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Let $H$ be a Hilbert space, and $H'$ the dual space. Here are 2 different versions of the Riesz Representation theorem:

Every $F \in H'$ has the form $F(X) = \langle y_F, x \rangle$ for some $y_F \in H$. Moreover, $\|F\|_{H'} = \|y_F\|_H$.

The map $J: H \to H'$ defined by $(Jy) (x) : = \langle x, y \rangle$ is an isometric isomorphism.

I can't quite show the equivalence of the two, particularly how the second would imply the first. I have a feeling it's quite straightforward but I'm just missing it. Thanks!

P.S, I am also curious about why it follows that for all $A \in \mathcal{L}(H,H)$ there is a unique linear $A^*$ with $\langle Ax, y \rangle = \langle x, A^* y \rangle$.

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In regards to the last part. Define, for each $y \in H$, the linear functional $l_y$, given by $l_y(x) = \langle Ax,y \rangle$. By Riesz, for each $y$, there is some $A^*y$ such that $l_y(x) = \langle x , A^*y \rangle$ for each $x \in H$. It's then an easy exercise to check that the mapping $y \mapsto A^*y$ is linear.

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Sorry, I realised this was a really stupid question. The second implies the first because $J$ is an isometric isomorphism, so in particular it is surjective, i.e., for any $F \in H'$ there is a unique $y_F \in H$ with $F = (Jy_F) = \langle \cdot, y \rangle$. Additionally, it is isometric so $\|F\| = \|Jy_f\| = \|y_f\|$

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    $\begingroup$ Surjective is enough $\endgroup$ – mathworker21 Jun 17 '18 at 7:45

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