To begin with we know about 3 convergences Normal (U), Strong (S), Weak (W), there are operators (A) and functionals (f). $U \rightarrow SA \rightarrow WA$ and $Sf \rightarrow Wf$, is something that I know for sure.
$U:$ $$A_m \rightarrow A \Leftrightarrow ||A_m-A||\rightarrow 0$$
$SA:$ $$A_m \underset{m \rightarrow \infty}{\stackrel s \longrightarrow} A \Leftrightarrow \forall x \in X: A_m x \underset{m \rightarrow \infty}{\longrightarrow} Ax$$
$WA:$ $$A_m \underset{m \rightarrow \infty}{\stackrel w \longrightarrow} A \Leftrightarrow \forall x \in X \; \forall f \in Y^*: f(A_m x) \underset{m \rightarrow \infty}{\longrightarrow} f(Ax)$$
$Sf:$ $$f_n \stackrel S \rightarrow f \Leftrightarrow ||f_n-f|| \underset{n \rightarrow \infty}{\longrightarrow} 0$$
$Wf:$ $$f_n \stackrel w \longrightarrow f \Leftrightarrow \forall x \in X: f_n(x) \underset{n \rightarrow \infty}{\longrightarrow} f(x)$$
Then as far as I understand, the functional is a specific type of operator that translates vector to scalar.
$X, Y$ - some vector, $x$ - some scalar
$$AX = Y$$
$$f(X) = x$$
So from all this I made an assumption that
$$ \leftrightarrow Sf \rightarrow Wf \\ U\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\downarrow\;\;\;\;\;\;\;\;\;\\ \rightarrow SA \rightarrow WA $$
Sf is basically the same as U Is this right? or are there any other connections that must be in that graph, also more mathematical explanation of what is going on here is appreciated.
UPD: After a talk with my teacher he corrected me in a few places, we assume that $A:X \rightarrow Y, Y=\mathbb{R}$($\mathbb{R}$ = Real numbers) In that case $U$ and $Sf$ is basically the same because $\underset{||x|| \leq 1}{sup}|f(x)|$ and $\underset{||x|| \leq 1}{sup}||Ax||$, if $Ax \in \mathbb{R}$ then $||Ax|| = |Ax|$
So $$ \rightarrow Wf \\ (U, Sf)\;\;\;\;\;\;\;\;\;\updownarrow\;\;\;\;\;\;\;\;\;\\ \rightarrow SA \rightarrow WA $$ I still dont understand why is there a full link between $Wf$ and $WA$ and as @s.harp said why is $Wf$ and $SA$ the same.
Ok, I figured it out thanks to @s.harp and my teacher, so. The answer depends on what operator we talk about, in case of $A: X \rightarrow Y, Y = \mathbb{R}$ the operator is basically the same as functional. (The functional is a special case of operator)
So right graph will look like $$(U, Sf) \rightarrow (SA, Wf, WA)$$
Why is that?
$U=Sf$:
It's actually an axiom in my book but, as I mentioned in a question norm of functional is $\underset{||x|| \leq 1}{sup}|f(x)|$, norm of operator is $\underset{||x|| \leq 1}{sup}||Ax||$.
So if $Ax = \mathbb{R}$ then $||Ax|| = |Ax|$ And as I already mention $Ax$ is a functional in case of $Y = \mathbb{R}$ So $$A_m \rightarrow A \Leftrightarrow ||A_m-A||\rightarrow 0$$ is the same as $$f_n \stackrel S \rightarrow f \Leftrightarrow ||f_n-f|| \underset{n \rightarrow \infty}{\longrightarrow} 0$$
$Wf=SA$:
It's the same as with $U=Sf$. $A_n x$, $Ax$ is the same as $f_n (x)$ and $f(x)$ for $A:X \rightarrow Y, Y = \mathbb{R}$
$WA=Wf$:
Let's take a look at the definition of $WA$: $$A_m \underset{m \rightarrow \infty}{\stackrel w \longrightarrow} A \Leftrightarrow \forall x \in X \; \forall f \in Y^*: f(A_m x) \underset{m \rightarrow \infty}{\longrightarrow} f(Ax)$$
$Y^*$ for $Y = \mathbb{R}$ is still $\mathbb{R}$.
Now for this part, $f(A_m x) \underset{m \rightarrow \infty}{\longrightarrow} f(Ax)$
The most basic functional in $\mathbb{R}$ is $f(x)=kx$, if we take $k=1$ then we keep the functional the same, so $f(A_n x)=f(f_n(x))=f_n(x)$ and $f(Ax)=f(f(x))=f(x)$.
So that is the same as definition of Wf.
If I manage to find this graph for $A: X \rightarrow Y, Y \neq \mathbb{R}$. I'll update this.