Exercise 1.3

Proof.

(a)

Pick an open set $A$ then let the variables $V_i$ ( $i=1,2,\dots$) run through all open subsets of $A$, so that $$\begin{array}{lll}\hfill A\subseteq \underset{t_i}{\bigcup }(t_1{V}_1+\cdots +t_i{V}_i+\cdots )\subseteq A& & \hfill \text{(1)}\end{array}$$

given all convex combinations $t_1{V}_1+\cdots +t_i{V}_i+\cdots$. We know from Section 1.7 of Functional Analysis that those sums are open; which achieves the proof.

(b)

Provided a bounded set $E$, pick $V$ a neighbourhood of $0$: By (b) of [1.14] in Functional Analysis, $V$ contains a convex neighbourhood of $0$, say $W$. There so exists a positive scalar $s$ such that $$\begin{array}{lll}\hfill E\subset \mathit{tW}\subset \mathit{tV}(t>s);& & \hfill \text{(2)}\end{array}$$

which yields

$$\begin{array}{lll}\hfill E\subset \mathit{tW}=\mathit{tW}=\mathit{tW}\subset \mathit{tV}.& & \hfill \text{(3)}\end{array}$$

So ends the proof.

(c)

At fixed $V$, neighbourhood of the origin, we combine the continuousness of $\text{+}$ with [1.14] of Functional Analysis to conclude that there exists $U$ a balanced neighborhood of the origin such that $$\begin{array}{lll}\hfill U+U\subset V.& & \hfill \text{(4)}\end{array}$$

Moreover, by the very definition of boundedness, $A\subset \mathit{rU}$ for some positive scalar $r$. Similarly, $B\subset \mathit{sU}$ for some positive $s$. Finally,

$$\begin{array}{lll}\hfill A+B\subset \mathit{rU}+\mathit{sU}\subset \mathit{tU}+\mathit{tU}\subset \mathit{tV}(t>r,s),& & \hfill \text{(5)}\end{array}$$

since $U$ is balanced. So ends the proof.

(d)

First, $A$ and $B$ are compact: So is $A\times B$. Next, $\text{+}$ maps continuously $A\times B$ onto $A+B$. In conclusion, $A+B$ is compact.

(e)

From now on, we assume that neither $A$ nor $B$ is empty, since otherwise the result is trivial. Now pick $c\in X$ outside $A+B$: The result will be established by showing that $c$ is not in the closure of $A+B$.

To do so, we let the variable $a$ range over $A$: Every set $a+B$ is closed as well; see [1.7] of Functional Analysis. Trivially, $a+B\ne c$: By [1.10] of Functional Analysis, there so exists $V=V(a)$ a neighborhood of the origin such that $$\begin{array}{lll}\hfill (a+B+V)\cap (c+V)=\varnothing .& & \hfill \text{(6)}\end{array}$$

Moreover, there are finitely many $a+V$, say $a_1+V_1,a_2+V_2,\dots$, whose union $U$ contains the compact set $A$. Therefore,

$$\begin{array}{lll}\hfill A+B\subset U+B.& & \hfill \text{(7)}\end{array}$$

Now define

$$\begin{array}{lll}\hfill W\triangleq V_1\cap V_2\cap \cdots ,& & \hfill \text{(8)}\end{array}$$

so that

$$\begin{array}{lll}\hfill (a_i+B+V_i)\cap (c+W)\stackrel{(6)}{=}\varnothing (i=1,2,\dots ).& & \hfill \text{(9)}\end{array}$$

As a conclusion, $c$ is not in the closure of $U+B$. Finally, (7 ) asserts that $c$ is not in $\overline{A+B}$ either; which achieves the proof.

Corollary: If $B$ is the closure of a set $S$, then

$$\begin{array}{lll}\hfill A+B\subset \overline{A+S}\subset \overline{A+B}=A+B& & \hfill \text{(10)}\end{array}$$

by (b) of [1.13] of Functional Analysis (since $A$ is closed; see [1.12] from the same source). The special case $A=\{x\}$, $B=X$ will occur in the proof of Exercise 15 in chapter 2.

(f)

The last proof will consist in exibhiting a counterexample. To do so, let $f$ be any continuous mapping of the real line such that

(i)

$f(x)+f(-x)\ne 0$ ( $x\in R$);

(ii)

$f$ vanishes at infinity.

For instance, we may combine (ii) with $f$ even and $f>0$ by setting $f(x)=2^{-|x|}$, $f(x)=e^{-x^2}$, $f(x)=1∕(1+|x|)$, …, and so on.

As a continous function, $f$ has closed graph $G$; see [2.14] of Functional Analysis. Moreover, (i) implies that the origin $(0,0)\ne \left(x-x,f(x)+f(-x)\right)$ is not in $G+G$. On the other hand,

$$\begin{array}{lll}\hfill \{\left(0,f(n)+f(-n)\right):n=1,2,\dots \}\subset G+G.& & \hfill \text{(11)}\end{array}$$

Now the key ingredient is that

$$\begin{array}{lll}\hfill \left(0,f(n)+f(-n)\right)\stackrel{(\mathit{ii})}{\underset{n\to \infty }{\text{→}}}(0,0).& & \hfill \text{(12)}\end{array}$$

We have so constructed a sequence in $G+G$ that converges outside $G+G$. So ends the proof.