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fixed lemma about orthogonal projection

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Josia Pietsch 2023-07-15 22:19:33 +02:00
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2
inputs/lecture_14.tex
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@ -101,7 +101,7 @@ and then do the harder proof.
i.e.~$H$ is a vector space with an inner product $\langle \cdot, \cdot \rangle_H$ which defines a norm by $\|x\|_H^2 = \langle x, x\rangle_H$ i.e.~$H$ is a vector space with an inner product $\langle \cdot, \cdot \rangle_H$ which defines a norm by $\|x\|_H^2 = \langle x, x\rangle_H$
making $H$ a complete metric space. making $H$ a complete metric space.
For any $x \in H$ and $K \subseteq H$ closed, For any $x \in H$ and $K \subseteq H$ closed and convex,
there exists a unique $z \in K$ such that the following equivalent conditions hold: there exists a unique $z \in K$ such that the following equivalent conditions hold:
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item $\forall y \in K : \langle x-z, y\rangle_H = 0$, \item $\forall y \in K : \langle x-z, y\rangle_H = 0$,

3
inputs/lecture_19.tex
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@ -148,6 +148,7 @@ However, some subsets can be easily described, e.g.
\end{proof} \end{proof}
\begin{theorem} \begin{theorem}
\label{thm:l1iffuip}
Assume that $X_n \in L^1$ for all $n$ and $X \in L^1$. Assume that $X_n \in L^1$ for all $n$ and $X \in L^1$.
Then the following are equivalent: Then the following are equivalent:
\begin{enumerate}[(1)] \begin{enumerate}[(1)]
@ -189,7 +190,7 @@ However, some subsets can be easily described, e.g.
(1) $\implies$ (2) (1) $\implies$ (2)
$X_n \xrightarrow{L^1} X \implies X_n \xrightarrow{\bP} X$ $X_n \xrightarrow{L^1} X \implies X_n \xrightarrow{\bP} X$
by Markov's inequality. by Markov's inequality (see \autoref{claim:convimpll1p}).
Fix $\epsilon > 0$. Fix $\epsilon > 0$.
We have We have

7
inputs/prerequisites.tex
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@ -108,7 +108,8 @@ from the lecture on stochastic.
Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$. Hence $\bE[|X_n - X|^q] \xrightarrow{n\to \infty} 0 \implies \bE[|X_n - X|^p] \xrightarrow{n\to \infty} 0$.
\end{subproof} \end{subproof}
\begin{claim} \begin{claim}
$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$ \label{claim:convimpll1p}
$X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$.
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
Suppose $\bE[|X_n - X|] \to 0$. Suppose $\bE[|X_n - X|] \to 0$.
@ -151,7 +152,9 @@ from the lecture on stochastic.
\end{subproof} \end{subproof}
\begin{claim} \begin{claim}
\label{claim:convimplpl1} \label{claim:convimplpl1}
$X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$ $X_n \xrightarrow{\bP} X \notimplies X_n\xrightarrow{L^1} X$.%
\footnote{Note that the implication holds under certain assumptions,
see \autoref{thm:l1iffuip}.}
\end{claim} \end{claim}
\begin{subproof} \begin{subproof}
Take $([0,1], \cB([0,1 ]), \lambda)$ Take $([0,1], \cB([0,1 ]), \lambda)$