# 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.

## 2 Answers

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!

• Wow, that's bad. The problem is that I am trying to develop the theory of bras in quantum mechanics. I tried presenting them as elements of the dual of the vector space of kets (which I called $H$), but since $H$ can in general be infinite dimensional, then is there no way out of the problem? Is there any case when even infinite-dimensional vector spaces have the property I am asking for? – Iván Mauricio Burbano May 28 '15 at 23:55
• @IvánMauricioBurbano If you consider continuous linear functionals, then for Hilbert spaces, you have self-(anti-)duality, every continuous linear functional is given by taking the inner product with a vector from the space (that's one of the Riesz representation theorems). – Daniel Fischer Jun 1 '15 at 19:23

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 $\sum_j |\phi(e_j)|^2 = \infty$"