A question about the uniqueness of Riesz representation theorem I am sorry this question may be too math related. However, I come from a physics background and I would like to ask for a physicist's explanation.
As far as I know, the Riesz representation theorem guarantees the unique correspondence between ket and bra vectors.
The proof can be found at http://planetmath.org/proofofrieszrepresentationtheorem.
The logic of the proof goes like following:
suppose the correspondence is not unique, we have $u_1$ and $u_2$ for every $ | x \rangle \in H$, where $H$ is a separable Hilbert space. We have:
$$ \langle u_1 | x \rangle =   \langle u_2 | x \rangle  $$
hence
$$ \langle u_1 - u_2 | x \rangle =0 $$
since vector $| x \rangle$ is arbitrary, we can set $| x \rangle = |u_1 - u_2 \rangle$
therefore
$$ \langle u_1 - u_2 | u_1 - u_2  \rangle =0   $$
hence $| u_1 - u_2 \rangle = 0$ and $|u_1 \rangle = | u_2 \rangle$, which is against our assumption.
My question is not about the possible extension to rigged Hilbert space, but about the proof, why do I require $u_1$ and $u_2$ for every $ | x \rangle  \in H$, rather than $\exists  | x \rangle $? If there $\exists  | x \rangle  \in H$ for two distinct bra vectors, the Riesz representation theorem will not guarantee the unique correspondence between ket and bra vectors.
 A: Dealing with rigged Hilbert spaces, the space of bras and the space of kets are not isomorphic, though the space of ket vectors is identified with a subspace of bra vectors.
Let us focus on the simplest case: ${\cal H} = L^2(\mathbb R)$. Here, the space of bra vectors is the space of Schwartz distributions ${\cal S}'(\mathbb R) \supset {\cal H}$, whereas the space of kets is the space of Schwartz test functions ${\cal S}(\mathbb R)\subset {\cal H}$.
It is possible to extend a bit the latter when dealing with distributions in ${\cal S}'(\mathbb R)$ with compact support, like $\delta(x-x_0)$, since they work on the whole space of smooth functions $C^\infty(\mathbb R)\supset {\cal S}(\mathbb R)$, which includes the exponentials $e^{ikx}$ in particular.
There is a continuous  embedding (see below) of the space of kets into the space of bras ${\cal S}(\mathbb R) \subset {\cal S}'(\mathbb R)$, but it is false that a bra vector is also a ket vector. Thus there is no  full version of Riesz' lemma.
Your uniqueness issue is easy. If $u_1,u_2 \in {\cal S}'(\mathbb R)$ are such that $\langle u_1|v\rangle = \langle u_2|v\rangle $ for every $v \in {\cal S}(\mathbb R)$, then $u_1=u_2$. This fact is trivially definitory because distributions are (continuous) linear functionals on ${\cal S}(\mathbb R)$ and two function(al)s coincide if and only if they coincide when applied to every element of their domain.
A bit more difficult issue is to prove that the map ${\cal S}(\mathbb R) \ni u \mapsto T_u \in {\cal S}'(\mathbb R)$ where
$$T_u(v) := \int \overline{u(x)}~v(x)~ dx \quad, \forall v \in {\cal S}(\mathbb R)$$
identifying ket vectors with bra vectors is injective.
This is the continuous embedding mentioned above. However, since  ${\cal S}(\mathbb R)$ is dense in $L^2(\mathbb R)$, $T_u(v)=T_{u'}(v)$ for every $v\in {\cal S}(\mathbb R)$ implies $u-u'=0$ almost everywhere with respect to the Lebesgue measure on $\mathbb R$. Since a continuous function is everywhere zero if it is almost everywhere zero with respect to the Lebesgue measure, we have that $u-u'=0$ everywhere and thus $u=u'$.
