Exercise 3.5

Question

Assume, in addition to the hypotheses of Exercise 3.4, that
    $$ \mu(X) = 1. $$
    \begin{enumerate}[label=(\alph*)]
      \item Prove that $\lVert f \rVert_r \le \lVert f \rVert_s$ if $0 < r < s
        \le \infty$.
      \item Under what conditions does it happen that $0 < r < s \le \infty$ and
        $\lVert f \rVert_r = \lVert f \rVert_s < \infty$?
      \item Prove that $L^r(\mu) \supset L^s(\mu)$ if $0 < r < s$. Under what
        conditions do these two spaces contain the same functions?
      \item Assume that $\lVert f \rVert_r < \infty$ for some $r > 0$, and prove
        \begin{equation*}
          \lim_{p \to 0} \lVert f \rVert_p = \exp\!\left\{ \int_X \log
          \lvert f \rvert\,d\mu \right\}
        \end{equation*}
        if $\exp\{-\infty\}$ is defined to be $0$.
    \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$]
      By H"older's inequality using the conjugate exponents $s/r$, $s/(s-r)$, 
      \begin{equation*}
        \int_X \lvert f vert^r\,d\mu \le \left(\int_X \lvert f
        vert^s\,d\muight)^{r/s}\left(\int_X d\muight)^{(s-r)/s} = \lVert f
        Vert_s^r
      \end{equation*}
      and taking $r$th roots we get the claim.
    \end{proof}
    \begin{proof}[Proof of $(b)$]
      By the remark on p.\ 65, we see that equality holds if and only if $\alpha
      f^s = \beta$ almost everywhere, for some constants $\alpha,\beta$.
    \end{proof}
    \begin{proof}[Proof of $(c)$]
    The first claim is just part $(a)$. We now claim that $L^r(\mu) \subset
      L^s(\mu)$ if and only if there exists $a$ such that $\mu(E) > a$ for any
      measurable set $E$ of positive measure.
      \par $\Leftarrow$. Let $f \in L^r(\mu)$, and let $E_n = \{x \colon \lvert
      f vert \ge n\}$. We claim there exists $N$ such that $E_{n} = 0$ for
      all $n \ge N$. For, if not, then $\lVert f vert_\infty = \infty$, hence
      $f 
otin L^\infty$, contradicting that $f \in L^r$ and part $(a)$.
      \par $\Rightarrow$. We show the contrapositive.
      Suppose there is a sequence of measurable sets
      $\{E_n\}$ such that $0 < \mu(E_n) < 3^{-n}$; we can assume without of
      generality that the $E_n$ are disjoint. Then, if $s < \infty$, define
      \begin{equation*}
        f = \begin{dcases*}
          \sum_{n=1}^\infty \frac{\chi_{E_n}}{\mu(E_n)^{1/s}} & if $s <
          \infty$,\
          \sum_{n=1}^\infty \frac{\chi_{E_n}}{\mu(E_n)^{1/2r}} & if $s =
          \infty$.
        \end{dcases*}
      \end{equation*}
      Then, $f \in L^r \setminus L^s$.
    \end{proof}
    \begin{proof}[Proof of $(d)$]
      We first have
      \begin{align*}
        \lVert f Vert_p &= \exp(\log(\lVert f Vert_p))\
        &= \exp\!\left\{ \frac{1}{p} \log \int_X \lvert f vert^p\,d\mu ight\}\
        &\ge \exp\!\left\{ \int_X \log \lvert f vert\,d\mu ight\}
      \end{align*}
      and letting $p 	o \infty$ gives us $\ge$ in the equation desired.
      Conversely, sine $x \ge \log x + 1$ on $[0,\infty)$, we have $(\lVert f
      Vert_p^p - 1)/p \ge \log \lVert f Vert_p$, and so
      \begin{equation*}
        \log \lVert f Vert_p \le \int_X \frac{\lvert f vert_p^p -
        1}{p}\,d\mu.
      \end{equation*}
      Now split up the integral and take the limit as $p 	o 0$:
      \begin{align*}
        &\lim_{p 	o 0}\int_{\lvert f vert \ge 1} \frac{\lvert f vert^p -
        1}{p}\,d\mu + \lim_{p 	o 0}\int_{\lvert f vert < 1} \frac{\lvert f
        vert^p - 1}{p}\,d\mu\
        &~= \int_{\lvert f vert \ge 1} \log \lvert f vert\,d\mu +
        \int_{\lvert f vert < 1} \log \lvert f vert\,d\mu = 
        \int_X \log \lvert f vert\,d\mu
      \end{align*}
      by dominating the left function by $(\lvert f vert^r - 1)/r$ and by
      using the monotone convergence theoreom for the right function. Combining
      the previous two equations, we get $\lVert f Vert_p \le \exp\left\{
      \int_X \log \lvert f vert\,d\mu ight\}$.
    \end{proof}

Exercise 3.5

Answers

Comments

Exercise 3.5

Answers

Comments

Add answer