# About the proof of inner product space(uniqueness of y)

Riez representation theorem: Let V be a finite-dimensional inner product space over $$F$$, and let $$g:V \rightarrow F$$ be a linear mapping. There exists a unique vector y in V such that $$g(x)= \langle x,y\rangle$$ for all $$x \in V$$.

Proof: Let $$\beta={v_1,v_2,...,v_m}$$ be an orthonormal basis for V, and let $$y=\sum_{i=1}^n \overline{g(v_i)} v_i$$. Define $$h:v \rightarrow F$$ by $$h(x)=\langle x,y \rangle$$. Furthermore, for $$1 \leq j \leq n$$ we have \begin{align} &h(v_j) = \langle v_j,y \rangle \\ &= \left\langle v_j, \sum_{i=1}^n \overline{g(v_i)} v_i\right\rangle \\&= \sum_{i=1}^n g(v_i)\langle v_j,v_i \rangle \\&= \sum_{i=1}^n \overline{g(v_i)} \end{align}

since $$g$$ and $$h$$ both agree on $$\beta$$, we have that $$g=h$$.

to show y is unique, suppose $$g(x)=\langle x,y' \rangle$$ for all $$x$$. Then $$\langle x,y \rangle = \langle x,y'\rangle$$ for all $$x$$. Hence we have $$y=y'$$

My question: why do we let $$y=\sum_{i=1}^n \overline{g(v_i)} v_i$$, especially there is a bar over $$g(v_i)$$. Also can someone explain the general idea of the proof here, especially why we define another linear map $$h$$?

• Please write the complete problem statement. I think you're trying to prove the Ries Representation theorem for the finite-dimensional case. People can guess what you intended to ask, but I would recommend you write the precise problem statement. Mar 11, 2020 at 1:57

Let me refresh you with some results.

Recall that, as a simple consequence of the definition of inner product, the brackets $$\langle \cdot,\cdot \rangle$$ take out the scalars from the first entry, but they bring out the complex conjugates from the second, that is, $$\langle u,cv \rangle = \overline{c} \langle u,v \rangle. \tag{1}$$

Also, following the analogy that in $$\mathbb{R}^3$$ (to be illustrative, it doesn't really matter) every vector $$\mathbf{x} = (x,y,z)$$ can be written as \begin{align} \mathbf{x} &= xe_1 + ye_2 + ze_3 \\ &= \langle \mathbf{x},e_1 \rangle e_1 + \langle \mathbf{x},e_2 \rangle e_2 + \langle \mathbf{x},e_3 \rangle e_3 \end{align} where $$e_1$$, $$e_2$$ and $$e_3$$ are the standard basis vectors, then, any vector $$v$$ in an abstract inner product space can be written as $$v = \langle v,w_1 \rangle w_1 + \langle v,w_2 \rangle w_2 + \cdots + \langle v,w_n \rangle w_n \tag{2}$$ where $$w_1,w_2,\dots,w_n$$ must be form an orthonormal basis for the whole space (which is the case for our example in $$\mathbb{R}^3$$).

Now, we start with the given problem. We want to find some vector $$y \in \textsf{V}$$ such that $$g(x) = \langle x,y \rangle$$ for all $$x \in \textsf{V}$$, that is, the rule of correspondence of $$g$$ is completely determined by $$y$$.

First, choose an orthonormal basis for $$\textsf{V}$$, let's say, $$v_1,v_2,\dots,v_n$$ (this can be done since the space is finite-dimensional). Then, for any $$x \in \textsf{V}$$ we have \begin{align} g(x) &= g(\langle x,v_1 \rangle v_1 + \langle x,v_2 \rangle v_2 + \cdots + \langle x,v_n \rangle v_n) \tag{3} \\ &= \langle x,v_1 \rangle g(v_1) + \langle x,v_2 \rangle g(v_2) + \cdots + \langle x,v_n \rangle g(v_n) \tag{4} \\ &= \langle x,\overline{g(v_1)}v_1 \rangle + \langle x,\overline{g(v_2)}v_2 \rangle + \cdots + \langle x,\overline{g(v_n)}v_n \rangle \tag{5} \\ &= \langle x, \overline{g(v_1)}v_1 + \overline{g(v_2)}v_2 + \cdots + \overline{g(v_n)}v_n \rangle \tag{6} \end{align} where $$(3)$$ happens by our observation $$(2)$$, $$(4)$$ by the linearity of $$g$$, $$(5)$$ by $$(1)$$ since $$g(v_1),\dots,g(v_n)$$ are scalars, and $$(6)$$ follows from the linearity in the second entry of the inner product.

So, letting $$y := \overline{g(v_1)}v_1 + \overline{g(v_2)}v_2 + \cdots + \overline{g(v_n)}v_n$$ in the last step we have that $$g(x) = \langle x,y \rangle$$, as desired. To show the uniqueness, suppose there are $$y'$$ such that $$g(x) = \langle x,y' \rangle$$ for any $$x$$. Then, for any $$x \in \textsf{V}$$, $$\langle x,y-y' \rangle = \langle x,y \rangle - \langle x,y' \rangle = 0.$$ In particular, for $$x=y-y'$$ we see that $$\langle y-y',y-y' \rangle = 0$$, which only happens when $$y-y'=0$$, that is, $$y'$$ must be the same as $$y$$.

First of all, based on your proof someone can guess that your field $$F$$ is the complex number field $$\mathbb{C}.$$ Note that the inner product is a linear operator on its first component, and it's conjugate linear on its second component. Also, the following equation is absolutely incorrect.

$$\sum_{i=1}^n g(v_i) = \sum_{i=1}^n \overline{g(v_i)}$$

I think you've wanted to write $$\sum_{i=1}^n g(v_i) = g(v_j)$$ which is true because of choice of orthonormal basis. By the way, the bar is irrelevant if your field is real. The whole idea of this proof is to prove that any functional on a finite-dimensional vector space can be written as an inner product. In fact, it's true for any Hilbert space. I hope this helps. Also, there is an issue with indexing $$m$$ and $$n.$$ Please clearly state your problem to receive more help from other experts.

• You might want to use \langle and \rangle for the inner product, it looks better $\langle \cdot , \cdot \rangle$. Mar 11, 2020 at 3:03