Format grader comments

analysis--berkeley-202a-hw08.tex

@@ -1,16 +1,27 @@
 \section{Math 202A - HW8 - Dan Davison - \texttt{[email protected]}}
 
-% 7.17. You're on the right track but are missing a lot of details. You need to justify commuting the integral and the limit (dominated convergence??). I don't know where you were going with your m's and delta's either. (-5)
-% Aidan Backus, Nov 3 at 3:54pm
-% 7.9. This problem doesn't count towards your grade but it's good that you're thinking of a Fatou lemma for limsups. In fact, if f_n is *nonpositive* on a measurable set A, then \limsup_n \int_A f \leq \int_A \limsup_n f_n; you can prove this by just multiplying everything in Fatou's lemma by -1's and noting that when you commute a -1 with a liminf it turns into a limsup.
-% Aidan Backus, Nov 3 at 3:56pm
-% 7.11. Your argument can be patched to work but it is false that F is an increasing function; what if f = sin? (-3)
-% Aidan Backus, Nov 4 at 8:37am
-% 4. Your g(x) is not integrable, so you cannot apply DCT. The dominating function you should try would bound (1+x/n)^-n by exp(-x/2), or something similar. This "slows down" the decay of g(x) enough to be dominating, but without introducing a constant that makes g non-integrable. (-7)
-% Ian Francis, Nov 24 at 1:01am
-% 7.11) 7 7.13) 3 7.17) 5 7.21) 10
-% Ian Francis, Nov 24 at 1:01am
 
+\begin{verbatim}
+7.17. You're on the right track but are missing a lot of details. You need to justify commuting the integral
+and the limit (dominated convergence??). I don't know where you were going with your m's and delta's either.
+(-5) Aidan Backus, Nov 3 at 3:54pm
+
+7.9. This problem doesn't count towards your grade but it's good that you're thinking of a Fatou lemma for
+limsups. In fact, if f_n is *nonpositive* on a measurable set A, then \limsup_n \int_A f \leq \int_A
+\limsup_n f_n; you can prove this by just multiplying everything in Fatou's lemma by -1's and noting that
+when you commute a -1 with a liminf it turns into a limsup. Aidan Backus, Nov 3 at 3:56pm
+
+7.11. Your argument can be patched to work but it is false that F is an increasing function; what if f = sin?
+(-3) Aidan Backus, Nov 4 at 8:37am
+
+4. Your g(x) is not integrable, so you cannot apply DCT. The dominating function you should try would bound
+(1+x/n)^-n by exp(-x/2), or something similar. This "slows down" the decay of g(x) enough to be dominating,
+but without introducing a constant that makes g non-integrable. (-7)
+
+Ian Francis, Nov 24 at 1:01am
+7.11) 7 7.13) 3 7.17) 5 7.21) 10
+Ian Francis, Nov 24 at 1:01am
+\end{verbatim}
 
 \begin{comment}
   \begin{mdframed}
@@ -71,13 +82,14 @@ \section{Math 202A - HW8 - Dan Davison - \texttt{[email protected]}}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw08-3203.png}
 \end{mdframed}
 
-% 7.9. This problem doesn't count towards your grade but it's good that you're thinking of a Fatou lemma for
-% limsups. In fact, if f_n is *nonpositive* on a measurable set A,
-% then $\limsup_n \int_A f \leq \int_A \limsup_n$ f_n; you can prove this by just multiplying everything in
-% Fatou's lemma by -1's and noting that when you commute a -1 with a liminf it turns into a limsup.
-
-% Aidan Backus , Nov 3 at 3:56pm
+\begin{verbatim}
+7.9. This problem doesn't count towards your grade but it's good that you're thinking of a Fatou lemma for
+limsups. In fact, if f_n is *nonpositive* on a measurable set A,
+then $\limsup_n \int_A f \leq \int_A \limsup_n$ f_n; you can prove this by just multiplying everything in
+Fatou's lemma by -1's and noting that when you commute a -1 with a liminf it turns into a limsup.
 
+Aidan Backus , Nov 3 at 3:56pm
+\end{verbatim}
 
 \begin{comment}
   \begin{remark*}
@@ -134,7 +146,10 @@ \section{Math 202A - HW8 - Dan Davison - \texttt{[email protected]}}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw08-d0c0.png}
 \end{mdframed}
 
-% 7.11. Your argument can be patched to work but it is false that F is an increasing function; what if f = sin? (-3)
+\begin{verbatim}
+7.11. Your argument can be patched to work but it is false that F is an increasing function; what if f = sin?
+(-3)
+\end{verbatim}
 
 \begin{comment}
   \begin{intuition*}
@@ -177,9 +192,11 @@ \section{Math 202A - HW8 - Dan Davison - \texttt{[email protected]}}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw08-9931.png}
 \end{mdframed}
 
-% 4. Your g(x) is not integrable, so you cannot apply DCT. The dominating function you should try would bound
-% (1+x/n)^-n by exp(-x/2), or something similar. This "slows down" the decay of g(x) enough to be dominating,
-% but without introducing a constant that makes g non-integrable. (-7)
+\begin{verbatim}
+4. Your g(x) is not integrable, so you cannot apply DCT. The dominating function you should try would bound
+(1+x/n)^-n by exp(-x/2), or something similar. This "slows down" the decay of g(x) enough to be dominating, but
+without introducing a constant that makes g non-integrable. (-7)
+\end{verbatim}
 
 \begin{proof}
   Let $f(n) = (1 + x/n)^{-n} \log(2 + \cos(x/n))$.
@@ -206,12 +223,13 @@ \section{Math 202A - HW8 - Dan Davison - \texttt{[email protected]}}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw08-6e60.png}
 \end{mdframed}
 
-% 7.17. You're on the right track but are missing a lot of details. You need to justify commuting the integral
-% and the limit (dominated convergence??). I don't know where you were going with your m's and delta's either.
-% (-5)
-
-% Aidan Backus , Nov 3 at 3:54pm
+\begin{verbatim}
+7.17. You're on the right track but are missing a lot of details. You need to justify commuting the integral
+and the limit (dominated convergence??). I don't know where you were going with your m's and delta's either.
+(-5)
 
+Aidan Backus , Nov 3 at 3:54pm
+\end{verbatim}
 
 \begin{remark*}
   The integrand may be unbounded. For example, take $g(x) = |x|^{-1/2}$ and $f = 1$. Then the integrand is

analysis--berkeley-202a-hw10.tex

@@ -1,22 +1,32 @@
 \section{Math 202A - HW10 - Dan Davison - \texttt{[email protected]}}
 
-% 12.3. You have all the tools you need to prove the converse; try using g. (-4)
-% Aidan Backus, Nov 18 at 6:18pm
-% Review 1: You need to prove that your formula for \limsup_{n \to \infty} A_n is valid. (-1) It is not true that a summable sequence x_i has compact support; x_i = i^{-2} is a counterexample. This invalidates your bound on \mu(E) and hence your proof of the Borel-Cantelli lemma. (-3)
-% Aidan Backus, Nov 21 at 1:24pm
-% 12.3) 6 12.4) 0
-% Ian Francis, Nov 24 at 5:07am
-% Review 2: You are correct to guess that f bar, a continuous function with compact support, is not necessarily lipschitz. Try \sqrt{x} on [0,1]. It's continuous with compact support, but not Lipschitz. However, you don't need Lipschitz. Simply note that the result holds for f bar (using uniform convergence or DCT or whatever), and then use the triangle inequality and translation invariance of Lebesgue measure to get the result for f. (-6)
-% Ian Francis, Dec 2 at 8:02pm
+\begin{verbatim}
+12.3. You have all the tools you need to prove the converse; try using g. (-4)
+Aidan Backus, Nov 18 at 6:18pm
 
+Review 1: You need to prove that your formula for \limsup_{n \to \infty} A_n is valid. (-1) It is not true
+that a summable sequence x_i has compact support; x_i = i^{-2} is a counterexample. This invalidates your
+bound on \mu(E) and hence your proof of the Borel-Cantelli lemma. (-3) Aidan Backus, Nov 21 at 1:24pm
+
+12.3) 6 12.4) 0
+Ian Francis, Nov 24 at 5:07am
+
+Review 2: You are correct to guess that f bar, a continuous function with compact support, is not necessarily
+lipschitz. Try \sqrt{x} on [0,1]. It's continuous with compact support, but not Lipschitz. However, you don't
+need Lipschitz. Simply note that the result holds for f bar (using uniform convergence or DCT or whatever),
+and then use the triangle inequality and translation invariance of Lebesgue measure to get the result for f.
+(-6) Ian Francis, Dec 2 at 8:02pm
+\end{verbatim}
 
 \begin{mdframed}
   \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw10-7127.png}
 \end{mdframed}
 
-% Review 1: You need to prove that your formula for \limsup_{n \to \infty} A_n is valid. (-1) It is not true
-% that a summable sequence x_i has compact support; x_i = i^{-2} is a counterexample. This invalidates your
-% bound on \mu(E) and hence your proof of the Borel-Cantelli lemma. (-3)
+\begin{verbatim}
+Review 1: You need to prove that your formula for \limsup_{n \to \infty} A_n is valid. (-1) It is not true
+that a summable sequence x_i has compact support; x_i = i^{-2} is a counterexample. This invalidates your
+bound on \mu(E) and hence your proof of the Borel-Cantelli lemma. (-3)
+\end{verbatim}
 
 \begin{enumerate}[label=(\alph*)]
 \item ~\\
@@ -160,11 +170,12 @@ \section{Math 202A - HW10 - Dan Davison - \texttt{[email protected]}}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw10-5e9e.png}
 \end{mdframed}
 
-% Review 2: You are correct to guess that f bar, a continuous function with compact support, is not necessarily
-% lipschitz. Try \sqrt{x} on [0,1]. It's continuous with compact support, but not Lipschitz. However, you don't
-% need Lipschitz. Simply note that the result holds for f bar (using uniform convergence or DCT or whatever),
-% and then use the triangle inequality and translation invariance of Lebesgue measure to get the result for f.
-% (-6)
+\begin{verbatim}
+Review 2: You are correct to guess that f bar, a continuous function with compact support, is not necessarily
+lipschitz. Try \sqrt{x} on [0,1]. It's continuous with compact support, but not Lipschitz. However, you don't need
+Lipschitz. Simply note that the result holds for f bar (using uniform convergence or DCT or whatever), and then
+use the triangle inequality and translation invariance of Lebesgue measure to get the result for f. (-6)
+\end{verbatim}
 
 \begin{proof}
 
@@ -209,8 +220,9 @@ \section{Math 202A - HW10 - Dan Davison - \texttt{[email protected]}}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw10-551e.png}
 \end{mdframed}
 
-% 12.3. You have all the tools you need to prove the converse; try using g.
-
+\begin{verbatim}
+12.3. You have all the tools you need to prove the converse; try using g.
+\end{verbatim}
 
 \begin{proof}
   Let $X = P \cup N$ be a Hahn decomposition of $X$, such that $P$ is a $\mu$-positive set and $N$ is

analysis--berkeley-202a-hw11.tex

@@ -1,18 +1,23 @@
 \section{Math 202A - HW11 - Dan Davison - \texttt{[email protected]}}
 
-% Review 1. Dirichlet's test is not valid here, because assumption (2) fails if \kappa = 0. (-2) You do have a lower bound on the integral based on the harmonic series but you should be more specific how you got it. If you're going to argue by picture you should probably draw a picture. (-2)
-% Aidan Backus, Nov 21 at 1:29pm
-% 3. What happens if \mu(X_i) = 0? (-2)
-% Aidan Backus, Dec 1 at 5:43am
+\begin{verbatim}
+Review 1. Dirichlet's test is not valid here, because assumption (2) fails if \kappa = 0. (-2) You do have a
+lower bound on the integral based on the harmonic series but you should be more specific how you got it. If
+you're going to argue by picture you should probably draw a picture. (-2) Aidan Backus, Nov 21 at 1:29pm
 
+3. What happens if \mu(X_i) = 0? (-2)
+Aidan Backus, Dec 1 at 5:43am
+\end{verbatim}
 
 \begin{mdframed}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw11-8650.png}
 \end{mdframed}
 
-% 1. Dirichlet's test is not valid here, because assumption (2) fails if \kappa = 0. (-2) You do have a lower
-% bound on the integral based on the harmonic series but you should be more specific how you got it. If you're
-% going to argue by picture you should probably draw a picture. (-2)
+\begin{verbatim}
+1. Dirichlet's test is not valid here, because assumption (2) fails if \kappa = 0. (-2) You do have a lower
+bound on the integral based on the harmonic series but you should be more specific how you got it. If you're
+going to argue by picture you should probably draw a picture. (-2)
+\end{verbatim}
 
 \begin{lemma}[Dirichlet's test for improper integrals]
   Let $I = \int_a^\infty f(x) g(x) \dx$ where $\int$ denotes the Riemann integral. Then $I$ converges if all
@@ -144,7 +149,9 @@ \section{Math 202A - HW11 - Dan Davison - \texttt{[email protected]}}
 \includegraphics[width=400pt]{img/analysis--berkeley-202a-hw11-3704.png}
 \end{mdframed}
 
-% 3. What happens if \mu(X_i) = 0? (-2)
+\begin{verbatim}
+3. What happens if \mu(X_i) = 0? (-2)
+\end{verbatim}
 
 \begin{proof}
   Since $\mu$ is $\sigma$-finite we can write $X$ as a countable disjoint union $X = \bigcup_{i=1}^\infty X_i$