HW7

Exercise 9

1. (i) Assume $0<{\int }_X{f}^{p}d\mu <\infty$ and $0<{\int }_X{g}^{q}d\mu <\infty$. Prove that equality holds in (10.3) if and only if there exists a constant $\alpha >0$ such that $g^q=\alpha f^p$ almost everywhere. (Hint. Use the proof of the Hölder inequality and the fact that equality holds in (9.2) if and only $p^p=b^q$.)

$${\int }_{X}fgd\mu \le {\left({\int }_X{f}^{p}d\mu \right)}^{1/p}{\left({\int }_X{g}^{q}d\mu \right)}^{1/q}\text{\hspace{1em}(10.3)}$$

(ii) Assume $0<{\int }_X{f}^{p}d\mu <\infty$ and $0<{\int }_X{g}^{q}d\mu <\infty$ Prove that equality holds in (9.4) if and only if there exists a constant $\beta >0$ such that $g=\beta f$ almost everywhere. Hint. Use part (i) and the proof of the Minkowski inequality.

$${\left({\int }_X(f+g{)}^{p}d\mu \right)}^{1/p}\le {\left({\int }_X{f}^{p}d\mu \right)}^{1/p}+{\left({\int }_X{g}^{q}d\mu \right)}^{1/q}\text{\hspace{1em}(10.4)}$$

\begin{proof} i) If $g^q=\alpha f^p$, then

$$\mathrm{LHS}={\left({\int }_X{f}^{p}d\mu \right)}^{1/p}{\left({\int }_X{g}^{q}d\mu \right)}^{1/q}={\left({\int }_X{f}^{p}d\mu \right)}^{1/p}{\left({\int }_X\alpha f^{p}d\mu \right)}^{1/q}={\alpha }^{1/q}{\int }_X{f}^p$$

and $\mathrm{RHS}={\int }_{X}f\cdot {\alpha }^{1/p}{f}^{p/q}d\mu ={\alpha }^{1/q}{\int }_X{f}^{p}d\mu$. Hence , the equality holds.

Conversely, assume that the inequality holds. In the proof of Holder inequality, take $f_1=c_{1}f$ and $g_1=c_{2}g$ such that $c_i>0$ and ${\int }_X{f}_1^{p}d\mu ={\int }_X{g}_1^{q}d\mu =1$, and then

$${\int }_X{f}_1{g}_{1}d\mu \le {\int }_X\left(\frac{f_1^p}{p}+\frac{g_1^q}{q}\right)d\mu =1={\left({\int }_X{f}_1^{p}d\mu \right)}^{1/p}{\left({\int }_X{g}_1^{q}d\mu \right)}^{1/q}$$

yields that $f_1{g}_1=f_1^p/p+g_1^q/q$ and Young’s inequality holds. Hence, $f_1^p=g_1^q$ almost everywhere. Therefore, there exists a constant $\beta >0$ such that $g=\beta f$.

ii) If $g=\beta f$, then

$$\mathrm{LHS}={\left({\int }_X(f+\beta f{)}^{p}d\mu \right)}^{1/p}=(1+\beta ){\left({\int }_X{f}^{p}d\mu \right)}^{1/p}$$

and

$$\mathrm{RHS}={\left({\int }_X{f}^{p}d\mu \right)}^{1/p}+{\left({\int }_X(\beta f{)}^{p}d\mu \right)}^{1/p}=(1+\beta ){\left({\int }_X{f}^{p}d\mu \right)}^{1/p}=\mathrm{LHS}.$$

On the other hand, assume that the inequality holds. Then in the proof of Minkowski inequality

$$\begin{array}{rl}& {\int }_{X}f(f+g{)}^{p-1}d\mu +{\int }_{X}g(f+g{)}^{p-1}d\mu \\ \le & {\left({\int }_X{f}^{p}d\mu \right)}^{1/p}{\left({\int }_X(f+g{)}^{(p-1)q}d\mu \right)}^{1/q}+{\left({\int }_X{g}^{p}d\mu \right)}^{1/p}{\left({\int }_X(f+g{)}^{(p-1)q}d\mu \right)}^{1/q}\end{array}$$

holds iff $\alpha f^p=(f+g{)}^{(p-1)q}=\beta g^p$. Therefore, there exists a constant $c$ such that $g=cf$. \end{proof}

3. Let $f\in {\mathcal{L}}^{\infty }(\mu )$. Show that there exists a measurable set $E\subset \mathcal{A}$ such that

$$\mu (E)=0\text{ and }\underset{X-E}{\sup }|f|=\Vert f{\Vert }_{\infty }\text{. }$$

\begin{proof} Define $E=\{x:|f(x)|>||f|{|}_{\infty }\}$, then $\mu (E)=0$ by lemma 10.6. By the definition of $E$, ${\sup }_{X-E}|f|⩽||f|{|}_{\infty }$. On the other hand, if ${\sup }_{X-E}|f|>||f|{|}_{\infty }$, there exists $x\in X\setminus E$ such that $|f(x)|⩾||f|{|}_{\infty }$ and so $x\in E$, contradiction. Hence, ${\sup }_{X-E}=||f|{|}_{\infty }$. \end{proof}

8. Let $\left\{f_n\right\}\subset L^1(\mu )$ be nonnegative and satisfying ${\mathrm{liminf}}_n{f}_n\ge f$ a.e. in $X$. Show that

$${\int }_X{f}_{n}d\mu ={\int }_{X}fd\mu =1\Rightarrow \underset{n}{\lim }{\int }_X\left|f_n-f\right|d\mu =0$$

(Hint: notice that the positive part and the negative part of $f-f_n$ have the same integral to obtain

$${\int }_X\left|f_n-f\right|d\mu =2{\int }_X{\left(f_n-f\right)}^{+}d\mu$$

Then, apply the dominated convergence theorem.)

\begin{proof} Since ${\int }_X{f}_{n}d\mu ={\int }_{X}fd\mu$, there is

$$0={\int }_X{f}_n-fd\mu ={\int }_X(f_n-f{)}^{+}d\mu -{\int }_X(f_n-f{)}^{-}d\mu$$

and so ${\int }_X(f_n-f{)}^{+}d\mu ={\int }_X(f_n-f{)}^{-}d\mu$. It deduces that ${\int }_X|f_n-f|d\mu =2{\int }_X(f_n-f{)}^{+}d\mu$. Since $\lim {\inf }_n{f}_n⩾f$ a.e. in $X$, there exists $N$ such that $f_n-f⩾0$ almost everywhere when $n⩾N$. Then by DCT, we have ${\lim }_n{\int }_X(f_n-f{)}^{+}=0$ and so for ${\lim }_n{\int }_X|f_n-f|d\mu$. Now we finish the proof. \end{proof}

9. Show that the following extension of Fatou’s lemma: if $f_n\ge -g_n$, with $g_n\in L^1$ nonnegative, $g_n\to g$ in $L^1$, then

$$\underset{n\to \infty }{\mathrm{liminf}}{\int }_X{f}_{n}d\mu \ge {\int }_X\underset{n\to \infty }{\mathrm{liminf}}{f}_{n}d\mu$$

whenever the right hand side is well defined. (Hint: prove first the statement under the additional assumption that $g_n\to g$ a.e. in $X$.)

\begin{proof} Since $f_n+g_n⩾0$, by Fatou’s lemma there is

$$\underset{n\to \infty }{\mathrm{liminf}}{\int }_X(f_n+g_n)d\mu ⩾{\int }_X\mathrm{liminf}(f_n+g_n)d\mu .\text{\hspace{1em}(*)}$$

To show ${\mathrm{liminf}}_{n\to \infty }{\int }_X{f}_{n}d\mu \ge {\int }_X{\mathrm{liminf}}_{n\to \infty }{f}_{n}d\mu$, it suffices to show ${\mathrm{liminf}}_n{\int }_X{g}_{n}d\mu ={\int }_X{\mathrm{liminf}}_n{g}_{n}d\mu$. Since $g_n\to g$ in $L^1$, we have ${\int }_X|g_n-g|d\mu \to 0$ as $n\to \infty$. Note that

$$0=\underset{n}{\lim }{\int }_X|g_n-g|d\mu ⩾\underset{n}{\lim }{\int }_X(g_n-g)d\mu ⩾\underset{n}{\mathrm{liminf}}{\int }_X{g}_{n}d\mu -{\int }_{X}gd\mu$$

yields that ${\mathrm{liminf}}_n{\int }_X{g}_{n}d\mu ⩽{\int }_X{\mathrm{liminf}}_n{g}_{n}d\mu$. By Fatou’s lemma there is ${\mathrm{liminf}}_n{\int }_X{g}_{n}d\mu ⩾{\int }_X{\mathrm{liminf}}_n{g}_{n}d\mu$ and so ${\mathrm{liminf}}_n{\int }_X{g}_{n}d\mu ⩽{\int }_X{\mathrm{liminf}}_n{g}_{n}d\mu$. Then $(\ast )$ deduces that ${\mathrm{liminf}}_{n\to \infty }{\int }_X{f}_{n}d\mu \ge {\int }_X{\mathrm{liminf}}_{n\to \infty }{f}_{n}d\mu$. \end{proof}

10. Let $\left\{f_n\right\}\subset L^1(\mu )$ be nonnegative functions. Show that the conditions

$$\underset{n\to \infty }{\mathrm{liminf}}{f}_n\ge f\text{ a.e., }\underset{n\to \infty }{\mathrm{limsup}}{\int }_X{f}_{n}d\mu \le {\int }_{X}fd\mu <\infty$$

imply the convergence of $f_n$ to $f$ in $L^1$. (Hint: use exercise 8.)

\begin{proof} The conditions and Fatou’s lemma yield that

$$\underset{n}{\mathrm{limsup}}{\int }_X{f}_{n}d\mu ⩽{\int }_{X}fd\mu ⩽{\int }_X\underset{n}{\mathrm{liminf}}{f}_{n}d\mu ⩽\underset{n}{\mathrm{liminf}}{\int }_X{f}_{n}d\mu .$$

It follows that ${\int }_{X}fd\mu ={\lim }_n{\int }_X{f}_{n}d\mu$. By exercise 8, we have ${\lim }_n{\int }_X|f_n-f|d\mu =0$, that is, $f_n$ converges to $f$ in $L^1$. \end{proof}

11. Let $(X,\mathcal{A},\mu )$ be a finite measure space. Show that $L^{\infty }\subset L^p$ for all $p\in (0,\infty )$. Also, show that, if $f$ is measurable, then

$$\underset{p\to \infty }{\lim }\Vert f{\Vert }_p=\Vert f{\Vert }_{\infty }$$

\begin{proof} To show $L^{\infty }\subseteq L^p$, it suffices to show for any $f$ satisfying $||f|{|}_{\infty }<\infty$, there is $||f|{|}_p<\infty$. By lemma 10.6, $|f|⩽||f|{|}_{\infty }$ a.e., it deduces that

$$||f|{|}_p=({\int }_X|f{|}^{p}d\mu {)}^{1/p}⩽(\mu (X)||f|{|}_{\infty }^p{)}^{1/p}<\infty .(\ast )$$

Therefore, $L^{\infty }\subseteq L^p$.

For any simple function $g={\sum }_{i=1}^n{a}_i{\chi }_{E_i}$, we have $||g|{|}_p=({\sum }_{k=1}^n\mu (E_i)a_i^p{)}^{1/p}$ and $||g|{|}_p\to {\max }_i{a}_i=||g|{|}_{\infty }$. Since $f$ is measurable, there exists a sequence of simple functions $\{g_k\}$ such that $|g_k|\uparrow |f|$. Then by MCT we have ${\lim }_p||f|{|}_p=||f|{|}_{\infty }$. \end{proof}

12. Let $0<p<1$ and $f,g$ be measurable functions on $X$ such that $f\ge 0,g\ge 0$. Then

$$\Vert f+g{\Vert }_p⩾\Vert f{\Vert }_p+\Vert g{\Vert }_p$$

(This is Minkowski’s Inequality for $0<p<1$.)

\begin{proof} Let $h(x)=x^p$. Then $h^{\prime \prime }(x)=p(p-1)x^{p-2}<0$ for any $x>0$ and $0<p<1$. If one of $||f|{|}_p$ and $||g|{|}_p$ equals $0$, then $f+g⩾f$ and $f+g⩾g$ yield that the inequality holds. Now assume that $||f|{|}_p>0$ and $||g|{|}_p>0$.

Define $t=||f|{|}_p/(||f|{|}_p+||g|{|}_p)$, then by Jensen’s inequality we have

$$(f+g{)}^p=(t\cdot \frac{f}{t}+(1-t)\cdot \frac{g}{1-t}{)}^p⩾t(f/t{)}^p+(1-t)(g/(1-t){)}^p=t^{1-p}{f}^p+(1-t{)}^{1-p}{g}^p.$$

It deduces that $||f+g|{|}_p^p⩾(||f|{|}_p^p+||g|{|}_p^p{)}^p$ and so $\Vert f+g{\Vert }_p⩾\Vert f{\Vert }_p+\Vert g{\Vert }_p$. \end{proof}

13. a) Let $p\ge 1$ and let ${\Vert f_n-f\Vert }_p\to 0$ as $n\to \infty$. Show that ${\Vert f_n\Vert }_p\to \Vert f{\Vert }_p$ as $n\to \infty$.

b) Suppose $f_n,f\in L^p,f_n\to f$ a.e and ${\Vert f_n\Vert }_p\to \Vert f{\Vert }_p$ as $n\to \infty$. Show that ${\Vert f_n-f\Vert }_p\to 0$ as $n\to \infty$.

\begin{proof} a) Note that

$$-||f_n-f|{|}_p⩽||f_n|{|}_p-||f|{|}_p⩽||f_n-f|{|}_p$$

and $||f_n-f|{|}_p\to 0$, then we have ${\lim }_n||f_n|{|}_p=||f|{|}_p$ by the sandwich theorem.

b) Since $L^p$ is a Banach space and $|f_n|,|f|\in L^p$, we have $|f_n|+|f|\in L^p$. Note that $|f_n-f{|}^p⩽(|f_n|+|f|{)}^p$ and ${\int }_X(|f_n|+|f|{)}^{p}d\mu <\infty$, then by DCT we have

$$\underset{n}{\lim }{\int }_X|f_n-f{|}^{p}d\mu =0.$$

Now we finish the proof. \end{proof}

15. Let $p$ and $q$ be conjugate indices and ${\lim }_{n\to \infty }{\Vert f_n-f\Vert }_p=0={\lim }_{n\to \infty }\Vert g_n-$ $g{\Vert }_q$, where $f_n,f\in L^p$ and $g_n,g\in L^q,n=1,2,\cdots$. Show that ${\lim }_{n\to \infty }\Vert f_n{g}_n-$ $fg{\Vert }_1=0$.

\begin{proof} Note that by the Holder inequality we have

$$0⩽||f_n{g}_n-fg|{|}_1⩽||f_n{g}_n-f_{n}g|{|}_1+||f_{n}g-fg|{|}_1⩽||f_n|{|}_p||g_n-g|{|}_q+||f_n-f|{|}_p||g|{|}_q\to 0$$

as ${\lim }_n||f_n-f|{|}_p={\lim }_n||g_n-g|{|}_q=0$ and $||f_n|{|}_p,||g|{|}_q<\infty$. Therefore, ${\lim }_n||f_n{g}_n-fg|{|}_1=0$. \end{proof}

17. Let $p,q\in (1,\infty )$. Show that the inclusion $L^p\subset L^q$ implies that the inclusion map $i:L^p\to L^q$ is continuous. (Hint: use closed graph theorem.)

\begin{proof}

Closed graph theorem: if a linear operator between Banach spaces has a closed graph, then the operator is continuous.

By the closed graph theorem, it suffices to show the graph $\{(x,x):x\in L^p\}\subseteq L^p\times L^q$ is closed. Suppose $\{f_n\}$ is a sequence in $L^p$ and $f_n\to f$ in $L^p$, $f_n\to g$ in $L^q$. Note that there exists a subsequence $f_{n_k}$ such that $f_{n_k}\to f$ pointwise a.e. In $L^q$, assume that $i(f_{n_k})=f_{n_k}\to g$ in $L^q$. Then there is a subsequence $f_{n_{m_k}}\to g$ pointwise a.e. It deduces that $f=g$ a.e. Therefore, $\{(f_n,f_n)\}\to \{(f,f)\}\in L^p\times L^q$ and so $\{(x,x):x\in L^p\}$ is closed in $L^p\times L^q$. By closed graph theorem, $i$ is continuous. \end{proof}

18. Let $(X,\mathcal{A},\mu )$ be a finite measure space. Assume that $f\in L^{\infty },\Vert f{\Vert }_{\infty }>0$. Let

$$a_n={\int }_X|f{|}^{n}d\mu ,n=1,\cdots .$$

Prove that

$$\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\Vert f{\Vert }_{\infty }.$$

\begin{proof} Note that $a_n^{1/n}=||f|{|}_n$ for all $n\in {\mathbb{N}}_{+}$. By exercise 11, we have ${\lim }_{p\to \infty }\Vert f{\Vert }_p=\Vert f{\Vert }_{\infty }$. Then

$$\frac{a_{n+1}}{a_n}={\left(\frac{a_{n+1}^{1/(n+1)}}{a_n^{1/n}}\right)}^n\cdot a_{n+1}^{1/(n+1)}\to ||f|{|}_{\infty },$$

because $a_n^{1/n}\to ||f|{|}_{\infty }$ as $n\to \infty$. \end{proof}

21. Let $(X,\mathcal{A},\mu )$ be a measure space, $2<p<\infty$, and $\left\{f_n\right\},f$ defined on $X$ with ${\Vert f_n\Vert }_p\to \Vert f{\Vert }_p$, such that $f_n\rightharpoonup f$ in $L^p$. Prove that $f_n\to f$ in $L^p$.

\begin{proof} We say $f_n\rightharpoonup f$ in $L^p$ if for all continuous linear functional $T$ on $L^p$, there is $T(f_n)\to T(f)$ as $n\to \infty$. The following proof is a proof of Radon-Riesz property.

WLOG we assume that $||f_n|{|}_p=1$, otherwise take $f_n=f_n/||f_n|{|}_p$. Recall that $L^p$ is uniformly convex, then it suffices to show

$$||\frac{f_n+f}{2}|{|}_p\to 1.$$

Since $T(f_n)\to T(f)$, we have $|T(f)|={\mathrm{liminf}}_n|T(f_n)|⩽{\mathrm{liminf}}_n||T||||f_n||$. Then by $||f||={\sup }_{T\in (L^p{)}^{\ast },||T||=1}|T(f)|$ we have $||f|{|}_{L^p}⩽{\mathrm{liminf}}_n||f_n||$. Since $\frac{f_n+f}{2}\rightharpoonup f$, there is

$$1=||f|{|}_p⩽\underset{n}{\mathrm{liminf}}||\frac{f_n+f}{2}|{|}_p.$$

Note that $(f_n+f)/2$ has norm at most $1$, then $||\frac{f_n+f}{2}|{|}_p\to 1$ and so $||f_n-f|{|}_p\to 0$. \end{proof}

A Banach space $X$ is said to satisfy the Radon-Riesz property (or the Kadec-Klee property) if: A sequence $\left\{x_n\right\}\subset X_{\text{converges strongly to }x}$ (i.e., $\Vert x_n-x\Vert \to 0$ ), if and only if:

• $\left\{x_n\right\}$ converges weakly to $x$, and
• $\Vert x_n\Vert \to \Vert x\Vert$.