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

$$+:X\times X\to X,\\(x,y)\mapsto +(x,y)=x+y$$ 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.

• What is $E$? What are all these $a$s and $b$s? which set do they come from? What are the $\phi_i$s? Why is the weak topology on $E$ and not on $X$? I'm very confused.
– jgon
Mar 8, 2019 at 2:37
• @jgon: Thanks a lot for the comment. Let me make some edits. Mar 8, 2019 at 2:41
• @jgon: I made some edits but I still feel it's not very fine yet. If you feel I should change anything, kindly let me know. Thanks! Mar 8, 2019 at 2:45
• Which definition of the weak topology are you most comfortable with? Mar 8, 2019 at 3:06
• @Theo Bendit: Composition of functions and continuous pre-image definition. Mar 8, 2019 at 3:08

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.

• But we are working on infinite dimensional spaces. $\Bbb{R}$ and $\Bbb{R}^2$ are finite dimensional. I'm I missing something? Mar 8, 2019 at 2:49
• @OmojolaMicheal Hm maybe I should be more clear. What's happening here is that the $+$ inside the $\phi$ is the $+$ on $X$, but the $+$ outside the $\phi$s is the $+$ in $\Bbb{R}$. Because linear functionals are linear, we can convert the addition on the infinite dimensional space into addition on $\Bbb{R}$, which we already know is nice and continuous.
– jgon
Mar 8, 2019 at 2:50
• @OmojolaMicheal Edited more detail in.
– jgon
Mar 8, 2019 at 2:54
• Thanks @jgon. I will go through it and feed you back. If I come across any challenge, I'll let you know. Mar 8, 2019 at 2:56
• +1. Sorry, the map $X\times X\to \Bbb{R}\times \Bbb{R}$, how does it work? Mar 8, 2019 at 3:05

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 $$X$$.

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 $$X$$.