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\title{Homework Assignment 1: Sequences, norms and $\lpspace{p}$.}

\date{Assigned Fri 04/06. Due Wed 04/11.}
\maketitle

The first few questions are a brief refresher on the basic concepts I need from analysis. They're stated in the context of normed spaces below, but are mostly valid in any metric space. I should mention that the standard theory (topology / metric spaces) deals with every thing in terms of open sets, and the Heine Borel form of compactness. However for us closed sets will be `more' important, as subspaces are typically closed. The version of compactness I will generally use is sequential compactness (Bolzano-Weierstrass). Thus the problems below are all formulated in terms of closed sets.

As the first four questions are `standard' facts which some of you might know already, I've marked two of them with a $\star$. This means that they are optional, and will not be graded should you turn them in. If you glance at them, and believe you have either seen it before, or know immediately how to solve it, then safely skip it. If not, solve it, look up the answer, or ask someone about it!

\begin{questions}
\item Let $V$ be a normed space, $(u_n), (v_n)$ sequences in $V$.
\begin{parts}
\item If $(u_n) \to u$ and $(v_n) \to v$, show that $(u_n + v_n) \to u + v$.
\item If $(\lambda_n) \to \lambda \in \C$, and $(u_n) \to u$, show that $(\lambda_n u_n) \to \lambda u$.
\item If $V$ is an inner product space, and $(u_n) \to u$, $(v_n) \to v$ then show that $\ip{u_n}{v_n} \to \ip{u}{v}$.
\end{parts}

\items Recall, $C \subset V$ is said to be closed if whenever $(c_n) \to c$, and $\forall n \in \N, \; c_n \in C$, then $c \in C$.
\begin{parts}
\item Show that $C$ is closed if and only if $\forall x \not\in C, \exists \epsilon > 0 \suchthat B_\epsilon(x) \cap C = \emptyset$. \noteb{Recall $B_\epsilon(x) = \{ y \in V \st \norm{x - y} < \epsilon \}$.}
\item If $C_1, \dots, C_n$ are closed sets, show that $\cup C_i$ is closed.
\item Give an example to show that the previous subpart is false when $n = \infty$.
\item Let $I$ be some (possibly infinite) index set. Given a family $C_\alpha$ (with $\alpha \in I$) of closed sets in $V$, show that $\cap_{\alpha_\in I} C_\alpha$ is closed.
\end{parts}

\items Recall if $S \subset V$, then we define the closure of $S$ (denoted by $\bar{S}$) to be the intersection of all closed subsets of $V$ sets containing $S$.
\begin{parts}
\item Show that $S\subset \bar{S}$ and $\bar{S}$ is closed.
\item Is $S$ is closed, show that $\bar{S} = S$. Conclude $\bar{\bar{S}} = \bar{S}$.
\item Show that $x \in \bar{S}$ if and only if there exists a sequence consisting of elements of $S$ converging to $x$.
\end{parts}

\item Let $V, V'$ be normed spaces.
\begin{parts}
\item Show that $f:V \to V'$ is continuous if and only if whenever $(v_n) \to v$, we have $(f(v_n)) \to f(v)$.
\item Show that $f:V \to V'$ is continuous if and only if for any closed set $C' \subset V'$, $f\inv(C')$ is a closed subset of $V$. \noteb{Recall $f\inv(C') = \{ v\in V \st f(v) \in C'\}$.}
\item Given an example of $f:V \to V'$ continuous and $C \subset V$ closed so that $f(C) \subset V'$ is not closed.
\end{parts}

\uplevel{\noindent The remainder of your homework is `fun', and not a `refresher mini-course'. These questions are quite important though, so if you can't do them yourself, look them up. You should be able to find them in any standard reference. \noteb{Beware the perils of looking it up before you burn midnight oil.}}

\item Let $V$ be a vector space. We say a set $S \subset V$ is convex if whenever $x,y \in S$ and $s \in [0,1]$ we have $sx + (1-s)y \in S$. \noteb{Geometrically, this means that the line segment $x$ and $y$ is completely contained in $S$. It might be illustrative to pick $x,y \in \R^2$, and sketch $\{sx + (1-s)y \st s\in [0,1] \}$.}
\begin{parts}
\item Let $S \subset V$ be convex, and $x_1, \dots, x_n \in S$. If $c_1, \dots, c_n \in [0,1]$ are such that $\sum c_i = 1$ show that $\sum c_i x_i \in S$. \noteb{When $c_i$'s are as above, $\sum c_i x_i$ is called a \textit{convex linear combination} of $x_1, \dots, x_n$.}
\uplevel{Let $S \subset V$ be a convex set. Recall that $\varphi:S \to \R$ is said to be convex if $\forall s \in [0,1]$ and $x, y \in \R$, $\varphi(s x + (1-s)y) \leq s\varphi(x) + (1-s)\varphi(y)$. \noteb{For those unfamiliar with this concept, if $S=[0,1]$, then a convex function is a function such that every chord lies above the function. It's easy to show that differentiable functions (of one variable) are convex if and only if their derivative is increasing.}}
\item {\em (Jensen's Inequality)} Let $\varphi: S \to \R$ be convex, and $x_1, \dots x_n \in S$. Let $c_1, \dots c_n \in [0,1]$ be such that $\sum c_i = 1$. Show that $\varphi(\sum_{i=1}^n c_i \cdot a_i) \leq \sum_{i=1}^n c_i \varphi(a_i)$. \hintb{Use induction. Note that we need the previous subpart, and the assumption that $S$ is convex to ensure $\sum c_i x_i$ is in the domain of $\varphi$.}
\item Do the previous subpart with $n = \infty$. You can assume that the series $\sum c_i x_i$ converges, and that $\varphi$ is positive.
\item Let $\varphi: \R \to [0,\infty)$ be convex, and $f:[0,1]\to\R$ some integrable function. Show that $\varphi( \int_0^1 f ) \leq \int_0^1 \varphi\circ f$. \noteb{This only works when the interval we integrate over has length $1$.}
\end{parts}

\item Let $p > 1$. We define $\lpspace{p}$ to be the set of all complex valued sequences $(x_n)$ such that $\sum \abs{x_n}^p < \infty \}$. If $x = (x_n) \in \lpspace{p}$, we define $\lpnorm{x}{p} = (\sum_{k=1}^\infty |x_k|^p)^{1/p}$. The aim of this problem is to show that $\lpnorm{\cdot}{p}$ defines a norm on $\lpspace{p}$.
\begin{parts}
\item \textit{(H\"older's inequality)} Let $p > 1$ and define $q$ such that $\frac{1}{p} + \frac{1}{q} = 1$. If $(x_n) \in \lpspace{p}$ and $(y_n) \in \lpspace{q}$, then show that $\sum |x_k y_k| \leq \lpnorm{(x_n)}{p} \lpnorm{(y_n)}{q}$. \hintb{The proof only uses Jensen's inequality. Once you figure it out, you should realize that exactly the same proof will show $\abs{\int_0^1 fg} \leq (\int_0^1 \abs{f}^p)^{1/p} (\int_0^1 \abs{g}^q)^{1/q}$.}
\item Show that $\lpnorm{\cdot}{p}$ defines a norm on $\lpspace{p}$. \noteb{The only thing that needs work is the triangle inequality. Here's a hint: $\lpnorm{x+y}{p}^p = \sum \abs{x_i + y_i} \abs{x_i + y_i}^{p-1}$ and use H\"older's inequality.}
\item \textit{(Convexity)} Let $x, y \in \lpspace{p}$ be such that $\lpnorm{x}{p} = \lpnorm{y}{p} = 1$. If $x \neq y$, and $s \in (0,1)$, show that $\lpnorm{s x + (1-s)y}{p} < 1$. \noteb{Geometrically this means that the line joining two points on the boundary of the unit ball in $\lpspace{p}$ lies strictly inside the unit ball. Good luck, you'll need it.}
\end{parts}

\item
\begin{parts}
\item \textit{(Parallelogram inequality)} If $V$ is an inner product space, and $x,y \in V$, show that $\norm{x + y}^2 + \norm{x - y}^2 = 2(\norm{x}^2 + \norm{y}^2)$. When $V = \R^2$ find a geometric interpretation.
\item Show that $\lpspace{p}$ is not an inner product space unless $p = 2$.
\end{parts}
\end{questions}
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