Inner Product of Dual Space
Introduction
For the vector space $V$, let’s denote $(V, \braket{\cdot, \cdot}_{V})$ as a Hilbert space. Let $V^{\ast}$ represent the dual space of $V$. By the Riesz Representation Theorem, any $f \in V^{\ast}$ can be expressed uniquely in terms of $\mathbf{v}_{f} \in V$ as follows:
$$ f = \braket{\cdot, \mathbf{v}_{f}}_{V}; \quad f(\mathbf{x}) = \braket{\mathbf{x}, \mathbf{v}_{f}}_{V} \tag{1} $$
That is, $f \in V^{\ast}$ and $\mathbf{v}_{f} \in V$ are in a one-to-one correspondence. Given that an inner product is well defined in $V$, and since $f, g \in V^{\ast}$ uniquely corresponds to an element in $V$, the inner product on $V^{\ast}$ can be naturally defined as follows.
Definition
The inner product for the dual space $V^{\ast}$ of the Hilbert space $(V, \braket{\cdot, \cdot}_{V})$ is defined as below.
$$ \braket{f, g}_{V^{\ast}} := \braket{\mathbf{v}_{f}, \mathbf{v}_{g}}_{V} , \qquad f, g \in V^{\ast} \tag{2} $$
Here, $\mathbf{v}_{f}, \mathbf{v}_{g} \in V$ refers to the vector corresponding to $f, g \in V^{\ast}$ according to the Riesz Representation Theorem.
Explanation
The definition $(2)$ can be re-expressed as follows due to $(1)$.
$$ \braket{f, g}_{V^{\ast}} = \braket{\mathbf{v}_{f}, \mathbf{v}_{g}}_{V} = g(\mathbf{v}_{f}) = \overline{f(\mathbf{v}_{g})} $$
Given an inner product, a norm is naturally defined from it. The norm of the dual space $V^{\ast}$ is as follows.
$$ \| f \|_{V^{\ast}} = \sqrt{\braket{f, f}_{V^{\ast}}} = \sqrt{\braket{\mathbf{v}_{f}, \mathbf{v}_{f}}_{V}} = \| \mathbf{v}_{f} \|_{V} $$