Proof of Banach Alaoglu theorem by Brezis I am trying to understand the proof of the Banach Alaoglu theorem in the functional analysis book by Brezis. These are the steps I don't quite follow.

*

*Let $Y = \mathbb{R}^E$, a set of functions from $E$ to $\mathbb{R}$. He says $K_1 = \{f:E \to \mathbb{R} : |f(x)| \le ||x||, \forall x \in E\}$ is compact on $Y$, by the Tychonoff's thm. To use Tychonoff's thm, given $x \in E$, we need to show $\{f:E \to \mathbb{R} : |f(x)| \le ||x||\}$ is compact. So my confusion is, what are the underlying topologies being used to say about compactness?


*It seems that he is also saying
$$
\{f: f(x+y)-f(x)-f(y)= 0\} ~\text{and}~ \{f: f(\lambda x)-\lambda f(x)= 0\}
$$
are closed in $Y$. Why is true?
 A: $\Bbb R^E$ is given the product topology as a product of $|E|$ copies of $\Bbb R$. If for each $x\in E$ we let $\Bbb R_x$ be a copy of $\Bbb R$, we can think of $\Bbb R^E$ as $\prod_{x\in E}\Bbb R_x$, and we then have
$$K_1=\prod_{x\in E}[-\|x\|,\|x\|]\;,$$
which is a product of compact spaces.
Fix $x,y\in E$, and let $F=\{f\in Y:f(x+y)=f(x)+f(y)\}$. Suppose that $g\in Y\setminus F$, so that $g(x+y)\ne g(x)+g(y)$. Let
$$\epsilon=\frac13\big|g(x+y)-\big(g(x)+g(y)\big)\big|\;,$$
and let
$$B_{x+y}=\{f\in Y:|f(x+y)-g(x+y)|<\epsilon\}\;,$$
$$B_x=\{f\in Y:|f(x)-g(x)|<\epsilon\}\;,$$
and
$$B_y=\{f\in Y:|f(y)-g(y)|<\epsilon\}\;.$$
These are open sets in the product topology on $Y$, so $U=B_{x+y}\cap B_x\cap B_y$ is open in $Y$, and clearly $g\in U$. Let $f\in U$; then
$$|f(x+y)-g(x+y)|<\epsilon$$
and
$$\big|\big(f(x)+f(y)\big)-\big(g(x)+g(y)\big)\big|\le|f(x)-g(x)|+|f(y)-g(y)|<2\epsilon\;,$$
so
$$\big|f(x+y)-\big(f(x)+f(y)\big)\big|>\big|g(x+y)-\big(g(x)+g(y)\big)\big|-3\epsilon=0\;,$$
and hence $f\in Y\setminus F$. That is, $U$ is an open nbhd of $g$ disjoint from $F$, and since $g\in Y\setminus F$ was arbitrary, $F$ is closed.
You can use similar reasoning to show that $\{f\in Y:f(\lambda x)=\lambda f(x)\}$ is closed in $Y$.
