Is this map continuous? Let $(E,\langle\cdot\;,\;\cdot\rangle)$ be a complex Hilbert space. For $M\in\mathcal{L}(E)^+$ (i.e. $M^*=M$ and $\langle M x\; |\;x\rangle\geq0$, for all $x\in E$). We define 
the semi-inner product $\langle x\;,\;y\rangle_M:=\langle Mx\;,\;y\rangle,\; \forall x,y\in E$. 

Assume that $\forall x,y\in E$, we have $x\cdot y\in E$ and $\langle x\cdot y\;,\;x\cdot y\rangle_M\leq \langle x\;,\;x\rangle_M\times \langle  y\;,\;y\rangle_M$. We consider the following map:
  \begin{eqnarray*}
              \psi
              :&(E,\langle\cdot\;,\;\cdot\rangle_M)\times (E,\langle\cdot\;,\;\cdot\rangle_M)&\longrightarrow (E,\langle\cdot\;,\;\cdot\rangle_M)\\
              &(x,y)&\longmapsto x\cdot y\;
              \end{eqnarray*}
  Is $\psi$ continuous?

Thank you.

Let's write $\|x\|_M=\langle Mx,x\rangle^{1/2}$ for $x\in E$.  Then for $x,y,x_0,y_0\in E$ we have 
$$\|x\cdot y-x_0\cdot y_0\|_M\leq\|x\|_M\|y-y_0\|_M+\|y_0\|_M\|x-x_0\|_M.$$
Now if $0<\varepsilon<1$, $\|x-x_0\|_M<\varepsilon$ and $\|y-y_0\|_M<\varepsilon$, then $\|x\|_M<\|x_0\|_M+1$.  Then we have 
$$\|x\cdot y-x_0\cdot y_0\|_M<(\|x_0\|_M+\|y_0\|_M+1)\varepsilon.$$
Thus if instead we suppose $\|x-x_0\|_M<\varepsilon/(\|x_0\|_M+\|y_0\|_M+1)$ and $\|y-y_0\|_M<\varepsilon/(\|x_0\|_M+\|y_0\|_M+1)$, then we have 
$$\|x\cdot y-x_0\cdot y_0\|_M<\varepsilon.$$
From this, continuity of the product is easily seen.
