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Merge pull request #1 from gilesgardam/master

Fix some typos
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commit be3ef158a48c420b7f312492aa5e63f9718e5c1b 2 parents 2c66213 + 6ba9eaa
Andrew Tulloch authored
Showing with 7 additions and 7 deletions.
  1. +7 −7 PMH3 - Functional Analysis/PMH3 Lecture Notes.tex
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14 PMH3 - Functional Analysis/PMH3 Lecture Notes.tex
@@ -775,7 +775,7 @@ \section{Lecture 8 - Wednesday 23 March} % (fold)
\end{defn}
\begin{thm}[Projection]
- Let $(\Hil, \langle \cdot, \cdot, \rangle )$ be a Hilbert space. Let $M \subseteq \Hil$ be closed and convex. Let $x \in \Hil$. THen there exists a unique point $m_x \in M$ which is closest to $x$, i.e. \[
+ Let $(\Hil, \langle \cdot, \cdot, \rangle )$ be a Hilbert space. Let $M \subseteq \Hil$ be closed and convex. Let $x \in \Hil$. Then there exists a unique point $m_x \in M$ which is closest to $x$, i.e. \[
\| x - m_x \| = \inf_{m \in M} \| x - m \| = d
\]
\end{thm}
@@ -1341,7 +1341,7 @@ \section{Lecture 12 - Wednesday 6 April} % (fold)
Let \[
\mathcal A_\R = \{ f \in \mathcal A \given \text{$f$ is real valued} \}.
- \] THen $\mathcal A_\R$ is an $\R$-subalgebra of $\mathcal C_\R(X)$. It is unital, as $1 \in \mathcal A$ and it is real valued.
+ \] Then $\mathcal A_\R$ is an $\R$-subalgebra of $\mathcal C_\R(X)$. It is unital, as $1 \in \mathcal A$ and it is real valued.
We now show $\mathcal A_\R$ separates points. If $x \neq y$, there is $f \in \mathcal A$ such that $f(x) \neq f(y)$. Write $f = u + iv$ with $u,v$ real valued. Either $u(x) \neq u(y)$ or $v(x) \neq v(y)$, and so $\mathcal A_\R$ separates points.
@@ -2214,7 +2214,7 @@ \section{Lecture 20 - Wednesday 11 May} % (fold)
\begin{proof}
$(a) \Rightarrow (b)$ by definition.
- $(b) \Rightarrow (a)$. Suppose $(b)$ holds. Let $B_1 \subseteq X$ be bounded. THen $B_1 \subseteq \alpha B$ for some $\alpha > 0$. So \[
+ $(b) \Rightarrow (a)$. Suppose $(b)$ holds. Let $B_1 \subseteq X$ be bounded. Then $B_1 \subseteq \alpha B$ for some $\alpha > 0$. So \[
\overline{T(B)} \subseteq \overline{T(\alpha B)} = \alpha \overline{T(B)}
\] which is a closed subset of a compact set, and hence compact.
@@ -2275,14 +2275,14 @@ \section{Lecture 21 - Monday 16 May} % (fold)
Consider a sequence of compact operators $T_n$. If $T_n$ is compact and $T_n \rightarrow T$, then $T$ is compact.
\begin{lem}[Riesz's Lemma]
- Let $X$ be a normed vector space. Let $Y \subsetneq X$ be a proper \textbf{closed} subspace. Let $\theta \in (0, 1)$ be given. THen there exists $x$ with $\| x \| = 1$ such that $\| x - y \| \geq \theta$ for all $y \in Y$.
+ Let $X$ be a normed vector space. Let $Y \subsetneq X$ be a proper \textbf{closed} subspace. Let $\theta \in (0, 1)$ be given. Then there exists $x$ with $\| x \| = 1$ such that $\| x - y \| \geq \theta$ for all $y \in Y$.
\end{lem}
\begin{proof}
- Pick any $z \in X \backslash Y$. Let $\alpha = \inf_{y \in Y} \| z - y \| > 0$ since $Y$ is closed. Then by the definition of the infimum, there is $y_0 \in Y$ with $\alpha \leq \| z - y_0 \| \leq \frac{\alpha}{\theta}$. Now let $x = \frac{x - y_0}{\| z - y_0 \|}$. Then $\| x \| = 1$.
+ Pick any $z \in X \backslash Y$. Let $\alpha = \inf_{y \in Y} \| z - y \| > 0$ since $Y$ is closed. Then by the definition of the infimum, there is $y_0 \in Y$ with $\alpha \leq \| z - y_0 \| \leq \frac{\alpha}{\theta}$. Now let $x = \frac{z - y_0}{\| z - y_0 \|}$. Then $\| x \| = 1$.
Now, \begin{align*}
\| x - y \| &= \left \| \frac{z - y_0}{\| z - y_0 \|} - y \right \| \\
- &= \frac{1}{\| z - y_0 \|} \left \| z - y_- + \| z - y_0 \| y \right \| \\
+ &= \frac{1}{\| z - y_0 \|} \left \| z - y_0 + \| z - y_0 \| y \right \| \\
&\geq \frac{\theta}{\alpha} \alpha = \theta
\end{align*}
\end{proof}
@@ -2321,7 +2321,7 @@ \subsection{Limits of compact operators} % (fold)
We now show $T x'_n$ is Cauchy, and hence converges. We have \begin{align*}
\| Tx'_m - T x'_n \| \leq \| T x'_m - T_k x'_m \| + \| T_K x'_m - T_k x'_n \| + T_k x'_n - T x'_n \|
-\end{align*} where $k$ is to be chosen. Suppose $\| x_n \| \leq M$ for all $n \geq 1$. THen \begin{align*}
+\end{align*} where $k$ is to be chosen. Suppose $\| x_n \| \leq M$ for all $n \geq 1$. Then \begin{align*}
\| Tx'_m - T x'_n \| \leq 2 M \| T - T_k \| + \| T_k x'+m - T_k x'_n \|
\end{align*} Let $\epsilon > 0$ be given. Since $\| T - T_k \| \rightarrow 0$ as $ k \rightarrow \infty$, fix a $k$ for which $\| T - T_k \| \leq \frac{\epsilon}{3M}$. For this fixed $k$, we know $(T_k x'_n)$ converges, and so is Cauchy. So there exists $N < 0$ such that $\| T_k x'm - T_k x'_n\| < frac{\epsilon}{3}$ for all $m,n < N$. Hence $\|T x'_m - Tx'_n \| \leq \frac{2M}{\epsilon}{3M} + \frac{\epsilon}{3} = \epsilon$ for all $m , n > N$, so is Cauchy, and so converges.
\end{proof}
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