update algebra and ana
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\documentclass[uebung]{../../../lecture}
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\title{Analysis 3: Übungsblatt 5}
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\author{Leon Burgard, Christian Merten}
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\begin{document}
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\punkte
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\begin{aufgabe}
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Beh.: $f_k g_k \to fg$ in $L^{1}(X, \mu)$.
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\begin{proof}
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\begin{enumerate}[(i)]
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\item Es ist $f$ integrabel, also $\int_{X}^{} f_+ \d{\mu} < \infty$
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und $\int_{X}^{} f_- \d{\mu} < \infty$, also
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folgt $\text{ess sup } f_+ < \infty$ und $\text{ess sup } f_- < \infty$, also
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folgt mit $f = f_+ - f_-$, auch $\text{ess sup } f = M < \infty$
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für ein $M \in \R$.
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\item Es ist $\sup_{k \in \N} \Vert g_k \Vert_{L^{\infty}} \le S < \infty$
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für ein $S \in \R$,
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also auch $\text{ess sup } g < \infty$ (Ang.: es gäbe
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ein $A \in \mathcal{E}$ mit $\mu(A) > 0$ und
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$g(x) = \infty$ für $x \in A$, dann folgt
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$g_k(x) \not\to g(x)$, da $g_k(x) \le S < \infty$ $\contr$).
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Also OE $\text{ess sup } g \le S$.
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\item Es ist $g_k \xrightarrow{k \to \infty} g$ $\mu$ fast überall, d.h.
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\[
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\int_{X}^{} \lim_{k \to \infty} |g_k - g| \d{\mu} = 0
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.\]
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\item Betrachte $h \coloneqq 2 S |f|$. Dann gilt $\forall k \in \N$:
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\[
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|f| |g_k - g| \le |f| (|g_k| + |g|) \stackrel{\text{(ii)}}{\le} 2 S |f| = h
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.\] $h$ ist außerdem integrabel, da $f$ integrabel, also $f_+$ und $f_-$ integrabel,
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insbesondere $|f| = f_+ + f_-$ integrabel.
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Damit folgt mit dominierter Konvergenz
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\begin{salign*}
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\lim_{k \to \infty} \int_{X}^{} |f| |g_k - g| \d{\mu}
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&\stackrel{\text{3.19}}{=} \int_{X}^{} \lim_{k \to \infty} |f| |g_k - g| \d{\mu} \\
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&\stackrel{\text{(i)}}{\le } \int_{X}^{} M \lim_{k \to \infty} |g_k - g| \d{\mu} \\
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&\stackrel{\text{(iii)}}{=} 0
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.\end{salign*}
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\item Da $f_k \to f$ in $L^{1}$ folgt
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$\int_{X}^{} |f_k - f| \d{\mu} = \Vert f_k - f \Vert_{L^{1}} \xrightarrow{k \to \infty} 0$.
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Damit folgt
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\begin{salign*}
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\Vert f_k g_k - fg \Vert_{L^{1}} &= \int_{X}^{} |f_k g_k - fg| \d{\mu} \\
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&= \int_{X}^{} |f_k g_k - g_k f + g_k f - fg| \d{\mu} \\
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&\le \int_{X}^{} |g_k| |f_k - f| \d{\mu}
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+ \int_{X}^{} |f| |g_k - g| \d{\mu} \\
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&\stackrel{\text{(ii)}}{\le} M
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\underbrace{\int_{X}^{} |f_k - f | \d{\mu} }_{\xrightarrow{k \to \infty} 0}
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+ \underbrace{\int_{X}^{} |f| |g_k - g| \d{\mu} }_{\xrightarrow{k \to \infty} 0}\\
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&\xrightarrow{k \to \infty} 0
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.\end{salign*}
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Also $f_k g_k \to fg$ in $L^{1}$.
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\end{enumerate}
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\end{proof}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}[a)]
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\item Beh.: $L^{1}(X, \mu) \cap L^{\infty}(X, \mu) \subseteq L^{p}(X, \mu)$ für
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$1 < p < \infty$ und
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\[
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\Vert f \Vert_{L^{p}} \le \Vert f \Vert_{L^{1}}^{\frac{1}{p}}
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\Vert f \Vert_{L^{\infty}} ^{\frac{p-1}{p}} \qquad
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\text{für alle } f \in L^{1}(X, \mu) \cap L^{\infty}(X, \mu)
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.\]
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\begin{proof}
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Sei $1 < p < \infty$ und $f \in L^{1}(X, \mu) \cap L^{\infty}(X, \mu)$. Dann
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ist
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\begin{salign*}
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\int_{X}^{} |f|^{p} \d{\mu} &=
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\int_{X}^{} |f|^{1} |f|^{p-1} \d{\mu} \\
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&\stackrel{\text{Hölder}}{\le } \Vert f \Vert_{1} \Vert f^{p-1} \Vert_{\infty} \\
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&= \Vert f \Vert_1 \text{ess sup}_{X} |f^{p-1}| \\
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&\stackrel{p-1>0}{=} \Vert f \Vert_1 \left( \text{ess sup}_X |f| \right)^{p-1} \\
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&= \Vert f \Vert_1 \Vert f \Vert_{\infty}^{p-1} \\
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&\stackrel{p < \infty}{<} \infty
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\intertext{Da beide Seiten nicht-negativ, folgt durch Potenzieren mit $\frac{1}{p}$:}
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\Vert f \Vert_p &\le \Vert f \Vert_1^{\frac{1}{p}} \Vert f \Vert_{\infty}^{\frac{p-1}{p}} < \infty
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.\end{salign*}
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Das zeigt beide Behauptungen.
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\end{proof}
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\item Sei $\mu(X) < \infty$ und $1 \le p \le q \le \infty$. Beh.:
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$L^{q}(X, \mu) \subseteq L^{p}(X, \mu)$ und $\exists C \in \R$, s.d. $\forall f \in L^{q}(X, \mu)$
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\[
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\Vert f \Vert_p \le C \Vert f \Vert_q
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.\]
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\begin{proof}
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Sei $f \in L^{q}(X, \mu)$. Dann wähle $r \coloneqq \frac{q}{p} \ge 1$ und
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$r' = \frac{q}{q-p}$. Damit folgt
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\[
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\frac{1}{r} + \frac{1}{r'} = \frac{p}{q} + \frac{q-p}{q} = 1
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.\] Dann betrachte
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\begin{salign*}
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\int_{X}^{} |f|^{p} \d{\mu} &= \int_{X}^{} |f|^{p}\cdot 1 \d{\mu} \\
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&\stackrel{\text{Hölder}}{\le } \Vert 1 \Vert_{r'} \Vert f^{p} \Vert_r \\
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&= \Vert 1 \Vert_{r'} \left( \int_{X}^{} (|f|^{p})^{r} \d{\mu} \right)^{\frac{1}{r}} \\
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&\stackrel{r = q / p}{=}
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\Vert 1 \Vert_{r'} \left( \int_{X}^{} |f|^{q} \d{\mu} \right)^{\frac{p}{q}} \\
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\intertext{Da beide Seiten nichtnegativ, folgt durch Potenzieren mit $\frac{1}{p}$}
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\Vert f \Vert_p &\le \Vert 1 \Vert_{r'}^{\frac{1}{p}} \Vert f \Vert_q
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\intertext{Für die Konstante folgt}
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\Vert 1 \Vert_{r'}^{\frac{1}{p}}
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&= \left( \int_{X}^{} 1^{r'} \d{\mu} \right)^{\frac{1}{pr'}} \\
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&\stackrel{r' = q / (q-p)}{=} \left( \int_{X}^{} 1 \d{\mu} \right)^{\frac{q-p}{pq}} \\
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&= \mu(X)^{\frac{q-p}{pq}} \eqqcolon C < \infty
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\intertext{Damit folgt insgesamt}
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\Vert f \Vert_p &\le C \Vert f \Vert_q < \infty
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.\end{salign*}
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Also insbesondere $f \in L^{p}(X, \mu)$.
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\end{proof}
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Beh.: Es ist $L^{4}((0,1), \lambda) \subsetneqq L^{1}((0,1), \lambda)$.
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\begin{proof}
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Betrachte $f = x^{-\frac{1}{2}}$. Dann ist
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\begin{salign*}
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\int_{0}^{1} |f|^{4} \d{\lambda} &= \int_{0}^{1} x^{-2} \d{\lambda} =
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- x^{-1}\Big|_{0}^{1} = -1 + \lim_{k \to \infty} \frac{1}{\frac{1}{k}} = \infty \\
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\int_{0}^{1} |f| \d{\lambda} &= \int_{0}^{1} x^{-\frac{1}{2}} \d{\mu}
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= 2 (\sqrt{1} - \sqrt{0}) = 2 < \infty
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.\end{salign*}
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Also ist $f \in L^{1}((0, 1), \lambda)$, aber $f \not\in L^{4}((0,1), \lambda)$.
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\end{proof}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}[a)]
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\item Beh.: $f_k \to f$ in $L^{1}(X, \mu)$ $\iff$
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\[
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\int_{X}^{} |f_k| \d{\mu} \xrightarrow{ k \to \infty} \int_{X}^{} |f| \d{\mu}
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.\]
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\begin{proof}
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\begin{itemize}
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\item ,,$\implies$''. Sei $\Vert f_k - f \Vert_{L^{1}} \xrightarrow{k \to \infty} 0$. Dann
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folgt
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\begin{salign*}
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0 &\le \left| \int_{X}^{} |f_k| \d{\mu} - \int_{X}^{} |f| \d{\mu} \right| \\
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&= \left| \int_{X}^{} (|f_k| - |f|) \d{\mu} \right| \\
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&\stackrel{\triangle}{\le } \int_{X}^{} | |f_k| - |f| | \d{\mu} \\
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&\stackrel{\triangledown}{=} \int_{X}^{} |f_k - f| \d{\mu}
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\xrightarrow{k \to \infty} 0
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.\end{salign*}
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\item ,,$\impliedby$''. Sei $\lim_{k \to \infty} \int_{X}^{} |f_k| \d{\mu}
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= \int_{X}^{} |f| \d{\mu} $ $(*)$.
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Es ist zunächst $|f_k - f| = |f_k + (-f)| \le |f_k| + |f|$. Dann
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betrachte $g_k \coloneqq |f_k| + |f| - |f_k -f| \ge 0$. Da
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$f_k \xrightarrow{ k \to \infty} f$ $\mu$-f.ü., folgt
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$g_k \xrightarrow{k \to \infty} 2 |f|$ $\mu$ f.ü. Damit folgt
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mit Lemma von Fatou
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\begin{salign*}
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\int_{X}^{} 2 |f| \d{\mu} &= \int_{X}^{} \liminf_{k \to \infty} g_k \d{\mu} \\
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&\stackrel{\text{3.18}}{\le } \liminf_{k \to \infty} \int_{X}^{} g_k \d{\mu} \\
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&= \liminf_{k \to \infty} \left( \int_{X}^{} 2|f_k| \d{\mu}
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+ \int_{X}^{} 2 |f| \d{\mu} - \int_{X}^{} |f_k - f| \d{\mu} \right) \\
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&\stackrel{(*)}{=} \int_{X}^{} |f| \d{\mu} + \int_{X}^{} |f| \d{\mu}
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- \limsup_{k \to \infty} \int_{X}^{} |f_k -f| \d{\mu}
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\intertext{Umstellen liefert}
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0 &\le \limsup_{k \to \infty} \int_{X}^{} |f_k -f| \d{\mu} \le 0
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\intertext{Wegen $0 \le \liminf_{k \to \infty} \int_{X}^{} |f_k - f|\d{\mu}
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\le \limsup_{k \to \infty} \int_{X}^{} |f_k -f| \d{\mu} \le 0$ folgt}
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0 &= \lim_{k \to \infty} \int_{X}^{} |f_k -f| \d{\mu}
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.\end{salign*}
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Also $f_k \to f$ in $L^{1}(X, \mu)$.
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\end{itemize}
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\end{proof}
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\item Beh.: $f_k \to f$ in $L^{1}(X, \mu)$.
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\begin{proof}
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Definiere $g_k := g \coloneqq 1$. Dann ist $g = g_k$ integrabel, da
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\[
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\int_{X}^{} g \d{\mu} = \int_{X}^{} 1 \d{\mu} = \mu(X) < \infty
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.\] Außerdem sind $f_k, f$ integrierbar, insbesondere messbar. Außerdem
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gilt $f_k g_k = f_k \cdot 1 = f_k$ und $f g = f \cdot 1 = g$. Dann folgt
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die Aussage aus Aufgabe 1.
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\end{proof}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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Beh.: $f \coloneqq \prod_{j=1}^{n} f_j \in L^{p}(X, \mu) $ und
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\[
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\Vert f \Vert_{L^{P}(X, \mu)} \le \prod_{j=1}^{n} \Vert f_j \Vert_{L^{p_j}(X, \mu)}
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.\]
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\begin{proof}
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per Induktion über $n$. $n=1$: Dann gilt $p_1 = p$ und $f = f_1$, also trivial.
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Sei nun $n \in \N$ mit Aussage gezeigt für $n$. Sei $1 \le p \le \infty$ und seien $1 \le p_j < \infty$
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für $j \in \{ 1, \ldots, n+1\} $ und
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\[
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\sum_{j=1}^{n+1} \frac{1}{p_j} = \frac{1}{p}
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.\] Dann ist $p \le p_j$ für $j \in \{1, \ldots, n+1\} $.
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Definiere $q \coloneqq \frac{p p_{n+1}}{p_{n+1}-p} \ge 1$. Dann ist
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\begin{salign*}
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\frac{1}{q} = \frac{p_{n+1} - p}{p p_{n+1}} = \frac{1}{p} - \frac{1}{p_{n+1}} =
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\sum_{j=1}^{n} \frac{1}{p_j}
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.\end{salign*}
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Außerdem definiere $r \coloneqq \frac{p_{n+1}}{p} \ge 1$. Dann ist
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$r' = \frac{p_{n+1}}{p_{n+1}-p}$, denn
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$\frac{1}{r} + \frac{1}{r'} = \frac{p_{n+1} -p}{p_{n+1}} + \frac{p}{p_{n+1}} = 1$.
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Seien nun $f_j \in L^{p_j}(X, \mu)$ für $j \in \{1, \ldots, n+1\} $.
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Damit betrachte
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\begin{salign*}
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\int_{X}^{} |f|^{p} \d{\mu} &= \int_{X}^{} \left| \prod_{j=1}^{n+1} f_j \cdot f_{n+1}\right| \d{\mu} \\
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&\stackrel{\text{Hölder}}{\le } \left\Vert \left( \prod_{j=1}^{n} f_j \right)^{p} \right\Vert_{r'}
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\left\Vert f_{n+1}^{p} \right\Vert_r \\
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&= \left( \int_{X}^{} \left| \prod_{j=1}^{n} f_j \right|^{\frac{p p_{n+1}}{p_{n+1}-p}} \d{\mu} \right)^{\frac{p_{n+1} -p}{p_{n+1}}} \left( \int_{X}^{} |f_{n+1}|^{p_{n+1}} \d{\mu} \right)^{\frac{p}{p_{n+1}}} \\
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\intertext{Da beide Seiten nicht-negativ sind, folgt durch Potenzieren mit $\frac{1}{p}$}
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\Vert f \Vert_p &\le \left\Vert \prod_{j=1}^{n} f_j \right\Vert_q \Vert f_{n+1} \Vert_{p_{n+1}} \\
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&\stackrel{\text{IV}}{=} \prod_{j=1}^{n} \Vert f_j \Vert_{p_j} \Vert f_{n+1} \cdot \Vert_{p_{n+1}} \\
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&= \prod_{j=1}^{n+1} \Vert f_j \Vert_{p_j}
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.\end{salign*}
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\end{proof}
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\end{aufgabe}
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\end{document}
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