On dual spaces and inner products Let V be a vector space over $\mathbb{C}$ equipped with an inner product $\langle\,  , \rangle:V\times V\mapsto\mathbb{C}$. I need to prove that any linear function $\phi:V\mapsto\mathbb{C}$ (element of the dual space) can be consider as the inner product of a vector $\vec{v}$ with the rest of the space. That is, for every $\phi \in V^*$, there exists a vector $\vec{v}\in V$ such that for every $\vec{u}\in V$ we have $\phi(\vec{u})=\langle\vec{v},\vec{u}\rangle$. I don't even know where to start! Any help s appreciated.  
 A: Hint: Consider the map $V\to V^\ast$ sending $\vec{v}$ to $\langle\vec{v},-\rangle$.  Show that it is injective, and conclude that it is surjective.
Alternatively, you could also try to explicitly construct such a vector.  To start, choose an orthonormal basis.
Note: This is false when $V$ is infinite-dimensional!
A: Not any linear function $\phi:V\to\mathbb{C}$ can be consider as the inner product of a vector v⃗ 
with the rest of the space. But, any bounded linear function (and only this) could be. 
This is the Riesz Representation Theorem. I can see this result more easily when I fix an orthonormal basis (total). Let be $\{e_j\}$ an orthonormal basis. The key is that 
"$\phi$ is bounded/continuous if, and only if $\sum_j |\phi(e_j)|^2 < \infty$" (the proof of this follows Cauchy-Schawarz inequality). 
In affirmative case ($\phi$ continous), the vector $v = \sum_j \overline{\phi(e_j)}e_j$ is in the space and then $\phi(w) = \langle w,v \rangle$. 
In negative case ($\phi$ not continous), if there exists a vector $v=\sum_j \langle v,e_j\rangle e_j$ such that $\phi(w) = \langle w,v \rangle$ than $\phi(e_j) = \langle e_j, v \rangle$ and then $v = \overline{\phi(e_j)}e_j$. But, this vector is impossible because $\sum_j |\phi(e_j)|^2 = \infty$"   
