# First Discussion

## Definitions
**Definition:**
A set $X$ is $F_\sigma$ iff
$$
X = \union_{i=1}^\infty F_i \quad \text{with each $F_i$ closed.}
$$

**Definition:**
A set $X$ is $G_\delta$ iff
$$
X = \intersect_{i=1}^\infty G_i \quad \text{with each $G_i$ open.}
$$

**Definition:**
A set $A$ is *nowhere dense* iff $(\overline{A})^\circ = \emptyset$ iff for any interval $I$, there exists a subinterval $S$ such that $S \intersect A = \emptyset$. 

Such a set is not dense in *any* (nonempty) open set. 

**Fact:**
If the closure of a subset of $\RR$ contains no open intervals, it will be nowhere dense.

**Definition**:
A set $A$ is *meager* or *first category* if it can be written as
$$
A = \union_{i\in \NN} A_i\quad \text{with each $A_i$ nowhere dense}
$$

**Definition**:
A set $A$ is *null* if for any $\varepsilon$, there exists a cover of $A$ by countably many intervals of total length less than $\varepsilon$, i.e. there exists $\theset{I_k}_{j\in\NN}$ such that $A\subseteq \union_{j\in \NN} I_j$ and $\sum_{j\in \NN}\mu(I_j) < \varepsilon$. 

If $A$ is null, we say $\mu(A) = 0$.

**Some facts:**

- If $f_n \to f$ and each $f_n$ is continuous, then $D_f$ is meager.

- If $f \in \mathcal{R}(a, b)$ and $f$ is bounded, then $D_f$ is null.

- If $f$ is monotone, then $D_f$ is countable.

- If $f$ is monotone and differentiable on $(a,b)$, then $D_f$ is null.

We define the *oscillation of $f$* as
$$
\omega_f(x) \definedas \lim_{\delta \to 0^+} \sup_{y,z \in B_\delta(x)} \abs{f(y) - f(z)}
$$

## Uniform Convergence

**Definition**:
We say that $f_n \to f$ *converges uniformly on $A$* if 
$$
\norm{f_n - f}_\infty \definedas \sup_{x\in A}\abs{f_n(x) - f(x)} \to 0
.$$ 

> Note that this defines a sequence of *numbers* in $\RR$.

This means that one can find an $n$ large enough that that for every $x\in A$, we have $\abs{f_n(x) - f(x)} \leq \varepsilon$ for any $\varepsilon$.

### Showing Uniform Convergence

Find some $M_n$, independent of $x$, such that $\abs{f_n(x) - f(x)} \leq M_n$ where $M_n \to 0$.

### Negating Uniform Convergence

Fix $\varepsilon$, let $n$ be arbitrary, and find a bad $x$ (which can depend on $n$) such that $\abs{f_n(x) - f(x)} \geq \varepsilon$.

*Example:* 
$\frac 1 {1 + nx} \to 0$ pointwise on $(0, \infty)$, which can be seen by fixing $x$ and taking $n \to \infty$. 

To see that the convergence is not uniform, choose $x = \frac 1 n$ and $\varepsilon = \frac 1 2$. 
Then
$$
\sup_{x > 0} \abs{\frac 1 {1+nx} - 0} \geq \frac 1 2 \not\to 0.
$$

Here, the problem is at small scales -- note that the convergence *is* uniform on $[a, \infty)$ for any $a > 0$. 

To see this, note that
$$
x > a \implies \frac 1 x < \frac 1 a \implies \abs{\frac 1 {1 + nx}} \leq \abs{\frac 1 {nx}} \leq \frac 1 {na} \to 0
$$
since $a$ is fixed.

### Uniformly Cauchy

Let $C^0( ( [a,b], \norm{\wait}_\infty))$ be the metric space of continuous functions of $[a,b]$, endowed with the metric
$$
d(f, g) = \norm{f - g}_\infty = \sup_{x\in [a,b]} \abs{f(x) - g(x)}
$$

**Proposition:**
This is a complete metric space, and
$$
f_n \to^U f \iff \forall\varepsilon \exists N \suchthat m \geq n \geq N \implies \abs{f_n(x) - f_m(x)} \leq\varepsilon \forall x\in X
$$

*Proof:*
$\implies$: 
Use the triangle inequality.

$\impliedby$: 
Find a candidate limit $f$: first fix an $x$, so that each $f_n(x)$ is just a number. 

Now we can consider the sequence $\theset{f_n(x)}_{n\in\NN}$, which (by assumption) is a Cauchy sequence in $\RR$ and thus converges. 

So define $f(x) \definedas \lim_n f_n(x)$. 

> Aside: we note that if $a_n < \varepsilon$ for all $n$ and $a_n \to a$, then $a\leq \varepsilon$.

Now take $m\to \infty$, we then have

\begin{align*}
\abs{f_n(x) - f_m(x)} < \varepsilon ~\forall x 
&\implies \lim_{m \to \infty} \abs{f_n(x) - f_m(x)} = \abs{f_n(x) - f(x)} \leq \varepsilon ~\forall x \\
&\implies f_n \to^U f
.\end{align*}

> Note: $f_n \to^U f$ does not imply that $f_n' \to^U f'$.

*Counterexample:*
Let $f_n(x) = \frac 1 n \sin(n^2 x)$, which converges to $0$ uniformly, but $f_n'(x) = n\cos(n^2 x)$ does not even converge pointwise.

To make this work, 

**Theorem**:
If $f_n' \to^U g$ for some $g$ and *for at least 1 point* $x, ~f_n(x) \to f(x)$, then $g = \lim f_n'$.

### Key Example
*Exercise:*
Let 
$$
f(x) = \sum_{n=1}^\infty \frac{nx^2}{n^3 + x^3}
.$$

Does it converge at all, say on $(0, \infty)$?

We can check pointwise convergence by fixing $x$, say $x=1$, and noting that
$$
x = 1 \implies \abs{\frac{nx^2}{n^3 + x^2}} \leq \abs{\frac n {n^3 + 1}} \leq \frac 1 {n^2} \definedas M_n,
$$

where $\sum M_n < \infty$. To see why it does not converge uniformly, we can let $x=n$. Then,
$$
x=n \implies \abs{\frac{nx^2}{ n^3 + x^2 }} = \frac{n^3}{2n^3} = \frac 1 2 \not\to 0,
$$

so there is a problem at large values of $x$.

However, if we restrict attention to $(0, b)$ for some fixed $b$, we have $x < b$ and so
$$
\abs{\frac{nx^2}{n^3 + x^2}} \leq
\frac{nb^2}{n^3 + b^2} \leq
b^2 \left( \frac{n}{n^3} \right) =
b^2 \frac 1 {n^2} \to 0.
$$

Note that this actually tells us that $f$ is *continuous* on $(0, \infty)$, since if we want continuity at a specific point $x$, we can take $b>x$. 

Since each term is a continuous function of $x$, and we have uniform convergence, the limit function is the uniform limit of continuous functions on this interval and thus also continuous here. 

Checking $x=0$ separately, we find that $f$ is in fact continuous on $[0, \infty)$.


## Series of Functions

Let $f_n$ be a function of $x$, then we say $\sum_{n=1}^\infty f_n$ converges uniformly to $S$ on $A$ iff the partial sums $s_n = f_1 + f_2 + \cdots$ converges to $S$ uniformly on $A$.

This equivalently requires that
$$
\forall\varepsilon \exists N \suchthat n\geq m \geq N \implies \abs{s_n - s_m} = \abs{\sum_{k=m}^n f_k(x)} \leq \varepsilon \quad \forall x\in A.
$$

> Showing uniform convergence of a series: **Always use the M-test**!!! I.e. if $\abs{f_n(x)} \leq M_n$, which doesn't depend on $x$, and $\sum M_n < \infty$, then $\sum f_n$ converges uniformly.


*Example:*
Let $f(x) = \sum \frac 1 {x^2 + n^2}$.

Does it converge at all? 
Fix $x\in \RR$, say $x=1$, then $\frac 1 {1+n^2} \leq \frac 1 {n^2}$ which is summable. 
So this converges pointwise. 

But since $x^2 > 0$, we generally have $\frac{1}{x^2 + n^2} \leq \frac{1}{n^2}$ for any $x$, so this in fact converges uniformly.

### Negating Uniform Convergence for Series
Todo


## Misc

A useful inequality:
$$
(1+x)^n = \sum_{k=1}^n {n \choose k}x^k = 1 + nx + n^2 x \geq 1 + nx + nx^2 > 1 + nx
$$


**Summary of convergence results:**

- Functions $f_n \to^U f$:
  
  - Showing:
    - $M$ test. Produce a bound $\norm{f_n - f}_\infty < M_n$ independent of $n$ where $M_n \to 0$.
  
  - Negating:
    - Each $f_n$ is continuous but $f$ is not,
    - Let $n$ be arbitrary, then find a bad $x$ (which can depend on $n$) and $\varepsilon$ such that 
    $$
    \sup{\abs{f_n(x) - f(x)}} \geq \varepsilon
    .$$

- Series of functions $\sum f_n \to^U f$:
  
  - Showing:
    - $M$ test. Produce a bound $\norm{f_n}_\infty < M_n$ where $\sum M_n < \infty$.
  
  - Negating:
    - Each partial sum is continuous but $f$ is not.
    - $f_n \not\to^U 0$.
    - Find a bad $x$? Work with the partial sums? 
    (Generally difficult?)