Let me first fix definitions and notation. Let $X$ be a normed space over $\mathbb{K}$ (either $\mathbb{R}$ or $\mathbb{C}$) and $X^*$ its dual space. Let $$J_X : X\to X^{**}, J_X(x)(f) = f(x)$$ denote the natural inclusion from $X$ to $X^{**}$.
The weak topology $\tau_W$ is the smallest topology on $X$ making all elements of $X^*$ continuous. The weak-* topology $\tau_W^*$ is the smallest topology on $X^*$ making all elements of $J_X(X)$ continuous. Another equivalent characterisation is that $\tau_W$ consists of arbitrary unions of finite intersections of sets of the form $f^{-1}(U)$ for $f\in X^*$, $U\subset \mathbb{K}$ open.
My question: Are these topologies "translation-invariant"? In other words, if $U$ is an open set, is $x+U$ open as well?
My motivation: In a functional analysis class I am attending, we always reduce proofs to neighbourhoods of $0$. For example, we prove that $J_X : (X,\tau_W) \to (X^{**},\tau_W^*)$ (where $\tau_W^*$ is the weak-* topology on $X^{**}$) is a continuous map, using that $\tau_W^*$-neighbourhoods of $0$ have the form $$V^{**} = \{x^{**}\in X^{**} : |x^{**}(x_i^*)| < \varepsilon_i, \forall i=1,\ldots,k\}$$ for fixed $x_1^*,\ldots,x_k^*\in X^*$ and $\varepsilon_1,\ldots,\varepsilon_k>0$. Then $$J_X^{-1}(V^{**}) =\{x\in X : |x_i^*(x)| < \varepsilon_i, \forall i=1,\ldots,k\}$$ which is an $\tau_W$-open neighbourhood of $0$ in $X$. Why is it enough to consider neighbourhoods of $0$?
My ideas: Purely intuitively, it makes sense to only consider neighbourhoods of $0$, since all functionals are linear and we can simply "shift back" to $0$. However, I am struggling to come up with a clean argument. We also have the equivalence that a linear functional is norm-continuous iff it is norm-continuous at $0$, but this has not much to do with the $\tau_W$ topology.
In fact the translation is a homeomorphism,which can be extended to more general situations.There are two ways to solve the question.One is to use the neighborhood basis and the other is use the Characteristic properties of the weak topology as follows:
$\quad$ Suppose $(X,\tau)$ is a topology space and $\{(Y_i,\tau_i)\}_{i\in I}$ is a family of topological spaces.$f_i:X\to Y_i$ is continuous.Then $\tau$ is the weak topoloy on X about $\{f_i\}_{i\in I}$ iff $$\forall g:Z\to X\ \text{is continuous},\forall i\in I,f_i\circ g\text{ is continous}, \implies g\text{ is continuous}$$
We can use this to prove a stronger proposition:
$\quad$ Suppose X is a normed linear space and $\mathscr{A}$ is the weak topology,then the addition $\phi:(X,\mathscr{A})\times(X,\mathscr{A})\to(X,\mathscr{A})$ is continuous.
$\quad$proof(way 2)$\quad\forall f\in X^*,f:(X,\mathscr{A})\to \mathbb{K}$ is continuous.Then $$(f,f):(X,\mathscr{A})\times(X,\mathscr{A})\to\mathbb{K}\times\mathbb{K},(x,y)\to(f(x),f(y))\text{ is continuous.}$$ $$\because \varphi :\mathbb{K}\times\mathbb{K}\to\mathbb{K},(a,b)\mapsto a+b\text{ is continuous}.$$
$$\therefore f\circ\phi=\varphi\circ(f,f)\text{ is continuous}.$$ According to the the Characteristic properties of the weak topology,$\phi$ is continuous.END.
$\quad$proof(way 1).If $x\in X,f_1,...,f_n\in X^*,\epsilon>0$, $$V_{x}(f_1,...,f_n;\epsilon):=\{y\in X:|f_i(x-y)|<\epsilon,i=1,...,n \},$$ then
$$\{V_{x}(f_1,...,f_n;\epsilon):f_1,...,f_n\in X^*,n\in\mathbb{N}_+,\epsilon>0\}$$
is a neighborhood basis of X at x.
Suppose $$\phi:X\times X\to X,(x,y)\mapsto x+y.$$
$$\forall x_0,y_0\in X,\forall V_{(x_0,y_0)}(f_1,...,f_n;\epsilon),$$
take $$V_1:=V_{x_0}(f_1,...,f_n;\epsilon/2),V_2:=V_{y_0}(f_1,...,f_n;\epsilon/2).$$
Then $\forall (x,y)\in V_1\times V_2,\forall i,$
$$||f_i(x-x_0)||<\epsilon/2,||f_i(y-y_0)||<\epsilon/2,$$ $$\implies ||f_i(x+y-x_0-y_0)||\le||f_i(x-x_0)||+||f_i(y-y_0)||<\epsilon.$$
So, $x+y\in V_{(x_0,y_0)}(f_1,...,f_n;\epsilon)$,namely $V_1\times V_2\subseteq\phi^{-1}(V_{(x_0,y_0)}(f_1,...,f_n;\epsilon)).$
Thereore $\phi$ is continuous at $(x_0,y_0)$.END
Similiar, we can prove the scalar multiplication is continuous under the weak topology or the weak * topology.