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fixed typo in orthogonal projection lemma

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Josia Pietsch 2023-07-20 14:02:11 +02:00
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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 and convex, For any $x \in H$ and closed, convex subspace $K \subseteq H$,
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 $\forall y \in K : \langle x-z, y\rangle_H = 0$,