\documentclass[a4paper]{article}
\title{Mods Analysis II --- Continuity \& Differentiability}
\author{v 0.1, \textsc{spw}}
\date{}
\usepackage{spwnotes}
\begin{document}
\begin{notes}
\modsnotesh
\part{Definitions \& Remarks}
+--- 95 lines: \section{Function limits}---------------------------------------------------------------------------------------
+--- 81 lines: \section{Continuity}--------------------------------------------------------------------------------------------
+--- 14 lines: \section{Uniform convergence}-----------------------------------------------------------------------------------
+--- 43 lines: \section{Differentiation}---------------------------------------------------------------------------------------
+--- 15 lines: \section{The Mean Value Theorem}--------------------------------------------------------------------------------
+--- 29 lines: \section{Taylor's Theorem}--------------------------------------------------------------------------------------
+--- 10 lines: \section{The Binomial Theorem}----------------------------------------------------------------------------------
\part{Theorems \& Propositions}
\setcounter{section}{0}
\section{Function limits}
\begin{propn}
A bounded sequence $(z_n)$ in $\mathbb{R}$ (or $\mathbb{C}$) converges to a
limit $L$ if and only if all convergent subsequences of $(z_n)$ also
converge to $L$.
\begin{reductio}
\proofLtR
See Proposition 96, Analysis I notes.
\proofRtL
Suppose that $(z_n)$ is divergent. By B-W, we have some $(z_{n_k}) \to L_1$, and then $\mathbb{N}\setminus\set{n_k}{k \in \mathbb{N}}$ is not finite because otherwise we get a common tail. So form another sequence with the remaining terms, $(z_{m_k})$, say. It takes little to show that $(z_{m_k}) \to L_1 \implies (z_n) \to L_1$.
So $\forall j \in \mathbb{N} \quad \exists r_j > j$ s.t. $|z_{m_{r_j}} - L_1| \geq \eps_0$. But then by B-W we can find a convergent subsequence of this, and by letting its index tend to infinity, we get that its limit is not $L_1$.
\end{reductio}
\end{propn}
The following is vital for our use of integration over intervals later.
\boxthm{\begin{thm}
$p \isreal$ is a limit point of $[a,b]$ (or $(a,b]$, $[a,b)$, $(a,b)$) if and only if $p \in [a,b]$.
\begin{proof}
Argue by cases by trichotomy. Choose a $\delta \defeq \frac{a-p}{2}$ for the $p < a$ case to get a contradiction. Set $x \defeq p + \nicefrac{\delta}{2}$ for the other direction.
\end{proof}
\end{thm}}
\begin{thm}[Uniqueness of Limits]
Suppose $f\colon E \to \mathbb{R}$ \orcomp is such that $\displaystyle \lim_{x\to p} f(x) = L_1$ and $\displaystyle \lim_{x\to p} f(x) = L_2$. Then $L_1 = L_2$.
\begin{reductio}
Suppose that $L_1 \ne L_2$, and then certainly $\frac{1}{2}|L_1 - L_2| > 0$, so we can use this for $\eps$.
\end{reductio}
\end{thm}
\begin{propn} \label{propn:leftrightlim}
Let $f\colon (a,b) \to \mathbb{R}$ \orcomp and let $p \in (a,b)$. Then $f \to L$ as $x \to p$ if and only if both $\displaystyle \lim_{x \to p+}f(x) = L$ and $\displaystyle \lim_{x\to p-}f(x) = L$.
\begin{proof}
Just take care with ends of intervals and precise definitions (Dyson's notes appear to falsify this proposition).
\end{proof}
\end{propn}
The following allows us to use many of our results from Analysis I. This is also useful for showing that limits of functions don't exist, by finding two sequences that, when wrapped in $f$, go to different limits.
\boxthm{\begin{thm}\label{thm:seqfunc}
Let $f\colon E \to \mathbb{R}$ \orcomp where $E \subr$ (or $\mathbb{C}$). Let $p$ be a limit point of $E$ and let $L \in \mathbb{C}$. Then
\begin{multline*}
f(x) \to L \text{ as } x \to p \iff \\ \forall (p_n) \in E \text{ s.t. } p_n \ne p \text{ and } \lim_{n \to \infty} p_n = p \\ \text{ we have } f(p_n) \to L \text{ as } n \to \infty.
\end{multline*}
\begin{proof}
\proofLtR
Fairly straightforward -- use $\delta$ for $\eps$ in the convergence of some $(p_n)$ with the appropriate properties, but since $p_n \ne p$ we then have $0 < |p_n - p| < \delta$ and we can then easily get the convergence of $f(p_n)$.
\proofRtL
Assume otherwise, and choose $\delta \defeq \nicefrac{1}{n}$. Then we can get $|f(x_n) - L| \geq \eps_0$, but then we're done as we have found $(x_n)$ such that $f(x_n) \not \to L$ but $x_n \to p$ --- contradiction.
\end{proof}
\end{thm}}
\begin{thm}[The Algebra of Limits (AOL)]
Let $E \subseteq \mathbb{C}$ and let $p$ be a limit point of $E$. Let $f,g : E \to \mathbb{C}$ and let $\alpha,\beta \in \mathbb{C}$. Suppose we have that $f(x) \to A$ and $g(x) \to B$ as $x \to p$. Then the following limits exist, and are such that
\begin{description}
\item[Linear combination] $\displaystyle\lim_{x\to p}(\alpha f + \beta g)(x) = \alpha A + \beta B$;
\item[Product] $\displaystyle\lim_{x\to p}(f(x)g(x)) = AB$;
\item[Quotient] if $B \ne 0$ then $\exists \delta > 0$ s.t. $g(x) \ne 0 \; \forall x \in E$ which are such that $0 < |x - p| < \delta$ and $\displaystyle\lim_{x\to p}(f(x)/g(x)) = A/B$; and
\item[Weak inequality] if $f(x) \geq 0 \; \forall x \in E$ then $A \geq 0$.
\end{description}
\begin{proof}
Mimic the proofs for the AOL of sequences (see Analysis I notes,
section 4.2 and Theorem 77), or use Theorem \ref{thm:seqfunc}.
\end{proof}
\end{thm}
\section{Continuity}
\begin{propn}
$f$ is continuous at any isolated point of $E$.
\begin{proof}
Inequality vacuously true.
\end{proof}
\end{propn}
\boxthm{\begin{propn}\label{propn:limeqlim}
If $p \in E$ is a limit point of $E$, then $f$ is continuous at $p$ if and only if $\displaystyle\lim_{x\to p}f(x)$ exists and
\[ \lim_{x\to p}f(x) = f(p). \]
\begin{proof}
\proofLtR
Trivial.
\proofRtL
We need that the inequality holds for $x = p$; obvious.
\end{proof}
\end{propn}}
\begin{propn}
Let $f\colon (a,b) \to \mathbb{R}$ and let $p \in (a,b)$. Then $f$ is continuous at $p$ if and only if $f$ if $f$ is both left-continuous at $p$ and right-continuous at $p$.
\begin{proof}
Proceed as in the proof of Proposition \ref{propn:leftrightlim}, or alternatively just make use of Proposition \ref{propn:limeqlim}.
\end{proof}
\end{propn}
\begin{thm}
Let $f\colon E \to \mathbb{R}$ \orcomp where $E\subr$ \orcomp and let $p \in E$. Then
\begin{multline*}
f(x) \text{ is continuous at } p \iff \forall (p_n) \in E \text{ s.t. } \lim_{n \to \infty} p_n = p \\ \text{ we have } f(p_n) \to f(p) \text{ as } n \to \infty.
\end{multline*}
\begin{proof}
Follows directly from Theorem \ref{thm:seqfunc} (p. \pageref{thm:seqfunc}) and Proposition \ref{propn:limeqlim}.
\end{proof}
\end{thm}
\begin{thm}[Algebraic Closure of Continuous Functions]
Let $E\subr$ \orcomp and let $p \in E$. Let $f,g : E \to \mathbb{R}$ \orcomp and let $\alpha,\beta \in \mathbb{C}$. Then if $f$ and $g$ are continuous at $p$, so are
\begin{description}
\item[Linear combination] $(\alpha f + \beta g)(x)$;
\item[Product] (f(x)g(x)); and
\item[Quotient] (f(x)/g(x)) provided $g(p) \ne 0$ (which ensures that $\exists \delta > 0$ s.t. $\forall x \in \varset{x \in E}{|x - p| < \delta},\ f(x)/g(x)$ is defined).
\end{description}
\begin{proof}
Get directly from AOL for functions, or, again, mimic the proofs of AOL for sequences.
\end{proof}
\end{thm}
\begin{eg}
Let $f \in \mathbb{R}[x]$. Then $f$ is continuous at every point of $\mathbb{R}$. Further, if $g(x) = \frac{r(x)}{q(x)}$ where $r,q \in \mathbb{R}[x]$ and $q(p) \ne 0$, then $g$ is continuous at $p$.
\begin{proof}
It is clear that $\iota : x \mapsto x$ is continuous, so this follows immediately from the above Theorem.
\end{proof}
\end{eg}
\begin{thm}
Let $f\colon E \to \mathbb{C}$ and $g : f(E) \to \mathbb{C}$, and define $h : E \to \mathbb{C}$, for $x \in E$, by
\[ h(x) \defeq (g \circ f)(x) \equiv g(f(x)). \]
Then if $f$ is continuous at $p \in E$ and $g$ is continuous at $f(p)$, $h$ is continuous at $p$.
\begin{proof}
The continuity of $g$ at $p$ gets us
\[ |g(f(x)) - g(f(p))| < \eps \quad \forall x \in E \text{ s.t. } |f(x) - f(p)| < \delta_1, \]
but the continuity of $f$ gets us
\[ |f(x) - f(p)| < \delta_1 \quad \forall x \in E \text{ s.t. } |x - p| < \delta \]
which allows us to get the required restriction on $x$ in the first line.
\end{proof}
\end{thm}
\subsection{Continuity on sets}
\begin{thm}
\textnormal{(Uniform continuity on $[a,b]$)}
Let $f\colon [a,b] \to \mathbb{R}$ \orcomp be continuous. Then $f$ is uniformly continuous.
\begin{reductio}
Suppose otherwise and choose $x_n$ and $y_n$ such that we have
\[ |x_n - y_n| < \nicefrac{1}{n} \qquad \text{but} \qquad |f(x_n) - f(y_n)| \geq \eps_0. \]
But the $x_n$s are bounded so we can find $x_{n_k} \to p$, making $p$ a limit point, so $p \in [a,b]$. Then we can show that $y_{n_k} \to p$ too. But then we can get a contradiction out of the second part of our contrapositive above by using the fact that $f$ is continuous, obtaining $0 < \eps_0 \leq 0$.
\end{reductio}
\end{thm}
\subsection{Closed and bounded intervals}
\begin{thm}
\textnormal{(Continuous functions on $[a,b]$ are bounded)}
Let $f\colon [a,b] \to \mathbb{R}$ \orcomp be continuous. Then $f$ is bounded.
\begin{reductio}
Suppose $f$ is unbounded. Then we have some $x_n \in [a,b]$ s.t. $|f(x_n)| \geq n$. $(x_n)$ is bounded so we can get a convergent subsequence to a limit point. Then we have
\[ \left|f\left(x_{n_k}\right)\right| \geq n_k \geq n \]
and
\[ f(p) = \lim_{k\to\infty}f\left(x_{n_k}\right) \]
from the continuity of $f$ at $p$. But then $\left(f\left(x_{n_k}\right)\right)$ is convergent and so bounded, but its $k$-th term exceeds $k$.
\end{reductio}
\end{thm}
\begin{thm}
\textnormal{(Continuous functions on $[a,b]$ attain their bounds)}
Let $f\colon [a,b]$ be continuous. Then $f$ achieves its supremum and infimum.
\begin{proof}
Consider $g : [a,b] \to \mathbb{R}$ given by
\[ g(x) \defeq \frac{1}{M - f(x)} \]
where $\displaystyle M \defeq \sup_{x \in E}f(x)$. But then we can apply the previous theorem to $g$, and get a contradiction to $M$ being the \textit{least} upper bound. \hfill \contra
Similarly for the infimum, or apply what we just did to $-f$ since $\inf\set{t}{t \in E} = -\sup\set{-t}{t \in E}$.\footnote{There is a (really nice) constructive proof in the lecture notes: \textsc{J. Dyson}, \textit{Analysis II --- Continuity and Differentiability} (2010), pp. 15--16}
\end{proof}
\end{thm}
\subsubsection{A generalisation (off-syllabus)}
In the proofs in this section we have only needed that $[a,b]$ is bounded, closed (i.e. contains all its limit points) and that $f$ is continuous. So we may define:
\begin{defn}
A subset $A$ of $\mathbb{R}$ \orcomp is \textbf{compact} if it is bounded, and if it contains all its limit points.
\end{defn}
Then we have\smallskip
\boxthm{\begin{thm}
Let $f\colon E \to \mathbb{R}$ be a real valued function on a compact subset $E$ of $\mathbb{R}$ (or $\mathbb{C}$). Then $f$ is bounded, uniformly continuous and attains its bounds.
\end{thm}}
\subsection{The Intermediate Value Theorem}
\boxthm{\begin{thm}[IVT]
Let $f\colon [a,b] \to \mathbb{R}$ be continuous, and let $c$ be a number between $f(a)$ and $f(b)$ (inclusive (not ``strictly'')). Then $\exists \xi \in [a,b]$ such that $f(\xi) = c$.
\begin{proof}
Wlog, we may set $g(x) \defeq f(x) - c$ and assume $g(a) < 0 < g(b)$. Let $x_1 \defeq a$ and $y_1 \defeq b$. Perform interval bisection to maintain this inequality, and get a pair of sequences satisfying:
\begin{enumerate}
\item[(i)] either $g\left(\frac{1}{2}\left(x_{n-1} + y_{n-1}\right)\right) = 0$ and we may set \\ $\xi \defeq \frac{1}{2}(x_{n-1} + y_{n-1})$, or $g(x_n)g(y_n) < 0$;
\item[(ii)] $[x_n,y_n] \subseteq [x_{n-1},y_{n-1}]$ for any $n \geq 2$; and
\item[(iii)] $|y_n - x_n| = \frac{1}{2}|y_{n-1} - x_{n-1}| = \dotsb = \frac{1}{2^{n-1}}|y_1 - x_1| = \frac{b-a}{2^{n-1}}$.
\end{enumerate}
Using monotonicity and boundedness and (iii), we can get that $(x_n)$ and $(y_n)$ both converge to $\xi$. Then using (i), the continuity of $g$, AOL and the preservation of weak inequalities, we get that $0 \geq [g(\xi)]^2$ and the result follows.
\end{proof}
\end{thm}}
\begin{cor}
Let $([x_n,y_n])$ be a decreasing net of closed intervals in $\mathbb{R}$ such that the length $y_n - x_n \to 0$. Then \[ \bigcap^\infty_{n=1}[x_n,y_n] \] contains exactly one point.
\begin{proof}
Extract relevant lines from IVT proof; trivial to show that there is at most one such point.
\end{proof}
\end{cor}
\begin{thm}[Closed bounded intervals map onto closed bounded intervals] \label{thm:closedtoclosed}
Let $f\colon [a,b] \to \mathbb{R}$ be a real-valued continuous map. Then, for some $m,M \in \mathbb{R}$, $f([a,b]) = [m,M]$.
\begin{proof}
Use the infimum of $f$ for $m$, and the supremum for $M$. $f([a,b]) \subseteq [m,M]$ is clear. The attainment of bounds means that $\exists \xi,\eta \in [a,b]$ s.t. $f(\xi) = m$ and $f(\eta) = M$. So $m,M \in f([a,b])$.
To complete the proof show that an arbitrary $y \in [a,b]$ is s.t. $y \in f([a,b])$ by applying IVT to $[\xi,\eta] \subseteq [a,b]$ (or $[\eta,\xi]$).
\end{proof}
\end{thm}
\subsection{Monotonic functions}
Here we require order, so we must restrict ourselves to $\mathbb{R}$ rather
than allowing $\mathbb{C}$.
\subsubsection{The Inverse Function Theorem}
\begin{thm}[Inverse Function Theorem]
Let $f\colon [a,b] \to \mathbb{R}$ be strictly increasing and continuous. Then $f$ has a well-defined continuous inverse on $[f(a),f(b)]$.
\end{thm}
This is contained in the following theorem.
\boxthm{\begin{thm}
Let $f\colon [a,b] \to \mathbb{R}$ be strictly increasing and continuous. Then
\begin{enumerate}
\item[(i)] $f([a,b]) = [f(a),f(b)]$;
\item[(ii)] $\exists ! g : [f(a),f(b)] \to \mathbb{R}$ s.t. $g(f(x)) = x\;\forall x \in [a,b]$ and $f(g(y)) = y\;\forall y \in [f(a),f(b)]$;
\item[(iii)] $g$ is strictly monotone increasing; and
\item[(iv)] $g$ is continuous.
\end{enumerate}
\begin{proof}
(i): See Theorem \ref{thm:closedtoclosed}. (ii): Get that $f$ is bijective from (i). Since we have a \textit{unique} $x \in [a,b]$ for every $y \in [f(a),f(b)]$, we can argue that $g$ so defined must be unique. (iii): Simple reductio.
(iv): Draw a graph and figure out the appropriate $\delta$: \[ \delta \defeq \min(f(g(y_0) + \eps) - y_0, y_0 - f(g(y_0) - \eps)) \] for $y_0 \in (f(a),f(b))$, and then one of these for each of $y_0 \defeq f(a)$ and $y_0 \defeq f(b)$.
Use the fact that $f$ and $g$ are strictly increasing to manipulate the inequality that gives a restriction on $\delta$ into one that gives the required restriction on $\eps$, $|g(y) - g(y_0)| < \eps$.
\end{proof}
\end{thm}}
\begin{rmk}
We saw on problem sheet three, question one that it is sufficient to assume for IFT that $f\colon [a,b] \to \mathbb{R}$ is continuous and injective.
\end{rmk}
\begin{propn}
$\displaystyle \lim_{x\to\infty}(1+\nicefrac{1}{x})^x = \lim_{x\to-\infty}(1+\nicefrac{1}{x})^x = e$.
\begin{proof}
\end{proof}<++>
\end{propn}<++>
\section{Uniform convergence}
\boxthm{\begin{thm}[Test for Uniform Covergence]
Let $\emptyset \ne E \subr$ \orcomp and let $f_n,f\colon E \to \mathbb{R}$ (or $\mathbb{C}$). Then
\begin{multline*} f_n \stackrel{u}{\to} f \iff \\ \exists N \in \mathbb{R} \quad \forall n > N \quad m_n \defeq \sup_{x\in E} |f_n(x) - f(x)| \isreal \text{ exists} \\ \text{and moreover } m_n \to 0 \text{ as } n \to \infty. \end{multline*}
\begin{proof}
\proofLtR
Get $\nicefrac{\eps}{2}$ to serve as an upper bound for $\varset{|f_n(x) - f(x)|}{x\in E}$, then $m_n \leq \nicefrac{\eps}{2} < \eps$ and we can get what is required.
\proofRtL
By definition, \[ \forall x \in E,\ \displaystyle |f_n(x) - f(x)| \leq \sup_{x\in E}|f_n(x) - f(x)|. \qedhere \]
\end{proof}
\end{thm}}
\begin{thm}[Cauchy's Criterion for Uniform Convergence]
Let $E\subr$ \orcomp and let $f_n : E \to \mathbb{R}$ (or $\mathbb{C}$). Then $f_n$ converges uniformly on $E$ if and only if
\[ \forall \eps > 0 \quad \exists N_\eps \in \mathbb{N} \quad \forall n,m \geq N_\eps \quad \forall x \in E \quad |f_n(x) - f_m(x)| < \eps. \]
\begin{proof}
\proofLtR
$\nicefrac{\eps}{2}$ and $\Delta$.
\proofRtL
For any fixed $x$, $(f_n(x))$ is Cauchy and thus convergent to $f(x)$, say. Then we use the preservation of weak inequalities under limiting processes to obtain
\[ |f_n(x) - f(x)| = \lim_{m\to\infty}|f_n(x) - f_m(x)| \leq \nicefrac{\eps}{2} < \eps. \qedhere \]
\end{proof}
\end{thm}
\begin{cor}[Cauchy's Criterion for the Uniform Convergence of Series]
The series $\sum f_n$ is uniformly convergent on $E$ iff
\[ \forall \eps > 0 \quad \exists N_\eps \in \mathbb{N} \quad \forall n > m \geq N_\eps \quad \forall n \in \mathbb{N} \quad \left|\sum_{k=m+1}^nf_k(x)\right| < \eps. \]
\end{cor}
\boxthm{\begin{thm}[The Weierstrass $M$-Test]
Let $(f_n)$ be a sequence of (real or complex) functions defined on $E\subr$ \orcomp, and suppose that there is a sequence $(M_n)$ of real numbers such that
\[ |f_n(x)| \leq M_n \qquad \forall x \in E. \]
Suppose further that $\sum M_n$ converges. Then $\sum f_n$ converges uniformly on $E$.
\end{thm}
Note that the $M_n$ must be independent of $x$.
\begin{proof}
Note that
\[ \left|\sum_{k=m+1}^nf_k(x)\right| \leq \sum_{k=m+1}^n|f_k(x)| \leq \sum_{k=m+1}^nM_k = \left|\sum_{k=m+1}^nM_k\right| < \eps. \qedhere \]
\end{proof}}
\begin{cor}
Suppose all the conditions of the $M$-test hold, including that $\sum M_n$ is convergent. Then
\[ \left|\sum_{n=0}^\infty f_n(x)\right| \leq \sum_{n=0}^\infty|f_n(x)| \leq \sum_{n=0}^\infty M_n \qquad \forall x \in E. \]
\begin{proof}
Preservation of weak inequalities on obvious inequalities from the above proof.
\end{proof}
\end{cor}
Examples show that the limit of a sequence of continuous functions need not be continuous. Uniformity gives us the extra condition needed for this continuity.
\boxthm{\begin{thm}
Let $f_n,f\colon E \to \mathbb{R}$ \orcomp and suppose $f_n \stackrel{u}{\to} f$. Suppose all $f_n$ are continuous at some $x_0 \in E$. Then the limit function $f$ is also continuous at $x_0$, so we have that
\[ \lim_{x\to x_0}\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty}f_n(x_0) = \lim_{n\to\infty}\lim_{x\to x_0}f_n(x). \]
\begin{proof}
Use the uniform convergence of $f_n$ and the continuity of $f_{N_\eps+1}$ to get
\begin{multline*} |f(x) - f(x_0)| \leq |f(x) - f_{N_\eps+1}(x)| + |f(x_0) - f_{N_\eps+1}(x_0)| \\ + |f_{N_\eps+1}(x) - f_{N_\eps+1}(x_0)| < \nicefrac{\eps}{3} + \nicefrac{\eps}{3} + \nicefrac{\eps}{3} = \eps. \qedhere \end{multline*}
\end{proof}
\end{thm}
Note that $N_\eps + 1$ must be fixed, so that $\delta$ does not depend on $n$ (\textbf{why?}).}
We have a version for series:
\begin{cor}\label{cor:uniseriescts}
If $\displaystyle \sum_{n=0}^\infty f_n$ converges uniformly on $E$ and every $f_n$ is continuous at $x_0 \in E$, then we have
\[ \lim_{x\to x_0}\sum_{n=0}^\infty f_n(x) = \sum_{n=0}^\infty f_n(x_0). \]
In particular, if $f_n$ is continuous on $E$ for all $n$ and $\displaystyle \sum_{n=0}^\infty f_n$ converges uniformly on $E$, then $\displaystyle\sum_{n=0}^\infty f_n$ is continuous on $E$.
\end{cor}
\subsection{Power series}
\boxthm{\begin{thm}[Continuity of power series]
Suppose the power series $\sum a_nx^n$ has radius of convergence $R$, where $0 \leq R \leq \infty$. Then for every $r$ s.t. $0 \leq r \leq R$, $\sum a_nx^n$ converges uniformly on the closed disk $\varset{x}{|x| \leq r}$, and therefore $\sum a_nx^n$ is continuous on the open disk $\varset{x}{|x| < R}$.
\begin{proof}
We have absolute convergence from the meaning of `radius of convergence,' so we can use $M_n \defeq |a_n|r^n$ to get the first claim.
For the second, apply Corollary \ref{cor:uniseriescts}, since $a_nx^n$ is a polynomial so is continuous $\forall n \in \mathbb{N}$, and for any $x$. The Corollary gets us continuity on $|x| \leq r$, but since $r$ was fairly arbitrary, we can extend to $|x| < R$.
\end{proof}
\end{thm}}
\begin{cor}[Continuity of the elementary functions]
The functions $\exp$, $\sin$, $\cos$, $\sinh$ and $\cosh$ can all be defined by power series with infinite radii of convergence. So all these functions are continuous on $\mathbb{C}$.
\end{cor}
\section{Differentiation}
\begin{propn}
Let $f\colon (a,b) \to \mathbb{R}$ (or $\mathbb{C}$). Then $f$ is differentiable at $x_0$ and $f'(x_0) = L$ if and only if $f$ has both left- and right-derivatives at $x_0$, and $f_-'(x_0) = L = f_+'(x_0)$.
\begin{proof}
Proceed as in Proposition \ref{propn:leftrightlim}.
\end{proof}
\end{propn}
\begin{propn}
Suppose that $f\colon (a,b) \to \mathbb{R}$ is differentiable at $x_0 \in (a,b)$ and that $f'(x_0) > 0$. Then there exists a positive $\delta$ such that
\[ f(x) > f(x_0) \qquad \forall x \in (x_0,x_0 + \delta), \]
and
\[ f(x) < f(x_0) \qquad \forall x \in (x_0 - \delta,x_0). \]
\begin{proof}
Choose $\eps = f'(x_0) > 0$.
\end{proof}
\end{propn}
\begin{rmk}
There are various corollaries of the above, including some concerning left and right derivatives.
\end{rmk}
\begin{thm}[Differentiability $\implies$ Continuity]
Let $f\colon (a,b) \to \mathbb{R}$ \orcomp be differentiable at $x_0 \in (a,b)$. Then $f$ is continuous at $x_0$.
\begin{proof}
\[ \lim_{x\to x_0}(f(x) - f(x_0)) = \lim_{x \to x_0}\frac{f(x) - f(x_0)}{x - x_0} \cdot \lim_{x\to x_0} (x - x_0) = 0. \qedhere \]
\end{proof}
\end{thm}
\begin{thm}[Algebraic properties of differentiation]
Let $f,g : (a,b) \to \mathbb{R}$ \orcomp both be differentiable at $x_0 \in (a,b)$, and let $\lambda, \mu \in \mathbb{R}$ (or $\mathbb{C}$). Then we have
\begin{enumerate}
\item[(i)] \textnormal{(Linearity of differentiation)} $\lambda f + \mu g$ is differentiable at $x_0$ and $(\lambda f + \mu g)'(x_0) = \lambda f'(x_0) + \mu g'(x_0)$;
\item[(ii)] \textnormal{(The Product Rule)} $fg$ is differentiable at $x_0$ and $(fg)'(x_0) = f(x_0)g'(x_0) + f'(x_0)g(x_0)$; and
\item[(iii)] \textnormal{(The Quotient Rule)} provided that $g(x_0) \ne 0$, $\nicefrac{f}{g}$ is differentiable at $x_0$ and
\[ \left(\frac{f}{g}\right)'(x_0) = \frac{f'(x_0)g(x_0) - f(x_0)g'(x_0)}{g^2(x_0)}. \]
\end{enumerate}
\begin{proof}
(i): Follows immediately from AOL. (ii): Let stuff tend in
\begin{multline*} \frac{f(x)g(x) - f(x_0)g(x_0)}{x - x_0} = f(x)\frac{g(x) - g(x_0)}{x - x_0} \\ + g(x_0)\frac{f(x) - f(x_0)}{x - x_0} \end{multline*}
where $f$'s continuity allows $f(x) \to f(x_0)$.
(iii): We must first confirm that $\nicefrac{f}{g}$ is well-defined on some subinterval of $(a,b)$ containing $x_0$; we can do this from the continuity of $g$ and the assumption that $g(x_0) \ne 0$. Then let stuff tend in
\begin{multline*}
\frac{\frac{f(x)}{g(x)} - \frac{f(x_0)}{g(x_0)}}{x - x_0} = \frac{1}{g(x)g(x_0)}\Biggl[g(x)\frac{f(x) - f(x_0)}{x - x_0} \\ - f(x)\frac{g(x) - g(x_0)}{x - x_0}\Biggr]. \qedhere
\end{multline*}
\end{proof}
\end{thm}
\begin{thm}[The Chain Rule]
Suppose $f\colon (a,b) \to \mathbb{R}$, and that $g\colon (c,d) \to \mathbb{R}$. Suppose that $f((a,b)) \subseteq (c,d)$, so that $g \circ f\colon (a,b) \to \mathbb{R}$ is defined.
Suppose further that $f$ is differentiable at $x_0 \in (a,b)$ and that $g$ is differentiable at $f(x_0)$. Then $g \circ f$ is differentiable at $x_0$ and \[ (g \circ f)'(x_0) = g'\bigl(f(x_0)\bigr)f'(x_0). \]
\begin{proof}
Let $y_0 \defeq f(x_0)$ and define
\[ h(y) \defeq \left\{ \begin{array}{lr}
\frac{g(y) - g(y_0)}{y - y_0} - g'(y_0) & \qquad \forall y \ne y_0; \\
0 & \qquad \text{if } y = y_0.
\end{array} \right.
\]
Note that $h(y) \to 0$ as $y \to y_0$, making $h$ continuous at $y_0$.
Turn this into an increment, substitute in $f(x)$, and divide through by $x - x_0$ to get something useful. Then we have the clever(ish) bit: $h$ is continuous at $y_0$, and $f$ is differentiable and thus continuous at $y_0$, so $h(f(x))$ is continuous at $x_0$ and thus $h(f(x)) \to 0$ as $x \to x_0$. Let stuff tend to obtain the result.
\end{proof}
\end{thm}
\begin{thm}[The Leibniz Formula]
Let $f,g\colon (a,b) \to \mathbb{R}$ \orcomp be $n$-times differentiable on $(a,b)$. Then $fg$ is $n$-times differentiable and we have
\[ (fg)^{(n)}(x) = \sum_{j = 0}^n {n \choose j} f^{(j)}(x)g^{(n - j)}(x). \]
\begin{proof}
Induction, making use of
\[ {n + 1 \choose j} = {n \choose j} + {n \choose j - 1}. \qedhere \]
\end{proof}
\end{thm}
\begin{lem}
The power series $\sum a_nx^n$ and $\sum(n + 1)a_{n + 1}x^n$ have the same radius of convergence.
\begin{proof}
Let the radii be $R$ and $R'$; we will show $R \geq R'$ and $R' \geq R$.
\end{proof}<++>
\end{lem}<++>
\begin{thm}[Term-by-term differentiation of power series]
The power series $f(x) \defeq a_nx^n$ and $g(x) \defeq \sum(n+1)a_nx^n$ have the same radius of convergence $R$, and $\forall x$ such that $|x| < R$ we have that $f$ is differentiable at $x$ and moreover $f'(x) = g(x)$.
\begin{proof}
Not examinable.
\end{proof}
\end{thm}
\begin{rmk}
So power series have derivatives of all orders within their radii of convergence. This means that all the elementary functions defined by power series are differentiable everywhere.
\end{rmk}
\begin{thm}[Extension to IFT]
Let $f \colon [a,b] \to [m,M]$ be strictly increasing and continuous, with inverse function $g \colon [m,M] \to [a,b]$. Suppose that $f$ is differentiable at $x_0 \in (a,b)$, and that $f'(x_0) \ne 0$. Then $g$ is differentiable at $f(x_0)$ and
\[ g'(f(x_0)) = \frac{1}{f'(x_0)}. \]
\begin{proof}
Let $y_0 = f(x_0)$. Then note that $x = g(y) \to g(y_0) = x_0$ as $y \to y_0$ by $g$'s continuity, and then use the fact that
\[ \lim_{y\to y_0}\frac{g(y) - g(y_0)}{y - y_0} = \lim_{y\to y_0}\frac{x - x_0}{f(x) - f(x_0)} = \lim_{y\to y_0}\frac{1}{\frac{f(x) - f(x_0)}{x - x_0}}. \qedhere \]
\end{proof}
\end{thm}
\begin{eg}
$\ln'(x) = \nicefrac{1}{x}.$
\begin{solution}
For any $y > 0$, find $A$ (by Archimedean Property) such that $\exp(-A) < y < \exp(A)$. Then using the IFT on the differentiable $\exp \colon [-A, A] \to [\exp(-A), \exp(A)]$ we will find
\[ \ln'y = \frac{1}{\exp'(\ln y)} = \frac{1}{\exp(\ln y)} = \nicefrac{1}{y}. \qedhere \]
\end{solution}
\end{eg}
\section{The Mean Value Theorem}
\begin{note}
If asked to prove the IVT, proofs of the following three results must be given.
\end{note}
\begin{propn}[Fermat]
Let $f \colon (a,b) \to \mathbb{R}$. Suppose that $f$ is differentiable at $x_0 \in (a,b)$, and that $x_0$ is a local extremum. Then $f'(x_0) = 0$.
\begin{proof}
For a local minimum, get $f'_+(x_0) \geq 0$ and $f'_-(x_0) \leq 0$ by the preservation of weak inequalities, and then since $f$ is differentiable at $x_0$, $f'(x_0) = f'_-(x_0) = f'_+(x_0) = 0$. Similarly for $x_0$ a local minimum.
\end{proof}
\end{propn}
\begin{rmk}
The interval must be open so that we may take limits from both sides.
\end{rmk}
\begin{thm}[Rolle, 1691]
Let $f \colon [a,b] \to \mathbb{R}$ be continuous, and suppose that $f$ is differentiable on $(a,b)$. Suppose further that $f(a) = f(b)$. Then $\exists \xi \in (a,b)$ s.t. $f'(\xi) = 0$.
\begin{proof}
$f$ attains its maximum and minimum on $[a,b]$, and either $f$ is constant (in which case $\xi \defeq \nicefrac{1}{2}(a + b)$ will do) or at least one of the maximum and minimum lie in $(a,b)$ --- apply Fermat to the one that does.
\end{proof}
\end{thm}
\boxthm{\begin{thm}[MVT]
Let $f \colon [a,b] \to \mathbb{R}$ be continuous, and suppose that $f$ is differentiable on $(a,b)$. Then $\exists \xi \in (a,b)$ s.t.
\[ f(b) - f(a) = f'(\xi)(b-a). \]
\begin{proof}
Check the conditions carefully, then apply Rolle to
\[ F(x) \defeq f(x) - k(x - a) \]
with
\[ k \defeq \frac{f(b) - f(a)}{b - a}. \qedhere \]
\end{proof}
\end{thm}}
\begin{cor}[Taylor Theorem, Mk. 1]
Let $f \colon [a,b] \to \mathbb{R}$ be continuous, and suppose that $f$ is differentiable on $(a,b)$, and let $x, x+h \in [a,b]$. Then
\[ f(x + h) - f(x) = f'(x + \theta h)h\qquad\text{for some } \theta \in (0,1). \]
\begin{proof}
Argue by trichotomy on $h$. In each case, apply MVT to the closed interval bordered by $x$ and $x + h$, set $\xi = x + \theta h$ and look at the inequalities of where $\xi$ is to get the appropriate restriction on $\theta$.
\end{proof}
\end{cor}
\boxthm{\begin{cor}[Identity Theorem]
Let $f \colon (a,b) \to \mathbb{R}$ be differentiable, and be such that $f'(t) = 0\;\forall t \in (a,b)$. Then $f$ is constant on $(a,b)$.
\begin{proof}
Wlog, pick $x,y \in (a,b)$ s.t. $x < y$. Apply MVT to $[x,y]$.
\end{proof}
\end{cor}}
\begin{cor}
Let $f \colon (a,b) \to \mathbb{R}$ be such that $f'(x) \geq 0\;\forall x \in (a,b)$. Then $f$ is increasing on $(a,b)$.
\begin{proof}
Pick $x,y \in (a,b)$ s.t. $x < y$. Apply MVT to $[x,y]$, and use fact that $y - x$ is non-negative.
\end{proof}
\end{cor}
Similar corollaries exist for decreasing, strictly increasing and strictly decreasing.
\subsection{Elementary applications of MVT}
\begin{rmk}
To prove identities like $\exp(x + y) = \exp(x)\exp(y)$ and the trig. addition formulae, fix $c \defeq x + y$ and use the Identity Theorem.
\end{rmk}
\begin{propn}
The function $\exp : \mathbb{R} \to (0,\infty)$ is strictly increasing.
\begin{proof}
As $\exp x > 0$, its derivative is positive.
\end{proof}
\end{propn}
\begin{propn}
The function $\cos x$ has a least positive zero, $\alpha$.
\begin{proof}
Epicness.
\end{proof}
\end{propn}
\begin{propn}
$\sin \alpha = 1$.
\begin{reductio}
By Pythagoras, $\sin \alpha = \pm 1$. Suppose $\sin \alpha = -1$ and apply MVT on interval $(0,\alpha)$ to get $\cos \xi < 0$ --- but $\cos 0 = 1$ and $\alpha$ is the first root, so IVT tells us the sign of $\cos \xi$ cannot be.
\end{reductio}
\end{propn}
\begin{propn}[Periodicity]
$\forall x \in \mathbb{R}$ we have that
\begin{enumerate}
\item[(i)] $\cos(x + \alpha) = -\sin x$ and $\sin(x + \alpha) = \cos x$;
\item[(ii)] $\cos(x + 2\alpha) = -\cos x$ and $\sin(x + 2\alpha) = -\sin x$;
\item[(iii)] $\cos(x + 4\alpha) = \cos x$ and $\sin(x + 4\alpha) = \sin x$.
\end{enumerate}
\begin{proof}
Apply addition formulae repeatedly, using $\cos \alpha = 0$ and $\sin \alpha = 1$.
\end{proof}
\end{propn}
\begin{propn}
The zeroes of $\cos x$ are precisely the points $\set{(k + \nicefrac{1}{2})\pi}{k \in \mathbb{Z}}$.
\begin{proof}
For existence, by periodicity, $\cos\left(\frac{1}{2}\pi + k\pi\right) = (-1)^k\cos\frac{1}{2}\pi = 0$.
For uniqueness, suppose that $\beta$ is a root; then so is $\beta + k\pi$ by the above. So there must be a $k$ s.t. $\beta_0 \defeq \beta + k\pi \in (0,\pi)$ is a zero of $\cos x$. Show that $\beta_0 = \nicefrac{\pi}{2}$ by trichotomy, using periodicity again.
\end{proof}
\end{propn}
\section{L'H$\hat{\text{o}}$pital's Rule}
\boxthm{\begin{thm}[Cauchy Generalised Mean Value Theorem]
Let $f,g \colon [a,b] \to \mathbb{R}$ be continuous, and differentiable on $(a,b)$. Suppose that $g'(x) \ne 0\;\forall x \in (a,b)$. Then $\exists \xi \in (a,b)$ such that
\[ \frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}. \]
\begin{proof}
Check that this makes sense: Rolle gets a contradiction from $g(b) = g(a)$. Then apply Rolle to
\[ F(x) \defeq \left| \begin{array}{ccc} 1 & 1 & 1 \\ f(x) & f(a) & f(b) \\ g(x) & g(a) & g(b) \end{array} \right|. \qedhere \]
\end{proof}
\end{thm}}
\begin{lem}
Let $f,g \colon [a, a + \delta] \to \mathbb{R}$ be continuous (for some $\delta > 0$), and differentiable on $(a, a + \delta)$, and suppose $f(a) = g(a) = 0$. Suppose further than $\displaystyle L \defeq \lim_{x\to a+} \frac{f'(x)}{g'(x)}$ exists. Then
\[ \lim_{x\to a+}\frac{f(x)}{g(x)} = \lim_{x\to a+}\frac{f'(x)}{g'(x)}. \]
\begin{proof}
The limit exists, so $\exists \delta' < \delta$ s.t. $g'(x) \ne 0$ on $(a, a + \delta']$. Apply CMVT to $f,g$ on $[a,x]$ for every $x \in (a, a + \delta')$, giving an $\xi_x \in (a,x)$ for each. But as $x \to a+$, $\xi_x \to a+$, and we can get the result by considering the fact that
\[ \lim_{\xi_x\to a+}\frac{f'(\xi_x)}{g'(\xi_x)} = \lim_{x\to a+}\frac{f'(x)}{g'(x)}. \qedhere \]
\end{proof}
\end{lem}
\begin{cor}
Let $f,g \colon [a - \delta, a] \to \mathbb{R}$ be continuous (for some $\delta > 0$), and differentiable on $(a - \delta, a)$, and suppose $f(a) = g(a) = 0$. Suppose further than $\displaystyle L \defeq \lim_{x\to a-} \frac{f'(x)}{g'(x)}$ exists. Then
\[ \lim_{x\to a-}\frac{f(x)}{g(x)} = \lim_{x\to a-}\frac{f'(x)}{g'(x)}. \]
\end{cor}
\boxthm{\begin{thm}[L'HR --- $\frac{0}{0}$ case]
Let $f,g \colon [a - \delta, a + \delta]$ (for some $\delta > 0$) be continuous, and differentiable on the \textbf{deleted interval} $(a - \delta, a + \delta) \setminus \{a\}$, and suppose $f(a) = g(a) = 0$. Suppose further than $\displaystyle L \defeq \lim_{x\to a}\frac{f'(x)}{g'(x)}$ exists. Then
\[ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}. \]
\end{thm}}
We can use this rule to show things like $\displaystyle \lim_{x\to 0} (1 + x)^{\frac{1}{x}} = e$ by removing the $\exp$, using L'HR, and then wrapping up in the continuous $\exp$.
\begin{cor}[L'HR --- infinite limits]
Let $f,g \colon [a - \delta, a + \delta]$ (for some $\delta > 0$) be continuous, and differentiable on the deleted interval $(a - \delta, a + \delta) \setminus \{a\}$, and suppose $f(a) = g(a) = 0$. Suppose further than $\frac{f'(x)}{g'(x)} \to +\infty$ as $x \to a$. Then
\[ \lim_{x\to a}\frac{g(x)}{f(x)} = 0. \]
\begin{proof}
Swap $f$ and $g$ and apply L'HR.
\end{proof}
\end{cor}
\begin{cor}[L'HR at $\infty$]
Let $f,g \colon (a,+\infty) \to \mathbb{R}$ be continuous and differentiable, with $f(x),g(x) \to 0$ as $x \to \infty$. Suppose that $g'(x) \ne 0\;\forall x \in (a,+\infty)$. Suppose further that $\frac{f'(x)}{g'(x)} \to L$ as $x \to \infty$. Then \[ \lim_{x\to\infty}\frac{f(x)}{g(x)} = L. \]
\begin{proof}
Apply L'HR to $F(x) \defeq f(\nicefrac{1}{x})$ and $G(x) \defeq g(\nicefrac{1}{x})$, setting $F(0) = 0 = G(0)$ and supposing they are continuous on an interval about zero, and then checking the conditions carefully.
\end{proof}
\end{cor}
\begin{thm}[L'HR --- $\frac{\infty}{\infty}$ case]
Let $f,g \colon (a, a + \delta) \to \mathbb{R}$ be differentiable for some $\delta > 0$, and suppose that $g'(x) \ne 0$. Suppose further that $f(x), g(x) \to \infty$ as $x \to a+$ and that $\displaystyle \lim_{x\to a+}\frac{f'(x)}{g'(x)}$ exists. Then
\[ \lim_{x\to a+}\frac{f(x)}{g(x)} = \lim_{x\to a+}\frac{f'(x)}{g'(x)}. \]
\begin{proof}
Not examinable.
\end{proof}
\end{thm}
\begin{eg}
$\displaystyle\lim_{x\to+\infty}\frac{\ln x}{x^\mu} = 0\qquad \forall \mu > 0$.
\begin{solution}
Apply $\frac{\infty}{\infty}$ case of L'HR.
\end{solution}
\end{eg}
\begin{rmk}
Similarly, $\qquad\displaystyle \lim_{x\to0+}x^\mu \ln x = 0\qquad\forall \mu > 0$.
\end{rmk}
These limits come up a lot and should be committed to memory.
\section{Taylor's Theorem}
\begin{thm}[Taylor's Theorem with Lagrange Remainder]\label{thm:lagrangetaylor}
Let $f \colon [a,b] \to \mathbb{R}$. Suppose that for some $n \geq 1$ we have that $f, f', f'', \dotsc, f^{(n-1)}$ exist and are continuous on $[a,b]$ and that $f^{(n)}$ exists on $(a,b)$. Then $\exists \xi \in (a,b)$ such that
\[ f(b) = \sum^{n - 1}_{k = 0}\frac{f^{(k)}(a)}{k!}(b - a)^k + \underbrace{\frac{f^{(n)}(\xi)}{n!}(b - a)^n}_{\text{Remainder; Lagrange's form}}. \]
\end{thm}
\begin{note}
Recall that at the end points of the interval, $a$ and $b$, `differentiable' means `left-' or `right-differentiable.'
\end{note}
\begin{proof}
Checking everything very carefully, apply CMVT to
\begin{align*}
F(x) &\defeq \sum^{n - 1}_{k = 0} \frac{f^{(k)}(x)}{k!}(b - x)^k \\
\intertext{and}
G(x) &\defeq (b - x)^n. \qedhere
\end{align*}
\end{proof}
\begin{rmk}
There are many other $G$s that satisfy the required hypotheses that we may use to get different forms of the remainder. One example is $G(x) \defeq x - a$. Another is $G(x) \defeq (x - a)^m$.
\end{rmk}
\boxthm{\begin{cor}[Classic Taylor's Theorem]
Let $f \colon (a - \delta, a + \delta) \to \mathbb{R}$ for some $\delta > 0$, and let $x \in (a - \delta, a + \delta)$. Suppose that for some $n \geq 1$ we have that $f',f'', \dotsc, f^{(n)}$ exist. Then $\exists \xi$ between $a$ and $x$ such that
\[ f(x) = \sum^{n - 1}_{k = 0}\frac{f^{(k)}(a)}{k!}(x - a)^k + \frac{f^{(n)}(\xi)}{n!}(x - a)^n. \]
\begin{proof}
If $x > a$, apply Theorem \ref{thm:lagrangetaylor}. If $x = a$, choose $\xi \defeq a$. If $x < a$, apply Theorem \ref{thm:lagrangetaylor} to $f(-x)$ and ``sort out the signs and inequalities.''
\end{proof}
\end{cor}}
\begin{cor}[The Error Estimate]
Let $f \colon [a,b] \to \mathbb{R}$ satisfy the hypotheses of Taylor's Theorem, and let $\displaystyle E_n \defeq \frac{|b - a|^n}{n!}\sup_{\xi\in(a,b)}f^{(n)}(\xi).$ Then
\[ \left| f(x) - \sum^{n - 1}_{k = 0}\frac{f^{(k)}(a)}{k!}(x - a)^k \right| \leq E_n \qquad \forall x \in [a,b]. \]
\end{cor}
\section{The Binomial Theorem}
\begin{lem}\label{lem:bin1}
\[ k{p \choose k} = p{p - 1 \choose k - 1}\qquad\forall k \geq 1. \]
\begin{proof}
Separate the case $k = 1$.
\end{proof}
\end{lem}
\begin{lem}\label{lem:bin2}
\[ {p \choose k} + {p \choose k - 1} = {p + 1 \choose k}\qquad\forall k \geq 1. \]
\begin{proof}
Separate the case $k = 1$.
\end{proof}
\end{lem}
\boxthm{\begin{thm}[The Binomial Expansion]\label{thm:realbin}
Let $p \in \mathbb{R}$, and suppose $|x| < 1$. Then
\[ (1 + x)^p = \sum_{k = 0}^\infty{p \choose k}x^k. \]
\end{thm}}
\begin{lem}\label{lem:bin3}
The function $f \colon (-1, 1) \to \mathbb{R}$ given by $f(x) \defeq (1 + x)^p$ is differentiable, is such that $f(0) = 1$ and satisfies
\[ (1 + x)f'(x) = pf(x). \]
\begin{proof}
Differentiable by Chain Rule.
\end{proof}
\end{lem}
\begin{lem}\label{lem:bin4}
The radius of convergence of
\[ \sum_p {p \choose k} x^k \]
is 1.
\begin{proof}
Ratio test.
\end{proof}
\end{lem}
\begin{lem}\label{lem:bin4}
The function $g \colon (-1, 1) \to \mathbb{R}$ given by $g(x) \defeq \sum_k {p \choose k} x^k$ is differentiable, is such that $g(0) = 1$ and satisfies
\[ (1 + x)g'(x) = pg(x). \]
\begin{proof}
Differentiable by chain rule; fiddle with sum indexes and use Lemmas \ref{lem:bin1}, \ref{lem:bin2} \& \ref{lem:bin3}.
\end{proof}
\end{lem}
\begin{improof}[Proof of Theorem \ref{thm:realbin}]
Noting that $f(x) > 0$, use the quotient rule on $\phi(x) \defeq \frac{g(x)}{f(x)}$. Then use the Identity Theorem together with Lemmas \ref{lem:bin3} \& \ref{lem:bin4}.
\end{improof}
\end{notes}
\end{document}