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\begin{document}

\centerline{\bf Homework }
\centerline{\bf
UTK -- M447 -- Honors Advanced Calculus I -- Fall 2015}
\centerline{\bf
Jochen Denzler}



\begin{enumerate} \relax\setcounter{enumi}{0}\relax
\item
Prove that for every $a\in\R_+$, there exists a unique $x\in\R_+$ 
for which $x^2=a$. Specifically:\\
(a) Show uniqueness of $x$; also obtain the lemma ``If $x,y>0$
and $x^2=a$ and $y^2=b$, and $a<b$, then $x<y$'' in this proof. 
\\
(b) Letting $S:= \{ y\in\R \mid y>0 \mbox{ and } y^2\le a\}$, 
show that $S$ is non-empty and bounded above. 
\\
(c) Defining $x:=\sup S$, show that for every $\eps>0$,
$a-\eps<x^2<a+\eps$, then conclude $x^2=a$. 
\item
For $z=x+iy\in\C$ with $x,y\in\R$, define $\abs{z}:=\sqrt{x^2+y^2}$. 
Prove that $\abs{z+w}\le \abs{z}+\abs{w}$. Make sure that any two
inequalities you write down are properly connected with a logical
direction of implication (verbally or in symbols: $\implies$ or
$\Longleftarrow$ or $\iff$), and that the logical direction of
implication matches the needs for the proof. 

{\em You may only use ordered field axioms, simple conclusions from them
(like subtraction on both sides of inequalities, or `$a<b$ and $c<d$
implies $a+c<b+d$' or `$0<a<b$ and $0<c<d$ implies $0<ab<cd$' and
their analogs with $\le$), the supremum axiom and simple conclusions
thereof, 
and results from the previous problem about the
square root. }
\item
Prove: If $\lim a_n=a_*$ and $\lim b_n=b_*$ and $b_*\neq0$, then
$\lim(a_n/b_n)=a_*/b_*$. 
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\item
Prove (directly from the axioms and consequences proved in class) 
that $\lim\frac1n=0$. (That should be just a few lines.)
\item 
Assuming $(a_n)$ and $(b_n)$ are bounded sequences of real numbers: 
Is the following statement true or false? ``>$\limsup (a_n+b_n) =
\limsup a_n + \limsup b_n$?''  If true, provide a proof; if false,
provide a counterexample, and if possible, prove an amendment replacing `$=$'
with either `$\le$' or `$\ge$' that results in a true statement. If no
such amendment can be proved, provide counterexamples against the
amended versions. 
\item
Prove by induction the lemma: If $y>-1$ and $n\in\N$, then
$(1+y)^n\ge 1+ny$. (You'll need it later.)
\item
{\sl Note: }The power laws $(ab)^n=a^nb^n$, $(a/b)^n=a^n/b^n$ and
$a^{n+m}=a^na^m$ are easy consequences from the field axioms, and
induction. You may use them without providing their proof.  
\\
(a) 
Given $x\in\R$, consider the sequence $(a_n)$ given by $a_n :=
(1+\frac xn)^n$. We will later  write $a_n(x)$ for $a_n$, when the dependence
on $x$ plays a role. Show that $a_{n+1}/a_n \ge1$ whenever $x\ge 0$ or
$n>\abs{x}+1$.  (The previous lemma may help in estimating $[(1+\frac
x{n+1})/(1+\frac xn)]^n$. )
\\
(b)  
Show that the sequence $(a_n)$ is bounded above. \\
{\em Hint: } There may be a variety of ways; but you could use 
  that $a_n(x)a_n(-x) <1$ for $\abs{x}<1$ and a lower bound of
  $a_n(x)$ for $-1<x<0$ to get an upper bound for $0<x<1$.  And maybe
  $a_n(x+y)<a_n(x)a_n(y)$ for $x>0$. 
\\
(c) Defining $\exp(x):= \lim_{n\to\infty} a_n(x)$, prove that, for all
$x,y\in\R$, it holds $\exp(x)\exp(-x) = 1$ and $\exp(x+y) =
\exp(x)\,\exp(y)$. Also prove $x<y\implies \exp(x)<\exp(y)$.  
\item
Let $(a_n)$ be a sequence in $\R$ that is bounded below and
`sub-additive', i.e., $\forall n,m: a_{n+m}\le a_n+a_m$. 
Prove that $\lim\frac{a_n}{n}$ exists. {\em Hints follow: }\\
(a) Prove for $k,n\in\N$ that $a_{kn}\le k a_n$.\\
(b) Prove for $n,r\in\N$ that $\limsup_{k\to\infty}
\frac{a_{kn+r}}{kn+r} \le \frac{a_n}{n}$. 
\\
(c) You may assume the `division with remainder theorem from
elementary number theory': $\forall m,n\in\N\;\exists k,r\in\N_0:
m=kn+r \mbox{ and } 0\le r\le n-1$. Use it to prove $\limsup \frac{a_m}{m}
\le \inf \frac{a_n}{n}$  and conclude the original claim. 
%
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\item 
Using the dot product in $\R^n$, namely $\vec{u}\cdot\vec{v} :=
\sum_{i=1}^n u_i v_i$,  we can write $\|\vec{u}\|_2^2 =
\vec{u}\cdot\vec{u}$. \\
(a) Prove the Cauchy-Schwarz inequality $\abs{\vec{u}\cdot\vec{v}}
\le \|\vec{u}\|_2\,\|\vec{v}\|_2$ by exploiting the fact that
$(\vec{u}+t\vec{v})\cdot(\vec{u}+t\vec{v})\ge0$ for all $t\in\R$ and
then choosing a special $t$. (Actually the one that, according to
Calculus 1, minimizes the given expression. But note that, logically,
you do not need to justify your choice of $t$.)
\\
(b) Use it to conclude $\|\vec{u}+\vec{v}\|_2 \le \|\vec{u}\|_2 +
\|\vec{v}\|_2$.\\ 
{\sl [too elementary: I won't collect this one for grading.]}
\item 
Prove the triangle inequality for $\|\cdot\|_\infty$ on ${\cal
BF}(X\to\R) = \{f: X\to\R: f \mbox{ bounded}\}$. Here $\|f\|_{\infty}
:= \sup\{ \abs{f(x)} : x\in X\}$ and $X$ is any set. 
\item
I mentioned the norms $\|u\|_p:= \left( \sum_{i=1}^n \abs{u_i}^p
\right)^{1/p}$ without proof of the norm properties, where $p\ge1$. 
Now try $p=\frac12$. Is it a norm? Proof or counterexample. 
\item
Assume $d$ is a distance function on~$X$. Show that $d_1:=\frac{d}{1+d}$ is
also a distance function. 
Show that a sequence is convergent with respect to $d$ if and only if
it is convergent with respect to $d_1$. 
\item
On $\N$ a distance function $d_1$ is given by $d_1(n,m):= \abs{n-m}$. 
Another distance function is given by $d_2(n,m):= \abs{\frac1n -
  \frac1m}$. Which are the convergent sequences in each case? Which
are Cauchy sequences in each case? 
\item
The following example is instructive, albeit not significant for
purposes of Calculus: You may and should use basic facts about prime
factor decompositions of rational numbers in this example.
Given a rational $q=\frac mn\in\Q$ with $m \in\Z, n\in\N$, we consider
the prime factor decomposition $q= \pm p_1^{k_1}\cdots p_r^{k_r}$. 
Choosing a particular prime $p$, we define $[0]_p:=0$; and for
$q\neq0$, we define 
$[q]_p:= 1/p^k$, where $p^k$ is the power of $p$ that occurs in the
prime factor decomposition of~$q$. 
{\sl Example: $[\frac{12}{25}]_3 = \frac13$, $[\frac29]_3 = 9$,
  $[\frac25]_3 = [1]_3 = 1$.\/} 

Prove the strengthened triangle inequality $[x+y]_p \le \max\{[x]_p,
[y]_p\}$ and show that $d_p(x,y) := [x-y]_p$ defines a metric on~$\Q$.  
(It is called the $p$-adic metric). 

Let $a_n:= \frac{2^n}{2^n+1}$. Find the limit $\lim a_n$ in $\Q$
with the metric $d(x,y):= \abs{x-y}$ (that is too easy). Next find
$\lim a_n$ with respect to the metric $d_2$. 
%
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\item
  Find an example of a function $f:\R\to\R$ that does not have a fixed
  point, but satisfies a weakened contraction condition
  $\abs{f(x)-f(y)} < \abs{x-y}$ whenever $x\neq y$. For this example,
  you may use Calculus 1 knowledge (even though I think it can also be
  done with the material constructed in this class already.)
\item
  Prove that $f:[0,\infty`[\to\R\,,\; x\mapsto \sqrt{x}$ is
  uniformly continuous.  
  (Of course continuity is understood here with respect to
  the usual distance on~$\R$.) 
\item
  Let $(X,d)$ be the metric space of $\R$ with the {\em discrete\/}
  metric $d(x,y):= 1$ whenever $x\neq y$. Let $(Y,d)$ be {\em any\/}
  metric space, and $\R$ be equipped with the usual metric.

  (a) Show that every function $f:X\to Y$ is continuous. Is the same
  true with `uniformly continuous'? \\
  (b) Conjecture which functions $f:\R\to X$ are continuous (we'll
  prove it in class) .
\item
  Suppose $(X, d_X)$ and $(Y, d_Y)$ are metric spaces: the set
  $X\times Y$ can be made into a metric space by defining 
  $d_\infty( (x,y) , (x',y') ) := \max\{ d_X(x,x') , d_Y(y,y') \}$.\\
  (a)  Prove this. 
  \\
  (b) With this definition of the metric on $X\times X$, write out
  what the statement ``$d_X: X\times X \to \R$ is continuous'' means
  and then prove it. Decide if the same statement is true with
  `uniformly continuous' instead of merely `continuous'. 
\item
  Let $g:X\to Y$ and $f:Y\to Z$ be functions, where $X$, $Y$, $Z$ are
  metric spaces; let $f\circ g:X\to Z\,,\; x\mapsto f(g(x))$ be their
  composition. If $g$ is continuous at $x_0$ and $f$ is continuous at
  $y_0:= g(x_0)$, show that $f\circ g$ is continuous at $x_0$. 
\item 
  Assume $X$ and $Y$ are metric spaces, where the metric on $Y$ is
  bounded. Per \#12, boundedness of the metric is no loss of
  generality. We can define ${\cal F}(X\to Y)$ to be the set of all
  functions from $X$ to $Y$ and make it into a metric space by
  defining $d_\infty(f,g):= \sup\{ d(f(x),g(x)) : x\in X\}$. (You don't need
  to prove these simple facts; they are similar to \#10.)

  Show that convergence $f_n\to f$ in the sense of the metric $d_\infty$
  is equivalent to uniform convergence $f_n\to f$. 

  We use the metric $d_\infty$ on $C^0(X\to Y)$, the subset of
  {\em continuous\/} functions from $X$ to $Y$. Show this metric space
  is complete, if $Y$ is complete. 
%
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\item
  (a) Prove that $\exp:\R\to\R_+$ as defined previously is bijective. 
  \\
  (b) We'll denote its inverse function as $\ln$, so $y=\ln x :\iff
  x=\exp(y)$. Prove  that $\ln:\R_+\to\R$ is increasing and
  continuous. Also prove $\ln(xy)=\ln x + \ln y$.
  \\
  (c) For $x>0$ and $q\in\R$, we define $x^q:= \exp(q \ln x)$. Prove
  for all $x>0$ and $q_1,q_2\in \R$
  that $x^{q_1+q_2} = x^{q_1}x^{q_2}$ and $x^{q_1q_2} =
  (x^{q_1})^{q_2}$. 
  Also prove $(xy)^q=x^q y^q$ for all $x,y>0$ and $q\in\R$. Note that
  $(x,q)\mapsto x^q$, $\R_+\times\R\to\R_+$ is continuous. 
  \\
  (d) letting $e:= \exp(1)$, show that $\exp(x) = e^x$
\item
  Having constructed arbitrary powers, it's time to introduce the
  inequality of the arithmetic and geometric mean (short: agm
  inequality): ``If $n\in \N$ and $x_1,x_2,\ldots,x_n\in\R_+$, then 
  $(x_1x_2\cdots x_n)^{1/n} \le \frac1n(x_1+x_2+\ldots+x_n)$. The
  inequality is strict unless $x_1=\ldots=x_n$.''

  (a) Prove the inequality for $n=2^k$ by induction over $k$. \\
  (b) Then prove it for arbitrary $n$ by choosing some $k$ such that
  $m:=2^k>n$ and defining $x_{n+1},\ldots,x_m$ wisely.

  (I won't collect this for grading, but if you're not familiar with
  it, you should do it, as it is a `must know'.)
\item
  Young's inequality frequently comes in handy: ``If $x,y>0$ and
  $p,q>1$ such that $\frac1p+\frac1q=1$, then $xy\le \frac1p x^p +
  \frac1q y^q$.'' (This is often proved as a single variable calculus
  minimization problem, using the derivative.)
  
  (a) Show that for $p>1$, $L>0$, the function $f:\R_+\to\R_+\,,\; x \mapsto  
      \frac1p x^p + (1-\frac1p) (\frac Lx)^{p/(p-1)}$ has a minimum. 
     (Hint: Start with some random~$x_0$ (eg., $x_0=1$), then show
      there exists $[a,b]\subset\R_+$ such that for
      $x\in\R_+\setminus[a,b]$, it holds $f(x)> f(x_0)$. Use the
      sequential compactness of $[a,b]$. )

  (b) Assuming for the moment Calculus-1 knowledge about derivatives
      without having proved it,
      you can now find the location and value of the minimum.

  (c) Alternatively, prove Young's inequality at least for rational
      $p,q$ writing $p=\frac{m+n}{n}$ with $m,n\in\N$, as a
      consequence of the agm inequality. 
      Then generalize to $p>1$ real by using a continuity and limit
      argument. 
\item
  Show that a sequentially compact metric space is complete. Also show
  by counterexample that  the  converse implication is false. 
\item
  Suppose $(X,d)$ is a sequentially compact metric space and assume
  $f:X\to X$ is a weak contraction, i.e., $d(f(x),f(y)) < d(x,y)$
  unless $x=y$. Prove that $f$ has a unique fixed point. 
  Hint for the existence part: Study the function $r:X\to\R\,,\;
  x\mapsto d(f(x),x)$ and show that it has a minimum. Assuming the
  minimum value is positive, obtain a contradiction. 
\item 
  Show that every continuous function $f:[a,b]\to[a,b]$ has a fixed
  point (not necessarily unique). 
\item
  H\"older inequality in $\R^n$: Prove, for $x_i,y_i\in\R$ with
  $i=1,2,\ldots,n$ and for $p,q>1$ subject to $\frac1p+\frac1q=1$,
  that $\sum_i \abs{x_i y_i} \le \Bigl(\sum_i \abs{x_i}^p\Bigr)^{1/p}
  \, \Bigl(\sum_i \abs{y_i}^q\Bigr)^{1/q}$. Hint: First prove it under
  the extra hypothesis $\sum_i\abs{x_i}^p=1=\sum_i\abs{y_i}^q=1$,
  using Young's inequality; then conclude the general case by
  replacing $x_i$ with $\lambda x_i$ and $y_i$ with $\mu y_i$ for
  appropriate $\lambda,\mu$. 
\item 
  Let $p>1$. For $x\in\R^n$, 
  define $\|x\|_p := \bigl(\sum_{i=1}^n \abs{x_i}^p\bigr)^{1/p}$. 
  Show that $\|\cdot\|_p$ is a norm. Hint: to prove the triangle
  inequality, observe that $\abs{x_i+y_i}^p \le \abs{x_i+y_i}^{p-1}
  \abs{x_i} + \abs{x_i+y_i}^{p-1} \abs{y_i}$ and use H\"older's
  inequality. {\sf Note: The triangle inequality for this norm is also
    known under the name Minkowski's inequality; the correct
    pronounciation of the name is minn-COUGH-skee, even though
    anglicized pronounciations are frequently encountered.}
\item
  (a) Show that ${\cal T}:= \{ `]-\infty,a`[ : a\in\R \} \cup \{\emptyset, \R\}$ 
  is a topology on~$\R$. Use the separation property (namely that for
  any two distinct points $x,y$ in a metric space one can find disjoint
  $r$-balls $U(x,r)$ and $U(y,r)$) to show that this topology is not
  generated by a metric (i.e., is not the family of open sets defined
  in terms of any metric on~$\R$). 
  \\
  (b) analogously, ${\cal T}' := \{`]a,\infty`[ : a\in\R \} \cup
      \{\emptyset, \R\}$ is a topology. No need to redo part (a) for
      this one; just mentioning it. 
\item
With the topology from the previous problem chosen on~$\R$, prove that 
$$f:(\R,\abs{\cdot}) \to (\R,{\cal T})$$ 
has the property ``$V$ open
$\implies f^{-1}(V)$ open'' if and only if $\limsup _{x\to x_0} f(x) \le
f(x_0)$ for all $x_0$. 
{\sf --- Def: We call such functions upper semicontinuous.}

{\bf Clarification: } The notation $f:(\R,\abs{\cdot}) \to (\R,{\cal
  T})$ means that the domain of $f$ is the metric space $\R$, equipped
with the usual metric defined by the absolute value. In contrast, the
target set is the set of real numbers with the topology ${\cal T}$
(i.e., it is not seen as a metric space). By definition, a set $V\in
(\R,{\cal T})$ will be called open, iff $V\in {\cal T}$. As set $U\in
(\R,\abs{\cdot})$ is called open according to the definition of open
in metric spaces. 
\item 
In a metric space $(X,d)$, let $A$ be a closed set and define
$f(x):= \dist(A,x):= \inf\{d(x,z) : z\in A\}$. Show that $f: X\to\R$ is
continuous and that $f^{-1}(\{0\}) = A$. (In particular, any closed set
can occur as the inverse image of a singleton under a continuous function.)
\item
Give an example of a continuous $f:\R\to\R$ (with the usual topology
defined by the metric), and a connected $C\subset \R$,  such that 
$f^{-1}(C)$ is not connected. 
\item
(a) Give an example of a function $f:\R\to\R$ that is continuous and a
compact set $K$ such that $f^{-1}(K)$ is not compact. 
\\
(b) Give an example of a function $g:\R\to\R$ which maps compact sets
into compact sets, but which is not continuous. 
\item
Show that every compact metric space $K$ contains a dense sequence,
i.e., a sequence $(a_n)$ such that the closure of the set
$\{a_n:n\in\N\}$ is $K$. 
\item
In a metric space $X$, 
let $C_1 \supset C_2\supset C_3\supset \ldots$ be a descending sequence of
connected and non-empty sets (the notion `descending' is defined by
the given set inclusions). Let $C_*:= \bigcap_{n=1}^\infty
C_n$. \\ (a) Show by counterexamples, eg., in~$\R^2$, that $C_*$ need not be
non-empty, and also that $C_*$ need not be connected. 
\\ 
(b) Show: if $(C_n)$ is a descending sequence of  compact connected
non-empty sets, then $C_*:= \bigcap_{n=1}^\infty C_n$ is also compact,
connected, and non-empty. {\sl Hint: for the connectedness, assume
  $C_* = \Omega^1_*\cup \Omega^2_*$, a disjoint union of non-empty
  open sets. Define $\Omega^1_n:= (X\setminus\Omega^2_*)\cap C_n$ and 
  $\Omega^2_n:= (X\setminus\Omega^1_*)\cap C_n$. }
\item
Show: If a function $f:\R^n\to\R^m$ has both the properties ``$K$
compact $\implies$ $f(K)$ compact'' and ``$C$ connected $\implies$
$f(C)$ connected'', then $f$ is necessarily continuous.\\ {\sl Hint: To show
  continuity at 0 (with no loss of generality), let $B_n:=
  B(0,\frac1n)$ and consider $f(B_n)=: D_n$. Use the previous hwk to
  learn about $\bigcap D_n =: D_*$. Explain why $D_*$ must consist of
  either one or infinitely many points. Then, assuming $D_*$ is
  infinite, construct a sequence $(y_n)$ of distinct points in
  $D_*\setminus\{f(0)\}$ 
  that converges to a point $y_*$ that is neither $f(0)$ nor any of
  the $y_n$. From $x_n$ satisfying $y_n=f(x_n)$, construct a compact
  set whose image under $S$ is not compact. Conclude continuity from
  $D_*$ being a singleton. }
\item
Show that $f:X\to\R$ is continuous exactly if it is both lower and
upper semi-continuous. 
\item
Assume $f: K\to\R$ is lower semicontinuous and $K$ is compact. Show
in two ways that $f$ takes on a minimum; namely using sequential
compactness, then using cover-compactness. --- Also give an example of
a lower semi-continuous function $f:K\to\R$ that does {\em not\/} take
on a maximum. 
\item
Assume $f:X\to Y$ is continuous and injective (not necessarily
surjective); here $X,Y$ are metric 
spaces. We wonder whether the inverse function $f^{-1}: f(X)\to X$ is
continuous as well:\\
(a) Show: If $X$ is compact, then we can conclude that $f^{-1}$ is
continuous. \\
(b) Show: If $X$ is an interval in $\R$, and $Y=\R$, 
then we can conclude that $f^{-1}$ is continuous. \\
(c) Give a counterexample where $X=[0,2\pi`[$ and $Y=\R^2$ where
    $f^{-1}$ is not continuous. \\
(d) Give a counterexample with $X\subset\R$ and $Y=\R$, where $f^{-1}$
    is not continuous. 
\item
Show that for a set $M$ in a metric space $X$, its interior $\oacc M$,
which we had defined as $M\setminus\partial M$ where $\partial M=\{
x\in X : \forall \eps>0 : U(x,\eps)\cap M\neq\emptyset \land
U(x,\eps)\cap (X\setminus M)\neq\emptyset\}$, is the union of all open
subsets of $M$, and thus is the largest open subset of $M$. 
Also, show that the closure $\bar M$, which we had defined as
$M\cup\boundary M$, is the intersection of all closed sets that
contain $M$. 
\item
Find a set $M \subset\R$ for which all of the following sets are
different: \\
$M$, $\oacc M$, $\bar M$, $\skew3\oacc{\bar M}$, $\skew3\bar{\oacc M}$, 
$\skew4\bar{\skew2\oacc{\bar M}}$, $\skew4\oacc{\skew2\bar{\oacc M}}$. 
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\end{enumerate}
\end{document}
