MA341 - Real Analysis - Spring 2016

Infimum and Supremum of sequences

Let \((x_n)\) be a monotonic sequence.

If \((x_n)\) is bounded then it is convergent

and \[\lim_{n\to\infty} x_n = \sup(x_n, n \in \mathbb{N})\] or \(\inf(x_n, n \in \mathbb{N})\)

Proof
  1. \(x_n = \frac{1}{n}\)

    \(x_n\) is decreasing and bounded, so it converges to \(\inf(\frac{1}{n}) = 0\)

  2. \(x_n = 1 + \frac{1}{2} + ... + \frac{1}{n}\)

    \((x_n)\) is increasing. We will show it is not bounded.

    \(x_n = 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + ... \\ \geq 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + ... = 1 + \frac{n}{2}\)

    So \(x_n\) is not bounded

  3. Let \(x_n\) be defined

    \(x_1 = 1\)

    \(x_{n+1} = \frac{1}{4}(2x + 3)\)

    First prove that \(x_n\) is increasing. (by induction)

    \(x_1 < x_2\)

    \(\frac{1}{4}(2x_n + 3) \leq \frac{1}{4}(2x_{n+1} + 3)\)

    so \(x_n \leq x_{n+1}\)

    Show bounded (by induction)

    \(x_1 \leq 2\)

    Assume \(x_n \leq 2\), then

    \(x_{n+1} = \frac{1}{4}(2x_n + 3) \leq \frac{7}{4} \leq 2\)

    so \(x_n\) is bounded.

    Now to find the limit. \(x_n = x_{n+1}\)

    so \(x = \frac{1}{4}(2x + 3) \Rightarrow x = \frac{3}{2}\)

Subsequences

subsequence

Let \((y_n)\) be a sequence.

If the sequence \((x_n) \subseteq (y_n)\)

Then \((x_n)\) is a subsequence of \((y_n)\)

\((2,4,6,8,...)\) is a subsequence of \((1,2,3,4,...)\)

\(x_n = (-1)^n\) \((-1,1,-1,1,...)\)

\(x_n = 1\) \((1,1,1,1,...)\)

\(x_n = -1\) \((-1,-1,-1,-1,...)\)

Convergence

Let \((x_n)\) be a convergent sequence

Then all subsequences are also convergent and the limits are the same.

Proof
  1. \(x_n = \frac{1}{n^4}\)

    Since \(n^4 \in \mathbb{N}\), \(x_n\) is a subsequence of \(\frac{1}{n}\).

    So \(x_n\) must also converge to 0.

  2. \(x_n = (-1)^n\)

    \[\lim_{n\to\infty} x_{2n} = 1\]

    \[\lim_{n\to\infty} x_{2n+1}= -1\]

    So \(x_n\) is divergent

Divergence

Similarly, if \((x_n)\) is a sequence and \((y_n)\) is a divergent subsequence

then \((x_n)\) is divergent

Proof

Monotonicity

Let \((x_n)\) be a sequence.

There exists a subsequence \((x_m)\) which is monotonic

Proof

Bolzano-Weierstrauss

Let \((x_n)\) be a bounded sequence

Then there exists a subsequence \((x_m)\) that is convergent

Proof
  1. \(x_n = (-1)^n\)

    \(x_{2n} \Rightarrow 1\)

    \(x_{2n+1} \Rightarrow -1\)

  2. Any subsequence of \((x_n) = \sin(n)\) can converge to any \(x \in [-1,1]\)

Cauchy Sequence

cauchy sequence

Let \((x_n)\) be a sequence.

\((x_n)\) is a cauchy sequence if \(\forall \varepsilon > 0\) \(\exists K(\varepsilon) \geq 0\)

\(\forall n,m \geq K(\varepsilon), | x_n - x_m | \leq \varepsilon\)

Cauchy and inverses

Let \((x_n)\) be a convergent sequence.

Then it is a Cauchy sequence

The inverse is not true

Proof
  1. Not a Cauchy sequence:

    \((x_n) = \sqrt{n}\)

    \((x_n)\) is Cauchy, but does not converge

Let \((x_n)\) be a bounded Cauchy sequence

Then \((x_n)\) is convergent

Proof
  1. \(x_n = 1 + \frac{1}{2} + ... + \frac{1}{n}\)

    \(x_{2n} - x_n = \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n} \geq \frac{1}{2n} + ... + \geq \frac{1}{2n}\)

    So \(x_{2n} - x_n\) does not converge to \(0\) therefore \((x_n)\) is not a Cauchy sequence

Contracting Sequences

contracting sequence

Let \((x_n)\) be a sequence

\((x_n)\) is contracting if

\[\exists C > 0, \forall n \in \mathbb{N}, | x_{n+1} - x_n | \leq C | x_n - x_{n-1} |\]

Convergence

Any contracting sequence is convergent

Proof
  1. xn = n

    Let \(M \geq 0\)

    for \(n \geq M, x_n \geq M\)

    So \((x_n)\) goes to infinity
  2. \(x_n = 1 - \frac{1}{n} + n \geq M\)
  3. \(x_n = n^2 - n = n(n-1) \geq (n-1)^2\)

Divergence

Let \((x_n)\) be increasing and not bounded. Then \[\lim_{n \to \infty} x_n = \infty\]

Let \((x_n)\) be decreasing and not bounded. Then \[\lim_{n \to -\infty} x_n = -\infty\]

Proof

Let \((x_n),(y_n)\) be sequences s.t. \((x_n) \geq (y_n)\)

\[\lim_{n \to \infty} (y_n) = \infty\] implies

\[\lim_{n \to \infty} (y_n) = \infty\]

Proving divergence

Proof

Series

series

A sequence generated by the sum of the first \(n\) terms of another sequence

The series \((s_n)\) generated by \((x_n)\) lookks like:

\(s_1 = x_1\)

\(s_2 = x_1 + x_2\)

...

\(s_n = x_1 + ... + x_n\)

limit of series

\[\lim_{n \to \infty} s_n = \sum_{k=1}^{\infty} x_k\]

  1. Let \((x_n) = \frac{1}{n(n+1)}\)

    Then

    \[ \begin{aligned} (s_n) & = \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \\ & = \sum_{k=1}^{\infty} \frac{1}{k} - \sum_{k=1}^{\infty} \frac{1}{k+1} \\ & = \sum_{k=1}^{\infty} \frac{1}{k} - \sum_{k=2}^{\infty} \frac{1}{k} \\ & = 1 - \frac{1}{n+1} \end{aligned} \]

  2. Let \((x_n) = r^{n-1}\)

    then

    \[ \begin{aligned} s_n & = \sum_{k=1}^{\infty} r^{n-1} \\ & = \frac{1-r^n}{1-r} \end{aligned} \]

    case \(|r| < 1\)

    \[\lim_{n \to \infty} r^n = 0\]

    So \[\sum_{k=1}^{\infty} r^{n-1} = \frac{1}{1-n}\]

    and \((s_n)\) is convergent

    case \(|r| = 1\)

    \[\sum_{k=1}^{\infty} x_k = \infty\]

    \(s_n = \frac{1}{2} (1 - (-1^n))\)

    So \((s_n)\) is divergent

  3. Let \(x_n = \frac{1}{2^n}\)

    then \[\lim_{n \to \infty} s_n = \sum_{k=1}^{\infty} \frac{1}{2^k} = 2\]

Let \((s_n)\) be a series generated by the sequence \((x_n)\)

If \((s_n)\) converges, then \((x_n)\) also converges

Proof

Convergence of \((x_n)\) does not imply convergence of \((s_n)\)

ex. harmonic series = \(x_n = \frac{1}{n}\)

If \((x_n)\) is a positive sequence

then \[0 < \lim_{n \to \infty} s_n \leq \infty\]

Proof

Let \((x_n)\) and \((y_n)\) be sequences s.t. \(0 \leq x_n \leq y_n\)

Let \((s_n)\) and \((t_n)\) be sequences of \((x_n)\) and \((y_n)\) respectively

If \((t_n)\) is convergent, then \((s_n)\) is convergent

If \((s_n)\) is divergent, then \((t_n)\) is divergent

Proof
  1. Let \(x_n = \frac{1}{n^r}, r \geq 0\)

    then

    \[ \begin{aligned} s_n = \sum_{k=1}^{\infty} \frac{1}{k^r} \ & = 1 + \frac{1}{2^r} + ... \end{aligned} \]

    case \(0 \leq r \leq 1\)

    \(\frac{1}{n^r} \geq \frac{1}{n}\)

    So \(s_n\) is divergent

    case \(r > 1\)

    For \(n \geq 2\) (by bernoulli inequality)

    \(\frac{1}{nr} \leq \frac{1}{r-1} (\frac{1}{(n-1)^{r-1}} - \frac{1}{n^{r-1}})\)

    Let \(y_n = \frac{1}{r-1} (\frac{1}{(n-1)^r} - \frac{1}{n^{r-1}})\)

    Then \[\sum_{k=1}^{\infty} y_k = \frac{1}{r-1}(1 - \frac{1}{n^{r-1}}) < \infty\]

    So \[\sum_{k=1}^{\infty} x_k < \infty\]

    We conclude \[\sum_{k=1}^{\infty} \frac{1}{n^r} < \infty\] if and and only if \(r > 1\)

Functions

cluster point

Let \(A \subseteq \mathbb{R}\) be a non empty set.

Then \(c \in \mathbb{R}\) is a cluster point of A if

\(\exists a \in A\), \(a \neq c\) s.t. \(\forall \varepsilon > 0\),

\(|a - c| < \varepsilon\)

\(c\) is a cluster point of the set \(A\) if the values of \(A\) get arbitrarily close to \(c\)

  1. Let \(A = \{ \frac{1}{n}, n \in \mathbb{N}\}\)

    0 is the only cluster point for the set \(A\)

  2. Let \(A = \{n | n \in \mathbb{N}\}\)

    There are no cluster points in the set. We can always find a smaller \(\varepsilon\) which excludes \(c\)

  3. Let \(A = \mathbb{Q} = \{\frac{x}{y} | x,y \in \mathbb{N}\}\)

    The set of cluster points is \(\mathbb{R}\)

    This is because \(\mathbb{Q}\) is dense in \(\mathbb{R}\)

function

Let \(A \subset \mathbb{R}\) be a non empty set.

Then the mapping \(f: A \rightarrow \mathbb{R}\) is a function

limit

\(f\) admits a limit \(L \in \mathbb{R}\) if:

\(\forall \varepsilon > 0\), \(\exists \zeta > 0\),

s.t. \(|x - a| \leq \zeta\) implies \(|f(x) - L| \leq \varepsilon\)

Or, as \(x\) converges to \(a\), \(f(x)\) converges to \(L\)

Let \(f(x) = x\)

Let \(a \in \mathbb{R}\), \(\varepsilon > 0\)

if \(|x - a| \leq \varepsilon\),

then \(|f(x) - a| \leq \varepsilon\)

Uniqueness of Limit

Let \(A \subseteq \mathbb{R}\) be a nonempty set

Let \(f: A \to \mathbb{R}\) and let \(c \in \mathbb{R}\) be a cluster point of \(A\)

Assume \(f\) has a limit \(L\) at \(c\) and also has limit \(L'\) at \(c\)

Then \(L = L'\)

and \(L = \lim_{x \to c} f(x)\) is called the limit of \(f\) at \(c\)

Proof
  1. find the limit of \(f(x) = a, a \in \mathbb{R}\)

    Let \(\varepsilon > 0\)

    For \(x \in \mathbb{R}\), s.t. \(|x - 1| \leq 1\)

    we have \(|f(x) - a| < \varepsilon\)

  2. Let \(f(x) = x\). Show \(\lim_{x \to 1} f(x) = 1\)

    Let \(\varepsilon > 0\)

    for \(x \in \mathbb{R}\) s.t. \(|x - 1| < \varepsilon\) we have \(|f(x) - 1| = |x - 1| < \varepsilon\)

  3. Let \(f(x) = x^2\) Show \(\lim_{x \to a} f(x) = a^2\)

    Let \(\varepsilon > 0\)

    \(|x^2 - a^2| = |x-a||x+a|\)

    If \(|x - a| \leq 1\), \(|x| \leq 1 + |a|\),

    so \(|x + a| \leq 1 + 2 |a|\)

    \(|x^2 - a^2| \leq |x-a|(1 + 2 |a|)\)

    Let \(\zeta = \min(1, \frac{\varepsilon}{1 + 2 |a|})\)

    For \(x \in \mathbb{R}\) s.t. \(|x - a| \leq \zeta\),

    we have \(|x^2 - a^2| < \varepsilon\)

  4. Let \(f(x) = \frac{1}{x}, x > 0\). Prove \(\lim_{x \to a} \frac{1}{x} = \frac{1}{a}\)

    \(|\frac{1}{x} - \frac{1}{a}| = \frac{|a - x|}{|a||x|}\)

    For \(|x - a| < \frac{a}{2}, x > \frac{a}{2}\)

    we have \(|\frac{1}{x} - \frac{1}{a}| \leq \frac{2}{a^2} |a - x|\)

    Let \(\zeta = \min(\frac{a}{2}, \frac{\varepsilon a^2}{2})\)

    if \(|x - a| \leq \zeta\)

    then \(|\frac{1}{x} - \frac{1}{a}| \leq \varepsilon\)

Let \(A \subset \mathbb{R}\) be a nonempty set and \(f: A \rightarrow \mathbb{R}\)

Let c be a cluster point of A. Then the conditions are equivalent:

  1. \(\lim_{x \to c} f(x) = L\)
  2. For every sequence \((x_n)\) in A where \(x_n \neq c\) and \(\lim_{n \to \infty} x_n = c\),

    then \(\lim_{n \to \infty} f(x_n) = c\)

Proof

Let \(A \subset \mathbb{R}\) be nonempty, \(f:A \rightarrow \mathbb{R}\), and \(c \in \mathbb{R}\) be a cluster point of A

  1. Assume that you can find a sequence \((x_n)\) in A, \(x_n \neq c\) and \(x_n \rightarrow c\), but \(f(x_n)\) is not convergent

    Then \(\lim_{x \ to c} f(x)\) does not exist

  1. Let \(f(x) = \frac{1}{x}, x > 0\)

    Let $xn = . We have $xn → 0, \(f(x_n)\) is convergent so

    \(\lim_{x \to 0} f(x)\) does not exist

  2. Show the function does not converge

    \[ f(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & \text{else} \end{cases} \]

    Let \((x_n) = \frac{1}{n}\). \(\lim_{n \to \infty} f(x_n) = 1\)

    Let \((y_n) = \frac{\sqrt{2}}{n}\). \(\lim_{n \to \infty} f(x_n) = 0\)

    Since the subsequences don't converge to the same value, \(f(x)\) does not converge

Limit Theorems

Let \(A \subset \mathbb{R}\) be a nonempty set. Let \(c \in \mathbb{R}\) be a cluster point of \(A\). Let \(f: A \longrightarrow \mathbb{R}\)

Then \(f\) is bounded in a neighborhood of c if there exists \(M > 0\) and \(\zeta > 0\)

s.t. if \(x\in[c - \zeta, c + \zeta] \cap A\)

\(|f(x) \leq M\)

  1. Let \(f(x) = x^2\). For \(x \in [-1,1]\), \(f(x) \leq 1\) so \(f\) is bounded in the neighborhood of 0

  2. Let \(f(x) = \frac{1}{x}\)

Let \(f: A \longrightarrow \mathbb{R}\) Let \(c\) be a cluster point of \(A\).

If \(\lim_{x \to c} f(x)\) exists, then \(f\) is a bounded in a neighborhood of \(c\)

Proof

Let \(f,g: A \longrightarrow \mathbb{R}\)

  • \((f + g)(x) = f(x) + g(x)\)
  • \((fg)(x) = f(x)g(x)\)
  • if \(\forall x \in A, g(x) \neq 0\)

    \(\frac{f}{g}(x) = \frac{f(x)}{g(x)}\)

Let \(f,g: A \longrightarrow \mathbb{R}\)

Let \(c \in \mathbb{R}\) be a cluster point of \(A\)

If \(\lim_{x \to c} f(x)\) and \(\lim_{x \to c} g(x)\) exist, then

\(\lim_{x \to c} f(x) + \lim_{x \to c} g(x) = \lim_{x \to c} f(x) + g(x)\)

Proof

Let \(f,g: A \to \mathbb{R}\). Let \(c \in \mathbb{R}\) be a cluster point of \(A\).

If \(\lim_{x \to c} f(x)\) and limx → c g(x)$ exist, then

\(\lim_{x \to c} f(x)g(x) = \lim_{x \to c} f(x) \lim_{x \to c} g(x)\)

  1. Find \(\lim_{x \to 1} \frac{x^2 + 1}{x}\)

    \(\lim_{x \to 1} \frac{x^2 + 1}{x} = \frac{\lim_{x \to 1} x^2 + 1}{\lim_{x \to 1} x} = 2\)

  2. Let \(f(x) = x + \mathbb{1}_{\mathbb{Q}}\)

    The limit of \(x\) exists, but not \(\mathbb{1}_{\mathbb{Q}}\), so the limit of \(f(x)\) does not exist

  3. Let \(f(x) = x \mathbb{1}_{\mathbb{Q}}\)

    Since \(\mathbb{1}_{\mathbb{Q}}\) is bounded and \(x\) is convergent, \(f(x)\) is convergent

Squeeze Theorem

Let \(f,g,h: A \to \mathbb{R}\). Let \(c \in \mathbb{R}\) be a cluster point of \(A\).

Assume \(f \leq g \leq h\) for all \(x\).

If \(\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} g(x) = L\)

Proof

M > 0. prop div if xn > M for all n > k

Proper Divergence

Let \(f: A \to \mathbb{R}\) Let \(c \in \mathbb{R}\) be a cluster point of \(A\)

We say that \(\lim_{x \to c} f(x) = \infty\)

if for \(\forall M \geq 0\), \(\exists \zeta > 0\), s.t. for \(x \in [c - \zeta, c + \zeta] \cap A, x \neq c\)

\(f(x) \geq M\)

  1. \(f(x) = \frac{1}{x}, x > 0\)

    \(\lim_{x \to 0} f(x) = \infty\)

  2. Let \(f(x) = \frac{1}{x}, x < 0\)

    \(\lim_{x \to 0} f(x) = -\infty\)

changing A can change the limit

Let \(f: A \to \mathbb{R}\). Let \(c \in \mathbb{R}\) be a cluster point of \(A\)

then \(\lim_{x \to c} f(x) = \infty\) iff

for every sequence \((x_n) \in A\) converging to \(c\), \(x_n \neq c\)

\(\lim_{n \to \infty} f(x_n) = \infty\)

Proof

Continuous Functions

Continuous at point

Let \(\varepsilon > 0\).

\(f\) is continuous at \(c\) if there exists \(\zeta > 0\) such that

\(\forall x \in A, |x - c| \leq \zeta\) implies \(|f(x) - f(c)| \leq \varepsilon\)

If \(c \in A\) is a cluster point of A,

then \(f\) is a continuous function at \(c\) iff \(\lim_{x \to c} f(x) = f(c)\)

If \(c \in A\) is not a cluster point, then \(f\) is automatically continuous at \(c\)

  1. Let \(f(x) = x, x \in \mathbb{Z}\).

    Then \(f\) is continuous at \(x\)

  2. Let

    \[ f(x) = \begin{cases} 1 & x \geq 0 \\ 0 & x < 0 \end{cases} \]

    Then \(f\) is not continuous at 0

  3. Let

    \[ f(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases} \]

    f(x) is nowhere continuous

  4. Let

    \[ f(x) = \begin{cases} 1 & x \in \mathbb{R} \\ 0 & x \notin \mathbb{R} \end{cases} \]

    from the squeeze theorem, \(f\) is continuous at 0

    but if \(x \neq 0\), then the limit does not exist and \(f\) is not continuous

Continuity of points

Let \(A \rightarrow \mathbb{R}\)

Let \(c \in A\), then \(f\) is continuous at \(c\) iff

for every \((x_n)\) s.t. \(\lim_{n\to \infty} x_n = c\), we have \(\lim_{n\to \infty} f(x_n) = f(c)\)

\(f\) is not continuous at \(c\), iff there exists \((x_n)\) that converges to c

where \(f(x_n)\) does not converge to \(f(c)\)

Continuity of functions

Let \(f:A \to \mathbb{R}\)

Let \(B \subset A\)

We say \(f\) is continuous on \(B\) if \(f(x)\) is continuous for every \(x \in B\)

  1. Let \(f(x) = 1, x > 0\)

    \(f\) is continuous on \((0, \infty)\)

    Let \(f:A \to \mathbb{R}\) and \(c \in \mathbb{R}\) be a cluster point of \(A\).

    Assume \(c \notin A\), but \(\lim_{x \to c} f(x)\) exists. Then we can define

    \(f'(x) = f(x), x in A, \lim_{x \to c}f(x), x = c\)

    f' fills in holes

Combinations of continuous functions

Let \(f,g:A \to \mathbb{R}\) be continuous functions at \(c\). Let \(c \in A\)

Then \(f+g\), \(fg\), and \(\frac{f}{g}\) (\(g \neq 0\)) are continuous at \(c\)

Proof
  1. Let \(f\) be any polynomial in \(\mathbb{R}\)

    Since \(f\) is a sum of products of \(x\) and \(x\) is continuous on \(\mathbb{R}\), \(f\) is also continuous on \(\mathbb{R}\)

Let \(f:A \to \mathbb{R},g:B \to \mathbb{R}\)

Assume for \(x \in B\), \(g(x) \in A\)

We define \((f \circ g)(x) = f(g(x))\)

If \(g\) is continuous at \(c\) and \(f\) is continuous at \(g(c)\), then \(f \circ g\) is continuous at \(c\)

Let \(A,B \subseteq \mathbb{R}\). Let \(f\) be continuous on \(A\), and \(g\) be continuous on \(B\)

If \(f(A) \subseteq B\), then \(g \circ f\) is continuous on \(A\)

closed set/function

Intervals

interval is a continuous subset of R

f has a supremum on a bounded set uses squeeze theorem

Assume that \(f\) is not bounded on \(I\). We can assume \(f\) has no upper bound.

\(n\) is not an upper bound for \(f\), so we can find \((x_n) \in I\) s.t. \(f(x_n) \geq n\)

\((x_n)\) is bounded, therefore it admits a convergent subsequence \((x_m)\)

Let \(x^* = \lim x_m\)

Since I is closed, \(x^* \in I\)

Since \(f\) is continuous at \(x^* \lim f(x_m) = \lim f(x^*)\)

This contradicts \(f(x_n) \geq n\)

So \(f\) admits a supremum on \(I\)

The proof is similar for infimum

The supremum/infimum of a function of closed set is also a min/max

Let \(I = [a,b]\). Let \(f:I \to \mathbb{R}\) be a continuous function.

Let \(M = \text{sup}{f(x), x \in I}\), \(S = \inf {f(x), x \in I}\)

\(\exists x_S, x_M \in I\) s.t. \(f(x_S) = S, f(x_M) = M\)

Proof
  1. Let \(f(x) = x^2, I = [-1, 1]\)

    \(\sup{F} = 1 = f(1)\)

    \(\inf{F} = 0 = f(0)\)

  2. Let \(f(x) = x, I = (0,1)\)

    \(\inf f = 0\), but there is no \(x \in I\) such that \(f(x) = 0\)

Intermediate Value Theorem

Let \(I\) be an interval and \(f:I \to \mathbb{R}\) be continuous

Let \(a,b,c \in I\) s.t. \(a \leq b\) \(f(a) \leq f(b)\)

Let \(\lambda \in [f(a), f(b)]\). We can always find \(c \in [a,b]\)

s.t. \(f(c) = \lambda\)

Proof

Zero crossing corollary

Let \(f:[a,b] \to \mathbb{R}\) be continuous

If \(f(a) < 0, f(b) > 0\) then \(\exists c\) s.t. \(f(c) = 0\)

Let \(f:\mathbb{R} \to \mathbb{R}\) be continuous. Let \(I \in \mathbb{R}\) be an interval.

Then \(f(I) = {f(x) | x \in I}\) is an interval

Proof

Uniform continuity

\(f\) is uniformly continuous if

for all \(\varepsilon\) there exists \(\delta\) such that

\(|x_1 - x_2| \leq \delta\) implies \(|f(x_1) - f(x_2)| \leq \varepsilon\)

Lipschitz

Let \(f:A \to \mathbb{R}\)

\(f\) is lipschitz if, there exists \(K \geq 0\) such that

for all \(x, y \in A\)

\(|f(x) - f(y)| \leq K|x - y|\)

  1. Let \(f(x) = x\).

    \(|f(x) - f(y)| \leq 1 * |x - y|\), so \(f\) is lipschitz

  2. Let \(f(x) = \sqrt{x}\)

    \(|f(x) - f(y)| = |\sqrt{x} - \sqrt{y}| = \frac{1}{\sqrt{x} + \sqrt{y}} |x - y|\)

    so \(f\) is not lipschitz on \([0, \infty)\).

    If we let \(x,y \in [a, \infty)\), then

    \(|f(x) - f(y)| \leq \frac{1}{2\sqrt{a}} |x - y|\)

    so \(f\) is lipschitz on \([a, \infty)\)

  3. Let \(f(x) = x^2\)

    \(|f(x) - f(y)| = |x^2 - y^2| = |x - y||x + y|\).

    \(f\) is not lipschitz on \(\mathbb{R}\)

    but it is for any bounded interval, since some \(K\) will be greater than \(|x - y|\)

Any lipschitz function is uniformly continuous

Proof

The converse is not true, as \(f(x) = \sqrt{x}\) is uniformly continuous, but not lipschitz.

ie. \(\sqrt{x}\) is continuous on a bounded interval \([0,1]\), but the derivative is not bounded (\(\lim_{x \to 0} f'(x) = \infty\))

The derivative of Lipschitz functions are bounded by K

Proof

Monotonic functions

*Increasing and decreasing functions

Let \(f:A \to \mathbb{R}\), and \(x,y \in A\), such that \(x \leq y\)

then \(f\) is increasing if \(f(x) \leq f(y)\)

the definition is similar for strictly increasing/decreasing

  1. Let \(f(x) = x\)

    \(f\) is strictly increasing

  2. Let \(f(x) = x^2\)

    \(f\) is strictly increasing on the domain \(x \geq 0\)

The left and right limits exist for all points in monotonic functions.

Let \(f:I \to \mathbb{R}\) be monotonic.

Let \(c \in I\) which is not an endpoint of \(I\).

Then \(\lim_{x \to +c} f(x)\) and \(\lim_{x \to -c} f(x)\) exist

Proof

Inverse functions

Inverse The function \(g:f(A) \to A\) such that \(g(f(a)) = a\) for \(a \in A\)

A function \(f\) has an inverse if \(f\) is injective. (\(f\) uniquely maps values in its domain)

ie. \(x \neq y\) implies \(f(x) \neq f(y)\)

Let \(f:A \to \mathbb{R}\) be a strictly increasing function.

Then \(f\) is injective and \(g:f(A) \to A\) is strictly increasing.

Proof

Let \(f\) be a strictly increasing and continuous function. Then \(f^{-1}\) is strictly increasing and continuous.

Proof

Derivatives

Differentiability

We say that a function \(f:I_1 \to \mathbb{R}\) is differentiable at \(c \in I\) if

\[\lim_{x \to c} \dfrac{f(x) - f(c)}{x - c}\]

exists. We call this \(f'(c)\)

  1. Let \(f(x) = c\) for \(c \in \mathbb{R}\)

    \(\frac{f(x) - f(c)}{x - c} = 0\)

    so \(f'(x) = 0\)

  2. Let \(f(x) = x\)

    \(\frac{f(x) - f(c)}{x - c} = \frac{x - c}{x - c} = 1\)

    so \(\lim_{x \to c} f'(x) = 1\)

  3. Let \(f(x) = x^2\)

    \(\frac{f(x) - f(c)}{x - c} = \frac{x^2 - c^2}{x - c} = \frac{(x - c)(x + c)}{x - c} = x + c\)

    so \(\lim_{x \to c} x + c = 2c\)

  4. Let \(f(x) = x^3\)

    \(\frac{f(x) - f(c)}{x - c} = \frac{x^3 - c^3}{x - c} = \frac{(x - c)(x^2 + xc + c^2)}{x - c} = x^2 + xc + c^2\)

    so \(\lim_{x \to c} x^2 + xc + c^2 = 3c^2\)

If a function is differentiable at c, it is continuous at c

Proof

Sum Rule

The derivative of a sum of functions is the sum of the derivative of the functions

\((f + g)'(x) = f'(x) = g'(x)\)

Proof

Product Rule

\((fg)'(c) = f'(c)g(c) = f(c)g'(c)\)

Proof

Quotient Rule

Let \(f\) be a differentiable function which is not zero on a neighborhood around \(c\)

Then \((\frac{1}{f})'(c) = \frac{f'(c)}{f(c)^2}\)

Proof

Let \(f,g\) be differentiable functions where \(g\) is not zero on a neighborhood around \(c\)

then \(\left(\frac{f}{g}\right)'(c) = \frac{f'g(c) - fg'(c)}{g(c)^2}\)

Proof

The Carathéoday Lemma is needed to prove the next theorem.

Carathéoday Lemma

Let \(f:I \to \mathbb{R}\) be continuous.

\(f\) is differentiable at \(c \in I\) iff,

there exists a continuous \(\phi:I \to \mathbb{R}\) such that

\(f(x) - f(c) = \phi(x)(x - c)\) for all \(x \in I\)

Proof

Chain rule

Let \(f:I \to \mathbb{R}\) and \(g:f(I) \to \mathbb{R}\) be continuous.

then \(g(f(c))' = g'(f(c))f'(c)\)

Proof

Derivative of inverse

Let \(f:I \to \mathbb{R}\) and \(f^{-1}:f(I) \to I\) be continuous.

then \(f'^{-1}(c) = \frac{1}{f'(g(f(c)))}\)

Proof

Relative Extrema

Let \(f: I \to \mathbb{R}\) be continuous.

We say \(f\) has a relative minimum at \(x_0 \in I\) if there exists \(\zeta\) s.t.

for \(x \in [x_0 - \zeta, x_0 + \zeta ] \cap I\)

\(f(x) \geq f(x_0)\)

Note that by the above definition, constant functions are maximum and minimums everywhere.

Derivative at relative extrema

Let \(f: I \to \mathbb{R}\), \(x_0 \in I\), where \(x_0\) is not a boundary point

Assume \(x_0\) is a relative extrema and \(f\) is differentiable at \(x_0\)

Then \(f'(x_0) = 0\)

Proof

Note that the converse is not true. \(\frac{d}{dx}\left[ x^3 \right]_{x = 0} = 0\), but \(x = 0\) is not an extremum

  1. Let \(f(x) = |x|\)

    There is a relative minimum at 0, but the function is not differentiable

Rolle's Theorem

Let \(f:I \to \mathbb{R}\) be a differentiable function.

If \(f(a) = f(b)\), where \(a < b\)

then \(f'(c) = 0\) for some \(c \in (a,b)\)

if abs(x) is not differentiable, then what is the integral of step function

Proof

Mean Value Theorem

Let \(f:I \to \mathbb{R}\) be differentiable over \([a,b]\), where \(a < b\)

then there is a point \(c \in (a,b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\)

Proof

Derivative is zero implies constant function

Let \(f:I \to \mathbb{R}\) be differentiable such that \(f'(x) = 0\) for all \(x\)

Then \(f(x)\) is constant on \(I\)

Proof

Positive derivatives implies increasing function

Let \(f:I \to \mathbb{R}\) be differentiable.

\(f\) is increasing iff \(f'(x) \geq 0\)

Proof

First Derivative Test for Extrema

Let \(f:I \to \mathbb{R}\)

If \(f'(c) = 0\) and \(f'(x) \geq 0\) for \(x \in (c - \delta, c]\)

and \(f'(x) \leq 0\) for \(x \in [c,c + \delta)\)

then \(c\) is a relative minimum.

L'Hôpital's Rule

Indeterminate limit

Let \(f,g:(a,b) \to \mathbb{R}\) be continuous.

If \(\lim_{x \to a+)} f(x) = 0\) and \(\lim_{x \to a+)} g(x) = 0\)

Then we say \[\lim_{x \to a+} \frac{f(x)}{g(x)}\]

is indeterminate

  1. If it exists, evaluate \(\lim_{x \to 0} \frac{sin(x)}{x}\)

    \(\frac{sin(x)}{x} = \frac{sin(x) - sin(0)}{x - 0}\)

    Since \(\sin(x)\) differentiable at 0, \(\lim_{x \to 0} \frac{\sin(x)}{x} = \cos(0) = 1\)

  2. If it exists, evaluate \(\lim_{x \to 0} \frac{x}{x^2}\)

    \(\lim_{x \to 0} \frac{x}{x^2} = \infty\)

    so it is indeterminate

L'Hôpital's Rule #1

Let \(f,g:I \to \mathbb{R}\) be continuous.

Assume \(f(a) = g(a) = 0\) and that \(f,g\) are differentiable at \(a\)

and \(g'(a) \neq 0\)

we have \[\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}\]

Proof
  1. If it exists, compute \(\lim_{x \to 0} \frac{1 - \cos(x)}{x}\)

    Let \(f(x) = 1 - \cos(x)\), \(g(x) = x\)

    Then \(f'(0) = 0\), \(g'(0) = 1\)

    so by L'Hôpital's rule, \(\lim_{x \to 0} \frac{1 - \cos(x)}{x}\)

  2. If it exists \(\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}\)

    We apply L'Hôpital's rule twice:

    \(f(x) = 1 - \cos(x)\), \(f'(x) = \sin(x)\), \(f''(x) = \cos(x)\)

    \(g(x) = x^2\), \(g'(x) = 2x\), \(g''(x) = 2\)

    so \(\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{f''(0)}{g''(0)} = \frac{1}{2}\)

L'Hôpital's Rule #2

Let \(f,g:I \to \mathbb{R}\) be continuous ad differentiable functions.

Assume \(\lim_{x \to a+} f(x),g(x) = 0\) and \(g'(x) \neq 0\) for \(x \in (a,b)\)

a) If \(\lim_{x \to a+} \frac{f'(x)}{g'(x)} = L\), then \(\lim_{x \to a+} \frac{f(x)}{g(x)} = L\) b) If \(\lim_{x \to a+} \frac{f'(x)}{g'(x)} = \infty\), then \(\lim_{x \to a+} \frac{f(x)}{g(x)} = \infty\)

This does not require that \(f,g\) be differentiable at \(a\)

To prove the above, we need the Cauchy Mean Value Theorem

Cauchy Mean Value Theorem

Let \(f,g:I \to \mathbb{R} be differentiable\)

Let \(a,b \in I, a < b\)

Assume \(g'(x) \neq 0\), for \(x \in (a,b)\)

You can find \(c \in (a,b)\) such that

\[\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}\]

Proof
Proof
  1. Evaluate \(\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\)

    \(f(x) = e^x - 1 - x\), \(f'(x) = e^x - 1\), \(f''(x) = e^x\)

    \(g(x) = x^2\), \(g'(x) = 2x\), \(g''(x) = 2\)

    \(\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{1}{2}\)

Riemann Integrals

Partition

Let \(I=[a,b]\) A partition of \(I\) is a sequence \(x_0,...,x_n\)

such that \(a = x_0 < ... < x_n = b\)

Tagged partition

A tagged partition is a partition where a tag \(t_i\) is defined for every interval.

where \(t_i \in [x_{i-1}, x_i]\)

Mesh

The mesh of a partition is defined as \(||\cal{P}|| = max(x_i - x_{i - 1} , 1 \leq i \leq n)\)

\[ \lim_{\Delta x \to 0} \sum_{n = 0}^{\frac{b - a}{\Delta x}} \Delta x (f(a + \Delta x n))\]

Riemann Sum

\(S(f, \bar{\cal{P}}) = \sum_{i = 1}{n} f(t_i)(x_i - x_{i-1})\)

Riemann Integrable We say a function \(f\) is Riemann integrable f \(\forall \varepsilon > 0\), you can find \(\delta > 0\) such that

for every tagged partition \(\bar{\cal{P}}\) with mesh \(||\bar{\cal{P}}|| \leq \delta\)

you have \(|S(f,\bar{\cal{P}}) - L| \leq \varepsilon\) and \(L\) is called the Riemann integral, where \(L = \int_a^b f(x) dx\)

  1. Let \(f(x) = 0, x \in [0,1]\)

    \(S(f,\cal{\bar{P}}) = 0\) for any tagged partition.

    Therefore \(\int_a^b f(x) dx = 0\)

  2. Let \(f(x) = 1, x \in \mathbb{R}\)

    \[ \begin{aligned} S(f,\bar{\cal{P}}) & = \sum_{i = 1}^n f(t_i)(x_i - x_{i-1}) \\ & = \sum_{i=1}^n (x_i - x_{i - 1}) \\ & = b - a \end{aligned} \]

    So \(\int_a^b f(x) dx = b - a\)

Properties of integrals Let \(f,g \in \cal{R}[a,b]\)

  1. \(\int_a^b f(x) + g(x) dx = \int_a^b f(x) dx + \int_a^b g(x) dx\)
  2. \(\int_a^b kf(x) dx = k\int_a^b f(x) dx\) for \(k \in \mathbb{R}\)
  3. if \(f\) is Riemann integrable, \(f\) must be bounded
Proof

If \(f:I \to \mathbb{R}\) is continuous, then \(f\) is Riemann integrable.

The converse is not true

If \(f\) is a monotonic function, then it is Riemann integrable