Suppose that $X$ is a set and $(f_n)_n$ is a sequence of functions $X \to \R$, so that each $f_n$ is bounded. Suppose further that $(f_n)_n$ converges uniformly to some $f : X \to \R$. Show that $(f_n)_n$ is uniformly bounded: that is, there is some $T \in \R$ so that for all $x \in X$ and all $n \in \N$, \[\abs{f_n(x)} \leq T.\]
In this problem, we will make the assumption that sequences are indexed beginning at $0$ rather than at $1$.
A function of the form above is called a power series. Since a power series is the uniform limit of polynomials—which are continuous—it is continuous.
Let $C : \R \to [-1, 1]$ be a continuous function with the properties that $C(2k) = 1$ and $C(2k+1) = -1$ for every $k \in \Z$, and that $\abs{C(x)-C(y)} \leq 4\abs{x-y}$ for all $x, y \in \R$.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment. They are coloured by approximate difficulty: easy/medium/hard.
For each $n \in \N$, let \begin{align*} f_n : \R &\longrightarrow \R \\ x &\longmapsto \frac{x}{1+nx^2}. \end{align*} Show that the sequence $(f_n)_n$ converges uniformly (on $\R$) to some function $f$, and that $(f_n')_n$ converges to $f'$ pointwise on $\R\setminus\set0$, but not at $0$.
Suppose $S$ is a set, and consider the set of functions from $S$ to $\R$, $\R^S = \set{f : S \to \R}$. Let us say that a metric $d$ on $\R^S$ gives rise to the topology of pointwise convergence if whenever $(f_n)_n$ is a sequence of functions in $\R^S$ we have that $(f_n)_n$ converges to a function $f$ with respect to $d$ if and only if $(f_n)_n$ converges pointwise to $f$.
Show that there is a metric on $\R^S$ giving rise to the topology of pointwise convergence if and only if there is a one-to-one function $\varphi : S \to \N$.
Let $R$ be the set of integrable functions on the interval $[a, b]$. Given $f \in R$, define \[\norm{f}_2 = \paren{\int_a^b\abs{f(t)}^2\,dt}^{\frac12}.\]
The function $d_2(f, g) = \norm{f-g}_2$ on $R\times R$ is a pseudo-metric on $R$: it satisfies the triangle inequality, $d_2(f,f) = 0$, and $d_2(f, g) = d_2(g, f)$, but $d_2(f, g)=0$ does not imply $f = g$. If we define $\sim$ by $f\sim g$ whenever $d_2(f, g) = 0$, then $R/\sim$ becomes a metric space; what we have done is show that (equivalence classes of) polynomials are dense in $R/\sim$.
State precisely what is meant by:
Provide examples to show that these three concepts are different.
Recall that for $x \in \R$, $\floor{x}$ (the floor of $x$) is the greatest integer $k \in \Z$ with $k \leq x$. For $n \in \N$, define \begin{align*} \varphi_n: \R &\longmapsto \R\\ t & \longmapsto 2^{-n}\floor{2^nt}. \end{align*} Finally, let $\mathscr{A}$ be the algebra of functions generated by $(\varphi_n)_n$. That is, $\mathscr{A}$ consists of all functions which are finite sums of scalar multiples of finite products of $\varphi_n$'s.
Let $\overline{\mathscr{A}}$ be the uniform closure $\mathscr{A}$. Show that $\overline{\mathscr{A}}$ is not an algebra. (Hint: $x\mapsto x$ is in $\overline{\mathscr{A}}$, but $x \mapsto x^2$ is not.)
Reflect on what this example says about trying to plot quadratic or linear functions on a pixelated display.
Prove that a power series is differntiable, and its derivative is given by differentiating the sum term-by-term. (This is not immediate and does take work, as in general, uniform limits of differentiable functions need not be differentiable, or may have derivatives unrelated to the derivatives of their approximants.) Conclude that a power series is infinitely differentiable.
Let $f : [0, 1] \to \R$ be continuous.
Show by example that there exists a function $f : \R \to \R$ such that $f$ takes on every real value on every non-empty open set. That is, for any $U \subseteq \R$ open and any $y \in \R$, there is some $x \in U$ with $f(x) = y$. (This isn't really related to this week's material in particular, it's just an interesting problem.)