r\in (1,p)$. When it comes to $0<p\le 1$, we have ${\left|\left|\left|\cdot \right|\right|\right|}_{L^{p,\infty }}^r$ to be the desired metric (but not a norm) which induces the same topology by observing the inequality after taking the power of $r$ on all terms. (d) Since $$\begin{array}{rl}\Vert \underset{n\to \infty }{\mathrm{liminf}}|g_n|{\Vert }_{L^{p,\infty }}& \le \underset{0<m(E)<\infty }{\sup }\mu (E{)}^{-\frac{1}{r}+\frac{1}{p}}({\int }_E\underset{n\to \infty }{\mathrm{liminf}}|g_n{|}^{r}d\mu {)}^{\frac{1}{r}}\\ & \stackrel{\text{Fatou}}{\le }\underset{0<m(E)<\infty }{\sup }\underset{n\to \infty }{\mathrm{liminf}}\mu (E{)}^{-\frac{1}{r}+\frac{1}{p}}({\int }_E|g_n{|}^{r}d\mu {)}^{\frac{1}{r}}\\ & \le \underset{n\to \infty }{\mathrm{liminf}}\underset{0<m(E)<\infty }{\sup }\mu (E{)}^{-\frac{1}{r}+\frac{1}{p}}({\int }_E|g_n{|}^{r}d\mu {)}^{\frac{1}{r}}\\ & \le (\frac{p}{p-r}{)}^{\frac{1}{r}}\underset{n\to \infty }{\mathrm{liminf}}\Vert g_n{\Vert }_{L^{p,\infty }}.\end{array}$$
1.1.13.	(a) By definition, $\Vert f_{\sigma }{\Vert }_{L^{1,\infty }}=\underset{\epsilon >0}{\sup }\mu \left(\right[\frac{{\sigma }^{-1}(1)-1}{N},\frac{{\sigma }^{-1}(1)}{N}\left)\right)(\frac{N}{1}-\epsilon {)}^1=1$. (b) Since $\sum _{\sigma \in S_N}{f}_{\sigma }=N!(1+\frac{1}{2}+\dots +\frac{1}{N}){\chi }_{[0,1)}$. (c) If not, suppose $\Vert \cdot \Vert$ is an equivalent norm, that is, $C_1\Vert f\Vert \le \Vert f{\Vert }_{L^{1,\infty }}\le C_2\Vert f\Vert$ for all $f$ and some $C_1,C_2>0$. Then we have $$\begin{array}{rl}{C}_{1}N!(1+\frac{1}{2}+\dots +\frac{1}{N})& =C_1\Vert \sum _{\sigma \in S_N}{f}_{\sigma }{\Vert }_{L^{1,\infty }}\le \Vert \sum _{\sigma \in S_N}{f}_{\sigma }\Vert \\ & \le \sum _{\sigma \in S_N}\Vert f_{\sigma }\Vert \le C_2\sum _{\sigma \in S_N}\Vert f_{\sigma }{\Vert }_{L^{1,\infty }}=C_{2}N!.\end{array}$$This contradicts with the fact that $1+\frac{1}{2}+\dots +\frac{1}{N}\to \infty$ when $N\to \infty$.
1.1.14.	(a) Since $$\begin{array}{r}{\int }_{|f|\le s}|f{|}^{q}d\mu =q{\int }_0^s{\alpha }^{q-1}{d}_f(\alpha )d\alpha \le q{\int }_0^s\Vert f{\Vert }_{L^{p,\infty }}^p{\alpha }^{q-p-1}d\alpha \le \frac{q}{q-p}{s}^{q-p}\Vert f{\Vert }_{L^{p,\infty }}^p.\end{array}$$ (b) Since $$\begin{array}{r}\Vert \underset{1\le j\le m}{\max }|f_j|{\Vert }_{L^{p,\infty }}^p=\sup \{{\gamma }^p{d}_{\underset{1\le j\le m}{\max }|f_j|}(\gamma )\}\le \sum _{j=1}^m\sup \{{\gamma }^p{d}_{f_j}(\gamma )\}\le \sum _{j=1}^m\Vert f_j{\Vert }_{L^{p,\infty }}^p.\end{array}$$
1.1.15.	Assume WLOG that $\Vert f_j{\Vert }_{L^{p_j,\infty }}=1$ for all $j$. Now the estimate follows $$\begin{array}{rl}{d}_{f_1\cdots f_k}(\alpha )& \le d_{f_1}\left(\frac{\alpha }{s_1}\right)+d_{f_2}\left(\frac{s_1}{s_2}\right)+\cdots +d_{f_k}\left(\frac{s_{k-1}}{1}\right)\\ & \le (\frac{s_1}{\alpha }{)}^{p_1}+\cdots +(\frac{1}{s_{k-1}}{)}^{p_k}.\end{array}$$Take $x_1=\frac{s_1}{\alpha },x_2=\frac{s_2}{s_1},\cdots ,x_k=\frac{1}{s_k}$, then we can rewrite it as $$\begin{array}{r}{d}_{f_1\cdots f_k}(\alpha )\le x_1^{p_1}+\cdots +x_k^{p_k},\end{array}$$where $x_1\cdots x_k=\frac{1}{\alpha }$. Set $x_1^{p_1}=\frac{\lambda }{p_1},\cdots ,x_k^{p_k}=\frac{\lambda }{p_k}$ for some constant $\lambda$ satisfying ${\prod }_j(\frac{\lambda }{p_j}{)}^{\frac{1}{p_j}}=\frac{1}{\alpha }$, and the right side becomes ${\sum }_j\frac{\lambda }{p_j}=p^{-1}({\alpha }^{-1}{\prod }_j{p}_j^{\frac{1}{p_j}}{)}^p$. Hence $\Vert f_1\cdots f_k{\Vert }_{L^{p,\infty }}\le p^{-\frac{1}{p}}{\prod }_j{p}_j^{\frac{1}{p_j}}$.
1.1.16.	We skip the cases when $\theta =0,1$ since there is nothing to prove. (a) If $p_1\ne \infty$, then $$\begin{array}{rl}({\int }_X|f{|}^{p}d\mu {)}^{\frac{1}{p}}& =({\int }_X|f{|}^{p(1-\theta )}|f{|}^{p\theta }d\mu {)}^{\frac{1}{p}}\le ({\int }_X|f{|}^{p(1-\theta )\frac{p_0}{p(1-\theta )}}d\mu {)}^{\frac{1-\theta }{p_0}}({\int }_X|f{|}^{p\theta \frac{p_1}{p\theta }}d\mu {)}^{\frac{\theta }{p_1}}\\ & =\Vert f{\Vert }_{L^{p_0}}^{1-\theta }\Vert f{\Vert }_{L^{p_1}}^{\theta }.\end{array}$$And if $p_1=\infty$, then $\frac{p_0}{p}=1-\theta$. Thus $$\begin{array}{rl}({\int }_X|f{|}^{p}d\mu {)}^{\frac{1}{p}}& \le \Vert f^{p-p_0}{\Vert }_{L^{\infty }}^{\frac{1}{p}}({\int }_X|f{|}^{p_0}d\mu {)}^{\frac{1}{p}}=\Vert f{\Vert }_{L^{p_0}}^{1-\theta }\Vert f{\Vert }_{L^{\infty }}^{L^{\theta }}.\end{array}$$ (b) Since $$\begin{array}{r}{t}^p{d}_f(t)=(t^{\frac{(1-\eta )p}{\theta }}{d}_f(t){)}^{\theta }(t^{\frac{\eta p}{1-\theta }}{d}_f(t){)}^{1-\theta },\end{array}$$we have by definition that $$\begin{array}{r}\Vert f{\Vert }_{L^{p,\infty }}^p\le \Vert f{\Vert }_{L^{\frac{\eta p}{1-\theta },\infty }}^{\eta }\Vert f{\Vert }_{L^{\frac{(1-\eta )p}{\theta },\infty }}^{1-\theta },\end{array}$$where $p_0(1-\theta )=\eta p$ and $p_1\theta =(1-\eta )p$.
1.1.17.	1. The $n=2$ case is nothing other than the $n=2$ Hölder’s inequality. For $n\ge 3$, suppose the proposition holds for $k\le n-1$, then for $k=n$, set $P(x_1,\dots ,x_{n-1})={\int }_{\mathbb{R}}|f_1({\pi }_1)(x)\cdots f_{n-1}({\pi }_{n-1})(x)|d{x}_n$, then $$\begin{array}{r}\Lambda (f_1,\dots ,f_n)\le {\int }_{{\mathbb{R}}^{n-1}}P(x_1,\dots ,x_{n-1})|f_n({\pi }_n)(x)|d{x}_1\cdots d{x}_{n-1}\le \Vert P{\Vert }_{L^{\frac{n-1}{n-2}}({\mathbb{R}}^{n-1})}\Vert f_n\circ {\pi }_n{\Vert }_{L^1({\mathbb{R}}^{n-1})}.\end{array}$$Note that by the induction hypothesis, we should have $$\begin{array}{rl}\Vert P{\Vert }_{L^{\frac{n-1}{n-2}}({\mathbb{R}}^{n-1})}& =\left({\int }_{{\mathbb{R}}^{n-1}}\right({\int }_{\mathbb{R}}|f_1({\pi }_1)(x)|\cdots |f_{n-1}({\pi }_{n-1})(x)|d{x}_n{)}^{\frac{n-1}{n-2}}d{x}_1\cdots d{x}_{n-1}{)}^{\frac{n-2}{n-1}}\\ & \stackrel{\text{Ho¨lder}}{\le }\left({\int }_{{\mathbb{R}}^{n-1}}\prod _{i=1}^{n-1}\right({\int }_{\mathbb{R}}|f_i({\pi }_i)(x){|}^{n-1}d{x}_n{)}^{\frac{1}{n-2}}d{x}_1\cdots d{x}_{n-1}{)}^{\frac{n-2}{n-1}}\\ & \stackrel{\text{induction}}{\le }\prod _{i=1}^{n-1}({\int }_{{\mathbb{R}}^n}|f_i(x){|}^{n-1}d{x}_1\cdots d{x}_n{)}^{\frac{1}{n-1}}=\prod _{i=1}^{n-1}\Vert f_i{\Vert }_{L^{n-1}({\mathbb{R}}^{n-1})}.\end{array}$$Thus, $$\begin{array}{r}\Lambda (f_1,\dots ,f_n)\le \Vert P{\Vert }_{L^{\frac{n-1}{n-2}}({\mathbb{R}}^{n-1})}\Vert f_n\circ {\pi }_n{\Vert }_{L^1({\mathbb{R}}^{n-1})}\le \prod _{i=1}^n\Vert f_i{\Vert }_{L^{n-1}({\mathbb{R}}^{n-1})}.\end{array}$$ (b) Take $f_j={\chi }_{{\pi }_j[\Omega ]}$, then the left side is no less than $$\begin{array}{r}{\int }_{{\mathbb{R}}^n}\prod _{i=1}^n|f_i\circ {\pi }_i|dx={\int }_{{\mathbb{R}}^n}{\chi }_{{\pi }_1[\Omega ]}\cdots {\chi }_{{\pi }_n[\Omega ]}dx\ge {\int }_{{\mathbb{R}}^n}{\chi }_{\Omega }dx=|\Omega |.\end{array}$$And the right side is no more than $$\begin{array}{rl}\prod _{i=1}^n\Vert f_j{\Vert }_{L^{n-1}({\mathbb{R}}^{n-1})}& =\prod _{i=1}^n({\int }_{{\mathbb{R}}^{n-1}}{\chi }_{{\pi }_i[\Omega ]}d{x}_1\cdots \widehat{d{x}_i}\cdots d{x}_n{)}^{\frac{1}{n-1}}=\prod _{i=1}^n|{\pi }_i[\Omega ]{|}^{\frac{1}{n-1}}\\ & \le \prod _{i=1}^n(\frac{1}{2}|\partial \Omega |{)}^{\frac{1}{n-1}}=2^{\frac{n}{1-n}}|\partial \Omega {|}^{\frac{n}{n-1}}.\end{array}$$

1.2 Convolution and Approximate Identities

1.2.1.	For $g\notin AB$, forbidding $h\in A$ along with $h^{-1}g\in B$, we have $$\begin{array}{r}{\int }_{G}f(h)g(h^{-1}g)dh=0\end{array}$$since at least one of $f(h)$ and $g(h^{-1}g)$ vanishes.
1.2.2.	Since $$\begin{array}{r}{}^{t}f\ast g(x)={\int }_{G}f(ty)g(y^{-1}x)dy={\int }_{G}f(y)g(y^{-1}tx)d(ty)={\int }_{G}f(y)g(y^{-1}tx)d(y){=}^t(f\ast g)(x).\end{array}$$and $$\begin{array}{r}f\ast g^t(x)={\int }_{G}f(y)g(y^{-1}xt)dy=(f\ast g{)}^t(x).\end{array}$$
1.2.3.
1.2.4.	(a) Since $f{\chi }_{B(0,n)}\stackrel{L^p}{⟶}f$. (b) Let $\phi \in C_c({\mathbb{R}}^n)$ be a positive function supported in $B(0,1)$ with integral $1$, then define ${\phi }_k(x)=k^n\phi (kx)$, which are compactly supported smooth functions with integral $1$ as well. Now we show ${\phi }_k\ast f\stackrel{L^p}{⟶}f$, which is because $$\begin{array}{rl}\Vert {\phi }_k\ast f-f{\Vert }_{L^p}^p& ={\int }_{\mathbb{R}}|{\int }_{\mathbb{R}}{\phi }_k(x-y)(f(x)-f(y))dy{|}^{p}dx\\ & \stackrel{\text{Jensen}}{\le }{\int }_{\mathbb{R}}{\int }_{|y-x|<\frac{1}{k}}{\phi }_k(x-y)|f(x)-f(y){|}^{p}dydx\\ & \le {\int }_{|t|<\frac{1}{k}}({\int }_{\mathbb{R}}|f(x)-f(x-t){|}^{p}dx){\phi }_k(-t)dt\\ & \le \underset{|t|<\frac{1}{k}}{\sup }{\int }_{\mathbb{R}}|f(x)-f(x-t){|}^{p}dx.\end{array}$$Then apply the average continuity.
1.2.5.	$\lambda$ is a Haar measure because for any $a>0$, $$\begin{array}{r}\lambda (aA)={\int }_{aA}\frac{dt}{t}={\int }_A\frac{d(at)}{at}={\int }_A\frac{dt}{t}=\lambda (A).\end{array}$$
1.2.6.	$\lambda$ is a left Haar measure because for any $(a,b)\in G$, $$\begin{array}{rl}\lambda ((a,b)A)& ={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{\chi }_{(a,b)A}(x,y)\frac{dxdy}{x^2}={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{\chi }_{(a,b)A}((a,b)(x,y))\det \left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right)\frac{dxdy}{(ax{)}^2}\\ & ={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{\chi }_A(x,y)\frac{dxdy}{x^2}=\lambda (A).\end{array}$$$\rho$ is a right Haar measure because for any $(a,b)\in G$, $$\begin{array}{rl}\rho (A(a,b))& ={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{\chi }_{A(a,b)}(x,y)\frac{dxdy}{|x|}={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{\chi }_{A(a,b)}((x,y)(a,b))\left|\det \left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)\right|\frac{dxdy}{|ax|}\\ & ={\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{\chi }_A(x,y)\frac{dxdy}{|x|}=\rho (A).\end{array}$$
1.2.7.	For the former, note that (here $G$ denotes the group $({\mathbb{R}}^{+},\frac{dt}{t})$) $$\begin{array}{rl}\Vert |f(x)|x^{\frac{1}{p}}\ast x^{-\frac{1}{p^{\prime }}}{\chi }_{[1,\infty )}{\Vert }_{L^p(G)}& =\left({\int }_0^{\infty }\right({\int }_0^{\infty }|f(t)|t^{\frac{1}{p}}(t^{-1}x{)}^{-\frac{1}{p^{\prime }}}{\chi }_{[1,\infty )}(t^{-1}x)\frac{dt}{t}{)}^p\frac{dx}{x}{)}^{\frac{1}{p}}\\ & =\left({\int }_0^{\infty }\right(\frac{1}{x}{\int }_0^x|f(t)|dt{)}^{p}dx{)}^{\frac{1}{p}}.\end{array}$$And $$\begin{array}{r}\Vert |f(x)|x^{\frac{1}{p}}{\Vert }_{L^p(G)}=({\int }_0^{\infty }(|f(x)|x^{\frac{1}{p}}{)}^p\frac{dx}{x}{)}^{\frac{1}{p}}=\Vert f{\Vert }_{L^p},\end{array}$$$$\begin{array}{r}\Vert x^{-\frac{1}{p^{\prime }}}{\chi }_{[1,\infty )}{\Vert }_{L^1(G)}={\int }_1^{\infty }{x}^{-\frac{1}{p^{\prime }}}\frac{dx}{x}=\frac{p}{p-1}.\end{array}$$Now combine with Minkowski’s inequality, $$\begin{array}{r}\Vert |f(x)|x^{\frac{1}{p}}\ast x^{-\frac{1}{p^{\prime }}}{\chi }_{[1,\infty )}{\Vert }_{L^p(G)}\le \Vert |f(x)|x^{\frac{1}{p}}{\Vert }_{L^1(G)}\Vert x^{-\frac{1}{p^{\prime }}}{\chi }_{[1,\infty )}{\Vert }_{L^p(G)}.\end{array}$$The latter and Exercise 1.2.8 follow in a rather similar manner, so we omit the procedure.
1.2.9.	By Young’s inequality, it’s quite straightforward to obtain $\Vert T{\Vert }_{L^p\to L^p}\le \Vert K{\Vert }_{L^1}$. To show the inverse, we use $K_R=K{\chi }_{B(0,R)}$ to approximate the result. Note that for $|x|\le R(1-\epsilon )$, we have $B(0,R\epsilon )\subset B(x,R)$. Thus $${\int }_{{\mathbb{R}}^n}{\chi }_R(x-y)K_{R\epsilon }(y)dy={\int }_{R^n}{K}_{R\epsilon }(y)dy=\Vert K_{R\epsilon }{\Vert }_{L^1}.$$We have $\Vert {\chi }_R{\Vert }_{L^p}^p=m(B(0,R){)}^p$, and $$\begin{array}{rl}\Vert K\ast {\chi }_R{\Vert }_{L^p}^p& \ge \Vert K_{R\epsilon }\ast {\chi }_R{\Vert }_{L^p(B(0,R(1-\epsilon )))}^p={\int }_{B(0,R(1-\epsilon ))}({\int }_{{\mathbb{R}}^n}{K}_{R\epsilon }(y){\chi }_R(x-y)dy{)}^{p}dx\\ & ={\int }_{B(0,R(1-\epsilon ))}\Vert K_{R\epsilon }{\Vert }_{L^1}^{p}dx=(1-\epsilon {)}^n\Vert K_{R\epsilon }{\Vert }_{L^1}^p.\end{array}$$This means $\Vert T{\Vert }_{L^p\to L^p}\ge (1-\epsilon {)}^{\frac{n}{p}}\Vert K_{R\epsilon }{\Vert }_{L^1}$ for each $\epsilon \in (0,1)$ and $R>0$. Let $R\to \infty$ and then $\epsilon \to 0^{+}$, we get $\Vert T{\Vert }_{L^p\to L^p}\ge \Vert K{\Vert }_{L^1}$.
1.2.10.	By Young’s inequality, we obtain $\Vert T{\Vert }_{L^p\to L^p}\le \Vert K{\Vert }_{L^1}$. For the inverse, take $f_n={\chi }_{[\frac{1}{n},n]}$ and $K_n=K{f}_n$, then for $n^{\epsilon -1}\le x\le n^{1-\epsilon }$, we have $$f_n\ast K_{n^{\epsilon }}\left(x\right)={\int }_0^{\infty }{K}_{n^{\epsilon }}\left(y\right)f_n\left(x-y\right)\frac{\text{d}y}{y}=||K_{n^{\epsilon }}|{|}_{L^1}$$Thus $$\frac{\Vert K\ast f_n{\Vert }_{L^p}^p}{\Vert f_n{\Vert }_{L^p}^p}\ge \frac{\Vert K_{n^{\epsilon }}\ast f_n{\Vert }_{L^p\left[n^{\epsilon -1},n^{1-\epsilon }\right]}^p}{2\log n}=\Vert K_{n^{\epsilon }}{\Vert }_{L^1}^p\left(1-\epsilon \right)$$Letting $n\to \infty$, we have $\Vert T{\Vert }_{L^p\to L^p}^p\ge \Vert K{\Vert }_{L^1}^p\left(1-\epsilon \right)$, then $\epsilon \to 0^{+}$, we get $\Vert T{\Vert }_{L^p\to L^p}\ge \Vert K{\Vert }_{L^1}$.
1.2.11.	(a) Since $$\begin{array}{r}{c}_k=({\int }_1^1(1-t^2{)}^{k}dt{)}^{-1}<({\int }_{-\frac{1}{\sqrt{k}}}^{\frac{1}{\sqrt{k}}}(1-t^2{)}^{k}dt{)}^{-1}<({\int }_{-\frac{1}{\sqrt{k}}}^{\frac{1}{\sqrt{k}}}(1-k{t}^2)dt{)}^{-1}=\frac{3}{4}\sqrt{k}<\sqrt{k}.\end{array}$$Thanks to the adjustment given by $c_k$, we have $$\begin{array}{r}(Q_k\ast f-f)(x)={\int }_{-1}^1(f(x-t)-f(x))Q_k(t)dt.\end{array}$$Thus $$\begin{array}{rl}\Vert Q_k\ast f-f{\Vert }_{L^1}& ={\int }_{\mathbb{R}}|{\int }_{-1}^1(f(x-t)-f(x))Q_k(t)dt|dx\\ & \le {\int }_{{\mathbb{R}}^2}|f(x)-f(x-t)|Q_k(t)dtdx\le {\int }_{\mathbb{R}}{Q}_k(t)\Vert f(x)-f(x-t){\Vert }_{L^1}dt.\end{array}$$Due to the average continuity, we have $\delta >0$ for each $\epsilon >0$ such that $\Vert f(x)-f(x-t){\Vert }_{L^1}<\epsilon$ for $|t|<\delta$. Plus, we have $\underset{k\to \infty }{\lim }\sqrt{k}(1-t^2{)}^k=0$ uniformly for $\delta <|t|<1$, thus take $k$ large so that $Q_k(x)<\epsilon$ for $n>k$, we have $$\begin{array}{rl}& {\int }_{\mathbb{R}}{Q}_k(t)\Vert f(x)-f(x-t){\Vert }_{L^1}dt\\ & \le {\int }_{|t|\le \delta }{Q}_k(t)\Vert f(x)-f(x-t){\Vert }_{L^1}dt+{\int }_{|t|>\delta }{Q}_k(t)\Vert f(x)-f(x-t){\Vert }_{L^1}dt\\ & \le \epsilon \Vert Q_k{\Vert }_{L^1}+{\int }_{|t|>\delta }{Q}_k(t)(\Vert f(x){\Vert }_{L^1}+\Vert f(x-t){\Vert }_{L^1})dt\\ & \le (1+2\Vert f{\Vert }_{L^1})\epsilon .\end{array}$$This ends the proof. (b) Due to the uniform continuity, we choose $\delta >0$ so that $|f(x)-f(x-t)|<\epsilon$ for $|t|\le \delta$. Now for large $k$, we have $$\begin{array}{rl}|(Q_k\ast f-f)(x)|& =|{\int }_{-1}^1(f(x-t)-f(x))Q_{k}dt|\\ & \le {\int }_{|t|\le \delta }|f(x)-f(x-t)|Q_k(t)dt+{\int }_{|t|>\delta }{Q}_k(t)|f(x)-f(x-t)|dt\\ & \le \epsilon \Vert Q_k{\Vert }_{L^1}+\epsilon \underset{x\in \mathbb{R}}{\sup }f(x)\\ & =\epsilon (1+\underset{x\in \mathbb{R}}{\sup }f(x)),\end{array}$$and note that the tending to $0$ does not depend on the choice of $\delta$. (c)&(d) Set $g(t)=f(t)-f(-1)-\frac{t+1}{2}(f(1)-f(-1))$, then $g(-1)=g(1)=0$. $g$ can be viewed as satisfying the condition in (c) by zero extension outside $[-1,1]$. Thus $g\ast Q_k\to g$ uniformly. However, also note that this means $$g\ast Q_k+f(-1)+\frac{x+1}{2}(f(1)-f(-1))\to f$$uniformly on $[-1,1]$, and $g\ast Q_k(x)$ is a polynomial in $x$, since $$\begin{array}{r}g\ast Q_k(x)={\int }_{-1}^{1}g(t)c_k(1-(x-t{)}^2{)}^{k}dt=\sum _{j=0}^{2k}({\int }_{-1}^{1}g(t)d_{j,k}{t}^{j}dt)x^j.\end{array}$$
1.2.12	Note that$$\begin{array}{rl}|L(f)(x){|}^2=|{\int }_0^{\infty }f(t)e^{-xt}dt{|}^2& \le {\int }_0^{\infty }|f(t){|}^2{e}^{-xt}\sqrt{t}dt{\int }_0^{\infty }\frac{e^{-xt}}{\sqrt{t}}dt\\ & ={\int }_0^{\infty }|f(t){|}^2{e}^{-xt}\sqrt{t}dt{\int }_0^{\infty }\frac{e^{-u}}{\sqrt{u/x}}\frac{du}{x}=\sqrt{\frac{\pi }{x}}{\int }_0^{\infty }|f(t){|}^2{e}^{-xt}\sqrt{t}dt.\end{array}$$Therefore $$\begin{array}{rl}\Vert L(f){\Vert }_{L^2}^2& \le \sqrt{\pi }{\int }_0^{\infty }{\int }_0^{\infty }|f(t){|}^2{e}^{-xt}\sqrt{\frac{t}{x}}dtdx\\ & =\sqrt{\pi }{\int }_0^{\infty }{\int }_0^{\infty }|f(t){|}^2{e}^{-u}\frac{t}{\sqrt{u}}\frac{du}{t}dt=\pi \Vert f{\Vert }_{L^2}^2.\end{array}$$ Another solution using the hint of convolution:$$\begin{array}{rl}L(f)(x)& ={\int }_0^{\infty }f(t)e^{-xt}\text{d}t={\int }_0^{\infty }f\left(\frac{1}{t}\right)e^{-\frac{x}{t}}\frac{\text{d}t}{t^2}\\ & ={\int }_0^{\infty }\sqrt{\frac{x}{t}}{e}^{-\frac{x}{t}}\cdot \frac{1}{\sqrt{t}}f\left(\frac{1}{t}\right)\frac{\text{d}t}{t}\cdot \frac{1}{\sqrt{x}}=(\sqrt{t}{e}^{-t}\ast \frac{1}{\sqrt{t}}f(\frac{1}{t}\left)\right)(x)\cdot \frac{1}{\sqrt{x}}.\end{array}$$Thus taking norm and using Minkowski’s inequality we obtain$$\begin{array}{rl}\Vert L(f){\Vert }_{L^2({\mathbb{R}}^{+},\text{d}t)}& =\Vert \sqrt{t}{e}^{-t}\ast \frac{1}{\sqrt{t}}f\left(\frac{1}{t}\right){\Vert }_{L^2({\mathbb{R}}^{+},\frac{\text{d}t}{t})}\\ & \le \Vert \sqrt{t}{e}^{-t}{\Vert }_{L^1({\mathbb{R}}^{+},\frac{\text{d}t}{t})}\Vert \frac{1}{\sqrt{t}}f\left(\frac{1}{t}\right){\Vert }_{L^2({\mathbb{R}}^{+},\frac{\text{d}t}{t})}=\sqrt{\pi }\Vert f{\Vert }_{L^2({\mathbb{R}}^{+},\text{d}t)}.\end{array}$$ Remark. The norm is exactly $\sqrt{\pi }$ indeed. In fact, consider a sequence of functions $f_n(t)=\frac{1}{\sqrt{t}}{\chi }_{[\frac{1}{n},n]}(t)$, we find $\Vert f_n{\Vert }_{L^2}=\sqrt{2\log (n)}$. Compute the norms of transformed functions:$$\begin{array}{r}\Vert L(f_n){\Vert }_2^2={\int }_{{\mathbb{R}}^{+}}({\int }_{\frac{1}{n}}^n\frac{1}{\sqrt{t}}{e}^{-st}dt{)}^{2}ds={\int }_{{\mathbb{R}}^{+}}\frac{1}{s}({\int }_{\frac{s}{n}}^{ns}\frac{1}{\sqrt{t}}{e}^{-t}dt{)}^{2}ds.\end{array}$$For arbitrary $\delta \in (0,1)$ fixed, we have$$\begin{array}{rl}{\int }_{{\mathbb{R}}^{+}}\frac{1}{s}({\int }_{\frac{s}{n}}^{ns}\frac{1}{\sqrt{t}}{e}^{-t}dt{)}^{2}ds& \ge {\int }_{(\frac{1}{n}{)}^{1-\delta }}^{n^{1-\delta }}\frac{1}{s}({\int }_{\frac{s}{n}}^{ns}\frac{1}{\sqrt{t}}{e}^{-t}dt{)}^{2}ds\\ & \ge {\int }_{(\frac{1}{n}{)}^{1-\delta }}^{n^{1-\delta }}\frac{1}{s}({\int }_{(\frac{1}{n}{)}^{\delta }}^{n^{\delta }}\frac{1}{\sqrt{t}}{e}^{-t}dt{)}^{2}ds.\end{array}$$For $n$ sufficiently large, we have by the famous integral that$$\begin{array}{r}{\int }_{(\frac{1}{n}{)}^{\delta }}^{n^{\delta }}\frac{1}{\sqrt{t}}{e}^{-t}dt\ge \sqrt{\pi }-\epsilon ,\end{array}$$thus,$$\begin{array}{rl}\Vert L(f_n){\Vert }_{L^2}^2& \ge {\int }_{(\frac{1}{n}{)}^{1-\delta }}^{n^{1-\delta }}\frac{\sqrt{\pi }-\epsilon }{s}ds=2\log (n)(1-\delta )(\sqrt{\pi }-\epsilon {)}^2\\ & =\Vert f_n{\Vert }_{L^2}^2(1-\delta )(\sqrt{\pi }-\epsilon {)}^2.\end{array}$$Finally, take $\delta \to 0^{+}$ and $\epsilon \to 0^{+}$.