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Inner Product of Dual Space 📂Hilbert Space

Inner Product of Dual Space

Introduction

For the vector space VV, let’s denote (V,,V)(V, \braket{\cdot, \cdot}_{V}) as a Hilbert space. Let VV^{\ast} represent the dual space of VV. By the Riesz Representation Theorem, any fVf \in V^{\ast} can be expressed uniquely in terms of vfV\mathbf{v}_{f} \in V as follows:

f=,vfV;f(x)=x,vfV(1) f = \braket{\cdot, \mathbf{v}_{f}}_{V}; \quad f(\mathbf{x}) = \braket{\mathbf{x}, \mathbf{v}_{f}}_{V} \tag{1}

That is, fVf \in V^{\ast} and vfV\mathbf{v}_{f} \in V are in a one-to-one correspondence. Given that an inner product is well defined in VV, and since f,gVf, g \in V^{\ast} uniquely corresponds to an element in VV, the inner product on VV^{\ast} can be naturally defined as follows.

Definition

The inner product for the dual space VV^{\ast} of the Hilbert space (V,,V)(V, \braket{\cdot, \cdot}_{V}) is defined as below.

f,gV:=vf,vgV,f,gV(2) \braket{f, g}_{V^{\ast}} := \braket{\mathbf{v}_{f}, \mathbf{v}_{g}}_{V} , \qquad f, g \in V^{\ast} \tag{2}

Here, vf,vgV\mathbf{v}_{f}, \mathbf{v}_{g} \in V refers to the vector corresponding to f,gVf, g \in V^{\ast} according to the Riesz Representation Theorem.

Explanation

The definition (2)(2) can be re-expressed as follows due to (1)(1).

f,gV=vf,vgV=g(vf)=f(vg) \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 VV^{\ast} is as follows.

fV=f,fV=vf,vfV=vfV \| f \|_{V^{\ast}} = \sqrt{\braket{f, f}_{V^{\ast}}} = \sqrt{\braket{\mathbf{v}_{f}, \mathbf{v}_{f}}_{V}} = \| \mathbf{v}_{f} \|_{V}