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I have this problem:

Let $(X, \langle\cdot,\cdot\rangle)$ a Hilbert Space on $\mathbb{R}$ with Riez map $\mathcal{R}:X^{\prime}\rightarrow X$, define $[\cdot,\cdot]:X^{\prime}\times X^{\prime}\rightarrow\mathbb{R}$ by $$[F,G]\ :=\ \langle\mathcal{R}(F),\ \mathcal{R}(G)\rangle,\;\;\; \forall\; F,G\in X^{\prime}$$ and show that $(X^{\prime},[\cdot,\cdot])$ is a Hilbert Space.

and I need some hints about how I can solve it.

So, I know that $(X^{\prime}, \|\cdot\|_{X^{\prime}})$ is a Banach space, because every $F\in X^{\prime}$ have codomain $\mathbb{R}$. Later, as $\langle\cdot,\cdot\rangle$ is a inner product and, then $[\cdot,\cdot]$ is also a inner product if I can prove that $\mathcal{R}(\cdot)$ is linear. And finally, I need to prove that $\|\cdot\|_{X^{\prime}}$ is subordinate by $[\cdot,\cdot]$. Am I correct ?

Thanks in advance.

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up vote 2 down vote accepted

The space $X'=\mathcal{B}(X,\mathbb{C})$ is complete. This follows from the fact that $\mathbb{C}$ is complete. For details see this answer.

Since $X$ is a Hilbert space then parallelogram law holds for elements of $X$. Recall that $\mathcal{R}:X'\to X$ is isometric, then parallelogram law holds for elements of $X'$. By Jordan von Neumann theorem $X'$ is an inner product space and the original norm coincide with the norm generated by inner product.

Since $X'$ is complete inner product space, then $X'$ is a Hilbert space.

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