$+:X\times X\to X,(x,y)\mapsto +(x,y)=x+y$ and $\cdot:\Bbb{R}\times X\to X,(\lambda,x)\mapsto \cdot(\lambda,y)=\lambda\cdot x$ are weakly continuous

Question

$$+:X\times X\to X,\\(x,y)\mapsto +(x,y)=x+y$$ and $$\cdot:\Bbb{R}\times X\to X,\\(x,y)\mapsto \cdot(\lambda,y)=\lambda\cdot x$$ are weakly continuous, where $X$ is an infinite dimensional normed linear space.

My trial

Define for $\;i=1,2,$ \begin{align}\phi_i:(X,&\omega)\to (X\times X,\tau_X\times \tau_X),\\&x\mapsto \phi_i(x)=(x,y) \end{align} where $\omega$ is the weak topology on $E$. By definition of product topology, $\phi_i$ for $\;i=1,2,$ is continuous. So, \begin{align}+\circ \phi_i:(X,&\omega)\to (X,\tau_X)\\&x\mapsto x+y \end{align} is weakly continuous.

Similarly, for $\;i=1,2,$ define \begin{align}\phi_i:(X,&\omega)\to (\Bbb{R}\times X,|\cdot|\times \tau_X),\\&x\mapsto \phi_i(x)=(x,y). \end{align} By definition of product topology, $\phi_i$ for $\;i=1,2,$ is continuous. So, \begin{align}\cdot\circ \phi_i:(X,&\omega)\to (X,\tau_X)\\&x\mapsto \lambda\cdot x \end{align} is weakly continuous.

Please, I'm I right? If yes, can you please explain it to me clearly? If I'm wrong, can you please, provide another proof? I'm new to weak topology.

Answer 1

The correct argument is as follows. Let $\phi\in X^*$. Then $\phi(x+y) = \phi(x)+\phi(y)$ is continuous, since $+ : \Bbb{R}^2\to\Bbb{R}$ and $\phi$ are continuous. Hence $+$ is weakly continuous, since by the characteristic property of the weak (initial) topology, to check that a function $g:Z\to X$ is continuous (for the weak topology on $X$), it suffices to show that $\phi\circ g$ is continuous for all $\phi\in X^*$.

To draw a commutative diagram, what is going on is that we are doing the following: $$\require{AMScd}\begin{CD} X\times X & @>+>> X \\ @V\phi\times \phi VV @VV\phi V\\ \Bbb{R}\times \Bbb{R}@>+>>\Bbb{R} \end{CD} $$ We use the commutativity of this square to convert the function we want to know about, $\phi\circ +$, into $+\circ \phi\times\phi$, which we already know is continuous.

Similarly, $\phi(\lambda x) =\lambda \phi(x)$, which is continuous since multiplication in $\Bbb{R}$ and $\phi$ are both continuous. Thus $\cdot : \Bbb{R}\times X\to X$ is weakly continuous.

For more detail on this method of checking weak continuity, see the wiki article for the initial topology. It might also be worth looking at the wiki article on the weak topology as well.

Answer 2

The most straightforward approach is using nets.

Assume that $((x_\lambda, y_\lambda))_{\lambda \in \Lambda}$ converges to $(x_0, y_0) \in X \times X$ in the product weak topology on $X \times X$. We claim that $+(x_\lambda, y_\lambda) \to +(x_0, y_0)$ in the weak topology on $X$.

$(x_\lambda, y_\lambda) \to (x_0, y_0)$ is equivalent to $x_\lambda \to x_0$ and $y_\lambda \to y_0$ in the weak topology on $X$. For any $\phi \in X^*$ we have $$\phi(+(x_\lambda, y_\lambda)) = \phi(x_\lambda + y_\lambda) = \phi(x_\lambda) + \phi(y_\lambda) \to \phi(x_0) + \phi(y_0) = \phi(x_0 + y_0) = \phi(+(x_0, y_0))$$ Since $\phi \in X^*$ is arbitrary, we conclude $+(x_\lambda, y_\lambda) \to +(x_0, y_0)$ in the weak topology on $X$. Hence $+$ is continuous w.r.t the product weak topology on $X \times X$ and the weak topology on $X$.

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Assume that $((\alpha_\lambda, x_\lambda))_{\lambda \in \Lambda}$ converges to $(\alpha_0, x_0) \in \mathbb{R} \times X$ in the product of the standard topology on $\mathbb{R}$ and weak topology on $X$. We claim that $\cdot(\alpha_\lambda, x_\lambda) \to \cdot(\alpha_0, x_0)$ in the weak topology on $X$.

$(\alpha_\lambda, x_\lambda) \to (\alpha_0, x_0)$ is equivalent to $\alpha_\lambda \to \alpha_0$ in $\mathbb{R}$ and $y_\lambda \to y_0$ in the weak topology on $X$. For any $\phi \in X^*$ we have $$\phi(\cdot(\alpha_\lambda, x_\lambda)) = \phi(\alpha_\lambda x_\lambda) = \alpha_\lambda \phi(x_\lambda) \to \alpha_0 \phi(x_0) = \phi(\alpha_0 x_0) = \phi(\cdot(\alpha_0, x_0))$$ Since $\phi \in X^*$ is arbitrary, we conclude $\cdot(\alpha_\lambda, x_\lambda) \to \cdot(\alpha_0, x_0)$ in the weak topology on $X$. Hence $\cdot$ is continuous w.r.t the product of the standard topology on $\mathbb{R}$ and weak topology on $X$, and the weak topology on $X$.