Prove a space is Hilbert I got stucked in this problem and get no clue to solve this. Can any one please help me? Thanks     Suppose $X$ is an inner product space. If for every bounded linear function $f$, there exists $z \in X$ such that $$f(x) = &lt;x, z&gt;$$ for all $x \in X$. Prove that $X$ is a Hilbert space.   I can't find any way to relate the complete of space $X$ with the existence of $z$. Please help me solve this. I really appreciate.
Show that if the Riesz map is surjective on $H$, then $H$ is a Hilbert space Let $H$ be a vector space equipped with an inner product $(\cdot, \cdot)$ and $f:H\to H',\ f(x)=(\cdot,x)$ surjective. Now, why $H$  is a Hilbert space? The other direction is clear by Riesz' representation theorem but what about this?
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