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.

  • $\begingroup$ What is $xy$ supposed to be? $\endgroup$ – Aweygan Nov 22 '17 at 14:31
  • $\begingroup$ is the product between $x$ and $y$ $\endgroup$ – Student Nov 22 '17 at 14:35
  • $\begingroup$ I ask because in general there is no product on a Hilbert space. $\endgroup$ – Aweygan Nov 22 '17 at 14:36
  • $\begingroup$ That isn't enough information to formulate an answer from. Is the product continuous with respect to the norm topology on $E$? Is it sub-multiplicative with respect to the norm? In any case, what are your thoughts? $\endgroup$ – Aweygan Nov 22 '17 at 14:42
  • $\begingroup$ I'm sorry, I have forget an hypothesis. I have edited my question $\endgroup$ – Student Nov 22 '17 at 14:45

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.


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