A few more minor typos fixed

ca.tex

@@ -13536,7 +13536,7 @@ \subsection{Real and imaginary parts of holomorphic functions}
 As in \exerciseref{exercise:Liouvilleharmonic},
 the analogue of \myquote{bounded} for holomorphic functions is
 \myquote{nonnegative}
-for harmonic functions.  Afterall, if $f$ is
+for harmonic functions.  After all, if $f$ is
 a bounded holomorphic function, then $\log \sabs{f(z)+M}$ or $\Re f(z) + M$
 is nonnegative for large enough $M$.  Conversely if $\log \sabs{f(z)} \geq
 0$, then $\frac{1}{f(z)}$ is bounded, and if $\Re f(z) \geq 0$, then
@@ -14190,7 +14190,7 @@ \subsection{The Dirichlet problem in a disc and the Poisson kernel}
 Define $\Phi(t) = \int_0^t \varphi(s)\, ds$, show that $\Phi$
 is increasing, continuous, but not differentiable
 on a dense set in $[0,1]$.  Use it to construct a
-$\psi(t)$ that is $2\pi$-periodic, continuous and 
+$\psi(t)$ that is $2\pi$-periodic, continuous, and 
 not differentiable on a dense subset of $\R$.
 \item
 Find a continuous $u \colon \overline{\D} \to \R$
@@ -15633,8 +15633,8 @@ \subsection{Applications, Rad\'o's theorem}
 for simply connected domains provided we know the Riemann map.  However,
 the kernel is difficult to compute in general, and it requires a very nice
 boundary to be able to integrate.  The Perron method works much more
-generally provided you can construct enough subharmonic functions (which can
-afterall be pieced together unlike harmonic functions).
+generally provided you can construct enough subharmonic functions (which
+can, after all, be pieced together unlike harmonic functions).
 
 If a solution exists, it clearly must equal to the Perron solution.