Exercise 3.4

Question

Suppose $f$ is a complex measurable function on $X$, $\mu$ is a positive
    measure on $X$, and
    \begin{equation*}
      \varphi(p) = \int_X \lvert f vert\,d\mu = \lVert f Vert_p^p \quad (0 <
      p < \infty).
    \end{equation*}
    Let $E = \{p : \varphi(p) < \infty\}$. Assume $\lVert f Vert_\infty <
    \infty$.
    \begin{enumerate}[label=(\alph*)]
      \item If $r < p < s$, $r \in E$, and $s \in E$, prove that $p \in E$.
      \item Prove that $\log \varphi$ is convex in the interior of $E$ and that
        $\varphi$ is continuous on $E$.
      \item By $(a)$, $E$ is connected. Is $E$ necessarily open? Closed? Can $E$
        consist of a single point? Can $E$ be any connected subset of
        $(0,\infty)$?
      \item If $r < p < s$, prove that $\lVert f Vert_p \le \max(\lVert f
        Vert_r,\lVert f Vert_s)$. Show that this implies the inclusion
        $L^r(\mu) \cap L^s(\mu) \subset L^p(\mu)$.
      \item Assume that $\lVert f Vert_r < \infty$ for some $r < \infty$ and
        prove that
        \begin{equation*}
          \lVert f Vert_p 	o \lVert f Vert_\infty \quad 	ext{as}\ p 	o
          \infty.
        \end{equation*}
    \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$] Write $p = (1-\lambda) r + \lambda s$ for $0 < \lambda < 1$. By H"older's inequality, \begin{align*} \int_X \lvert f vert^p\,d\mu &= \int_X \lvert f vert^{(1-\lambda)r} \lvert f vert^{\lambda s}\,d\mu\ &\le \left(\int_X vert f vert^r\,d\mu ight)^{1-\lambda} \left(\int_X vert f vert^s\,d\mu ight)^{\lambda} \end{align*} which is finite. \end{proof} \begin{proof}[Proof of $(b)$] We want to show the inequality $3.1(1)$ for all $r < s$ inside $E$. By our proof in $(a)$, we have the inequality \begin{equation*} \varphi((1-\lambda) r + \lambda s) \le \varphi(r)^{1-\lambda} \cdot \varphi(s)^{\lambda} \end{equation*} and taking logarithms gives \begin{equation*} \log \varphi((1-\lambda) r + \lambda s) \le (1-\lambda)\log \varphi(r) + \lambda \log \varphi(s) \end{equation*} which holds for all $\lambda \in [0,1]$. This implies $\log\varphi$ and also $\varphi$ are continuous on the interior of $E$ by Theorem $3.2$. $\varphi$ is also continuous on the boundary of $E$ since if $r$ is the lower limit of $E$, \begin{equation*} \lim_{p o r^+} \varphi(p) = \int_{\lvert f vert \ge 1} \lvert f vert^p\,d\mu + \int_{\lvert f vert < 1} \lvert f vert^p\,d\mu, \end{equation*} and then first (resp.\ second) term on the right decreases (resp.\ increases) monotonically to the corresponding integral for $p=r$, hence $\lim_{p o r^+} \varphi(p) = \varphi(r)$. The same argument works for $p o s^-$. \end{proof} \begin{proof}[Solution for $(c)$] We claim $E$ can be an arbitrary connected subset of $(0,\infty)$. Consider if $X = (0,\infty)$ with the Lebesgue measure. \par We first write down some functions. For $r > 0$, the function \begin{equation*} a_r = \frac{\chi_{[1,\infty)}}{x^{1/r}} \end{equation*} is in $L^p$ for all $p > r$ but not any $p \le r$. For $s < \infty$, the function \begin{equation*} b_s = \frac{\chi_{(0,1]}}{x^{1/s}} \end{equation*} is in $L^p$ for all $p < s$ but not any $p \ge s$. \par First, $f = \chi_{(0,1]}$ gives $E = (0,\infty]$. Next, $a_r$ gives $E = (r,\infty]$. Also, $b_s$ gives $E = (0,s)$. And $a_r + b_s$ gives $E = (r,s)$. So it suffices to show we can get $E$ that are closed on one side. \par For intervals of the form $[r,\infty]$, then function \begin{equation*} \sum_{n=0}^\infty 2^{-n} a_{r-1/n}(x-n) \end{equation*} works. For intervals of the form $(0,s]$, we can do a similar trick with the $b_s$: \begin{equation*} \sum_{n=0}^\infty 2^{-n} b_{r+1/n}(x-n). \end{equation*} Finally, for intervals of the form $(r,s],[r,s),[r,s]$, we can add the above examples together. \end{proof} \begin{proof}[Proof of $(d)$] By $(a)$, we have \begin{equation*} \lVert f Vert_p \le \lVert f Vert_r^{1 - \lambda s/p} \lVert f Vert_s^{1 - (1-\lambda)r/p}. \end{equation*} By dividing up into cases when the $L^r$ or $L^s$ norms are larger, we then have \begin{equation*} \lVert f Vert_r^{1 - \lambda s/p} \lVert f Vert_s^{1 - (1-\lambda)r/p} \le \max(\lVert f Vert_r,\lVert f Vert_s). \end{equation*} The inclusion is then trivial. \end{proof} \begin{proof}[Proof of $(e)$] For every $p$, we have \begin{align*} \lVert f Vert_p &= \left(\int_X \lvert f vert^{p-r}\lvert f vert^r\,d\mu ight)^{1/p}\ &\le \lVert \lvert f vert^{p-r} Vert_\infty^{1/p} \lVert f Vert_r^{r/p}\ &\le \lVert f Vert_\infty^{1-r/p} \lVert f Vert_r^{r/p}. \end{align*} Passing to limits, $\limsup_{p o \infty} \lVert f Vert_p \le \lVert f Vert_\infty$. \par On the other hand, let $\beta = \lVert f Vert_\infty - \epsilon$. Then, \begin{align*} \lVert f Vert_r = \left( \int_X \lvert f vert^r\,d\mu ight)^{1/r} &\ge \left( \int_{\{\lvert f vert \ge \beta\}} \lvert f vert^r\,d\mu ight)^{1/r}\ &\ge \beta \mu(\lvert f vert \ge \beta)^{1/r} \end{align*} for all $r$, and passing to limits, we have \begin{equation*} \liminf_{r o \infty} \lVert f Vert_r \ge \beta = \lVert f Vert_\infty - \epsilon \end{equation*} for arbitrary $\epsilon$, hence $\liminf_{p o \infty} \lVert f Vert_p \ge \lVert f Vert_\infty$. \end{proof}

Exercise 3.4

Answers

Comments

Exercise 3.4

Answers

Comments

Add answer