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\begin{flashcard}[holder-ineq-thm]
\begin{theorem*}[Holder Inequality]
  \label{holder_ineq}
  \cloze{
  Let $f, g$ measurable and $1 \le p \le g \le
  \infty$ be conjugate, i.e. $\frac{1}{p} +
  \frac{1}{q} = 1$, then
  \[ \mu(|fg|) \le \normp\|f\|_p \normp\|g\|_q .\]
  }
\end{theorem*}

\begin{proof}
  \cloze{
  If $p$ or $q = 1$ or $\infty$ then clear. So are
  the cases when $\normp\|f\|_p$ or
  $\normp\|g\|_q = 0$. Exclude these.

  Dividing both sides by $\normp\|f\|_p$, we may assume
  $\normp\|f\|_p = 1$, i.e. $\int |f|^p \dd \mu =
  1$. So, define a probability measure $\PP$ on
  $\mathcal{E}$ by
  \[ \PP(A) = \int_A |f|^p \dd \mu \]
  ($\PP$ has probability density $|f|^p$ with
  respect to $\mu$). Also, for $h \ge 0$
  measurable,
  \[ \label{lec15_l25_eq} \int h \dd\PP = \int
  h|f|^p \dd \mu \tag{$*$} \]
  Then,
  \begin{align*}
    \mu(|fg|)
    &= \mu(|fg| \indicator{\{|f| > 0\}}) \\
    &= \int \frac{|f|^p |g|}{|f|^{p - 1}}
    \indicator{\{|f| > 0\}} \dd \mu \\
    &= \int \frac{|g|}{|f|^{p - 1}}
    \indicator{\{|f| > 0\}} |f|^p \dd\mu \\
    &= \int \frac{|g|}{|f|^{p - 1}}
    \indicator{\{|f| > 0\}} \dd\PP \\
    &= \EE \left( \frac{|g|}{|f|^{p - 1}}
    \indicator{\{|f| > 0\}} \right) \\
    &\le \left( \EE \left( \frac{|g|^q}{|f|^{q(p -
    1)}} \indicator{|f| > 0} \right) \right)^{1/q}
    &&\text{(\nameref{jensens_ineq})} \\
    &= \left( \int \left( \frac{|g|^q}{|g|^p}
    \indicator{\{|f| > 0\}} \right) \dd\PP
    \right)^{1/q} \\
    &= \left( \int |g|^q \indicator{\{|f| > 0\}}
    \dd \mu \right)^{1/q}
    &&\text{(by \eqref{lec15_l25_eq})} \\
    &\le \left( \int |g|^q \dd \mu \right)^{1/q}
    \\
    &= \normp\|g\|_q &&\qedhere
  \end{align*}
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  $p = q = 2$ is the Cauchy-Schwarz inequality
  (see \es{3}).
\end{remark*}

\begin{flashcard}[minkowski-ineq-thm]
\begin{theorem*}[Minkowski inequality]
  \label{minkowski_ineq}
  \cloze{
  For $p \in [1, \infty]$ and $f, g$ measurable,
  $\normp\|f + g\|_p \le \normp\|f\|_p +
  \normp\|g\|_p$.
  }
\end{theorem*}

\begin{proof}
  \cloze{
  $p = 1, \infty$ obvious. Also obvious when
  $\normp\|f\|_p$ or $\normp\|g\|_p = 0$ or
  $\normp\|f + g\|_p$. Assume otherwise. Since
  \[ |f + g|^p \le 2^p (|f|^p + |g|^p) \]
  we have
  \[ \mu(|f + g|^p) \le 2^p (\mu(|f|^p) +
  \mu(|g|^p)) < \infty \]
  if $f, g \in \Lp$. So $f + g \in \Lp$. With $q$
  the conjugate of $p$,
  \begin{align*}
    \normp\|f + g\|_p^p
    &= \int |f + g|^p \dd \mu \\
    &= \int |f + g| |f + g|^{p - 1} \dd \mu \\
    &\le \int |f + g|^{p - 1} |f| \dd \mu + \int
    |f + g|^{p - 1} |g| \dd \mu \\
    &\le \normp\|f\|_p \normp\|(f + g)^{p -
    1}\|_q + \normp\|g\|_p \|(f + g)^{p - 1}\|_q
    &&\text{(\nameref{holder_ineq})} \\
    &= \left( \int |f + g|^{\ob{(p - 1)q}^{=p}} \dd \mu
    \right)^{1 / q} (\normp\|f\|_p +
    \normp\|g\|_p) \\
    &= \ub{\left( \int |f + g|^p \dd \mu \right)^{1 /
    q}}_{=\normp\|f + g\|_p^{p/q}} (\normp\|f\|_p + \normp\|g\|_p)
  \end{align*}
  Hence, dividing by $\normp\|f + g\|_p^{p/q}$, we
  get
  \[ \normp\|f + g\|_p \le \|f\|_p + \|g\|_p
  \qedhere \]
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[Lcurly-p-is-banach-space-thm]
\begin{theorem*}[$\Lpcal$ is a complete normed
  vector space (Banach space)]
  Let $p \in [1, \infty]$. Let $(f_n)_{n \in \NN}$
  be a sequence of functions in $\Lp$ such that
  \[ \forall \eps > 0 ~\exists N \in \NN ~\forall
  m, n \ge N \quad \normp\|f_m - f_n\|_p < \eps \]
  (i.e. $(f_n)$ is Cauchy in $\Lp$). Then
  $\exists$ $f \in \Lp$ such that $\normp\|f_n -
  f\|_p \to 0$ as $n \to \infty$.
\end{theorem*}

\begin{proof}
  \cloze{
  Assume $1 \le p < \infty$ ($p = \infty$ is an
  exercise).

  Choose a subsequence $(n_k)$ such that
  $\normp\|f_{n_{k + 1}} - f_{n_k}\|_p \le
  2^{-k}$. Then $S = \sum_{k = 1}^\infty
  \normp\|f_{n_{k + 1}} - f_{n_k}\|_p < \infty$.
  By \nameref{minkowski_ineq}, for any $K \in
  \NN$,
  \[ \left\| \sum_{k = 1}^K |f_{n_{k + 1}} -
  f_{n_k} \right\|_p \le \sum_{k = 1}^\infty
  \normp\|f_{n_{k + 1}} - f_{n_k}\|_p = S < \infty
  .\]
  So,
  \[ \int \left( \sum_{k = 1}^K |f_{n_{k + 1}} -
  f_{n_k}| \right)^p \dd \mu \le S^p < \infty .\]
  But
  \[ \left( \sum_{k = 1}^K |f_{n_{k + 1}} -
  f_{n_k} \right)^p \uparrow \left( \sum_{k =
  1}^K |f_{n_{k + 1}} - f_{n_k} \right)^p .\]
  So by \nameref{mconv_thm},
  \[ \left\|\sum_{k = 1}^\infty |f_{n_{k + 1}} -
  f_{n_k}| \right\|_p < \infty .\]
  So, in particular,
  \[ \sum_{k = 1}^\infty |f_{n_{k + 1}} - f_{n_k}|
  < \infty \text{$\mu$.\gls{al_ev}} .\]
  Let $A$ be the set where this is $< \infty$.
  Then $\mu(A^c) = 0$. For any $x \in A$,
  $(f_{n_k}(x))$ is Cauchy, and since $\RR$ is
  complete, it converges to $f(x)$ say. Define
  $f(x) = 0$ for all $x \in A^c$. Then $f$ is
  measurable and $f_{n_k} \to f$ as $k \to \infty$
  $\mu$ \gls{al_ev}. Then,
  \[ \normp\|f_n - f\|_p^p = \mu(|f_n - f|^p) =
  \mu(\liminf_k |f_n - f_{n_k}|_p^p)
  \stackrel{\text{\nameref{fatous_lemma}}}{\le}
  \liminf_k \mu(|f_n - f_{n_k}|^p) \le \eps^p \]
  and
  \[ \normp\|f\|_p
  \stackrel{\text{\nameref{minkowski_ineq}}}{\le}
  \normp\|f_N - f\|_p + \normp\|f_n\|_p \le \eps^p
  + \|f_N\|_p < \infty .\]
  Hence $f \in \Lp$ and $f_n \stackrel{\Lp}{\to}
  f$.
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  One can show that any choice of vector spaces
  \[ V = \mathcal{C}[0,1],
    \{\text{simple functions}\},
    \{\text{finite linear combination of
    indicators of intervals}\}
  \]
  are dense in $\Lp((0,1), \mathcal{B}, \lambda)$
  and so $(\mathcal{C}[0, 1],
  \normp\|\bullet\|_1)$ is $\Lp[1]$ space of
  Lebesgue integrable functions (exercise in
  \es{3}).
\end{remark*}
\vspace{-1em}

$\Lpcal[2]$ as Hilbert space: On a vector space
$V$, a symmetric bilinear form $V \times V \to
\RR$, $(u, v) \mapsto \langle u, v \rangle$ is
called an inner product if $\langle v, v \rangle
\ge 0 ~\forall v \in V$ and $\langle v, v \rangle
= 0$ if and only if $v = 0$.

Then $\sqrt{\langle v, v \rangle} = \|v\|$ is a
norm (Cauchy-Schwarz inequality gives the triangle
inequality for $\|\bullet\|$). If $(V,
\|\bullet\|)$ is complete, it is called a Hilbert
space.

\begin{corollary*}
  $\Lpcal[2]$ with the inner product $\langle f, g
  \rangle = \int fg \dd \mu$ is a Hilbert space.
\end{corollary*}

\subsubsection*{Basic Geometry}

\begin{enumerate}[(1)]
  \item Pythagoras theorem $\normp\|f + g\|_2^2 =
    \normp\|f\|_2^2 + \normp\|g\|_2^2 + 2 \langle
    f, g \rangle$.
  \item Parallelogram law: $\normp\|f + g\|_2^2 +
    \normp\|f - g\|_2^2 = 2(\normp\|f\|_2^2 +
    \normp\|g\|_2^2)$
\end{enumerate}
We say $f$ is \emph{orthogonal} to $f$ (written $f
\perp g$) if $\langle f, g \rangle = 0$. Then
$\normp\|f + g\|_2^2 = \normp\|f\|_2^2 +
\normp\|g\|_2^2$. For a subset $V \subseteq L^2$,
we define its orthogonal complement
\[ V^\perp = \{f \in \Lp[2] : \langle f, v \rangle
= 0 ~\forall v \in V\} \]
A subset $V$ is closed if $(f_n) \in V$ and $f_n
\stackrel{\Lp[2]}{\to} f$ implies $f \in V$.