If we are given a topological semigroup $G$ (i.e. it just has an associative operation "$\cdot$", without other properties). Now we want to verify this operation $"\cdot"$ is continuous. We denote a function $(x,\cdot)$ by $f_x(\cdot)$(i.e. $f_x(y)=x\cdot y$.)
I want to prove a claim: $(\cdot,\cdot):X\times X\rightarrow X$ is continuous if and only if $f_x:X\rightarrow X$ is continuous for all $x\in X$.
Is it right? How can we prove it? Or could you give some counterexample? Thank you very much.
This may fail even for groups.
Consider the set $B(H)$ of all bounded, linear transformations on the Hilbert space $H=\ell ^2({\mathbb Z})$. The so called weak operator topology (WOT) on $B(H)$ is defined to be the coarsest topology relative to which the maps $$ T\mapsto \langle T(x), y\rangle $$ are continuous, for every $x$ and $y$ in $H$.
Define $\text{GL}(H)$ to be the group of all invertible elements of $B(H)$ (the multiplication operation being the composition of operators), equipped with the induced WOT topology.
It is then an easy exercise to prove that multiplication is separately continuous. However it is not jointly continuous, and here is why. Consider the bilateral shift operator $B$, defined on the standard orthonormal basis $\{e_n\}_{n\in {\mathbb Z}}$ of $H=\ell ^2({\mathbb Z})$ by $$ B(e_n)=e_{n+1}, \quad \forall n\in {\mathbb Z}. $$
It may be easily proved that the powers $B^n$ converge to zero in the WOT as $n\to \pm \infty $. Consequently the invertible operators $$ T_n = B^n+2I $$ converge to $2I$ and hence $$ (\lim_{n\to \infty }T_n) (\lim_{n\to \infty }T_{-n}) = (2I)(2I) = 4I. $$ However $$ T_nT_{-n} = (B^n+2I)(B^{-n}+2I) = I + 2B_n + 2 B^{-n} + 4I \to 5I, $$ so multiplication is not jointly continuous.