Prove that the bilinear function $(l,x)$ gives a natural identification of $X$ with $X’’$.

Question

I’m attempting a proof of Theorem 3 below. I would like to know if my proof is valid. If it is, is there a “cleaner” proof? If it is not valid, can you push me in the right direction? Please first read Theorem 1 and Definition 1 below as they are used in my proof. Also, $X’$ denotes the dual of $X$, and $X’’$ denotes the dual of $X’$.

Theorem 1. Let $X$ be a linear space of dimension $n$. The elements $x$ of $X$ can be represented as arrays of $n$ scalars:

$$ x=(c_1,\ldots,c_n) $$

Addition and multiplication by a scalar is defined componentwise. Let $a_1,\ldots,a_n$ be any array of $n$ scalars; the function $l$ defined by

$$ l(x)=a_1c_1+\cdots+a_nc_n $$

is a linear function of $x$. Conversely, every linear function $l$ of $x$ can be so represented.

Definition 1. The right-hand side of the above equation depends symmetrically on the two arrays representing $x$ and $l$. Therefore we ought to write the left-hand side also symmetrically, we accomplish that by the scalar product notation

$$ (l,x)^{\text{def}}=l(x) $$

We call it a product because it is a bilinear function of $l$ and $x$.

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Theorem 3. The bilinear function $(l,x)$ defined above gives a natural identification of $X$ with $X’’$.

Proof. Let $X$ be a linear space; let $x_1,\ldots,x_n$ be a basis of $X$. For some field $K$, define a linear function $l:X\rightarrow K$ such that

$$ l_i(x)=l_i\left(\sum_{i=1}^{n} a_ix\right)=a_i $$

where $a_1,\ldots,a_n$ is any array of $n$ scalars. Then $l$ is in the dual $X’$, and $l_1,\ldots,l_n$ is a basis of $X’$.

By Theorem 3, Definition 1 and the above definition of $l_i(x)$, for a fixed $x=(c_1,\ldots,c_n)$ in $X$

\begin{align} l_i(x)=(l_i,x)&=a_1c_1+\cdots+a_nc_n \\&\equiv l_1c_1+\cdots+l_nc_n \end{align}

Since $(l,x) \in X’’$, we have shown that for a fixed $x \in X$, we can relate the elements of $X$ with the elements of $X’’$.

Answer 1

First of, as already has been noted, your notation isn't quite right. For fixed $x$, $(l,x)$ is just a number, it is just a rewrite of $l(x)$, the value of $l$ at $x$. This is not an element of $X''$. What you mean is that to any fixed $x$ you can assign a map, lets call it $\phi_x$, which is defined for any $l \in X'$ and acts by feeding $x$ to $l$, i.e. $$ \phi_x(l) := (l,x) = l(x) $$ I introduce this notation $\phi_x$ to distinguish between the roles of $x$ as a vector and the way $x$ corresponds to an element of $X''$. Now, proving your theorem 3 is equivalent to checking three properties of this correspondence:

• $\phi_x$ is a linear functional on $X'$, i.e. we have $\phi_x(l + \alpha m) = \phi_x(l) + \alpha \phi_x(m)$. This means that the $\phi_x$ is really an element of $X''$ and not just some random nonlinear function on $X'$
• The assignment $x \to \phi_x$ is surjective. This means that for every $\psi \in X''$ there is a $x$ s. t. $\psi = \phi_x$.
• The assignment $x \to \phi_x$ is injective. This means that $\phi_x = \phi_y$ only if $x = y$ for two vectors $x,y$.

These three conditions are a more precise statement of what it means that the assignment $x \to \phi_x$ is an identification of $X$ and $X''$. What does it mean that this is natural? I think this has two components:

1. The identification $x \to \phi_x$ is linear as a map between vector spaces
2. The construction of this identification contains no arbitrary choice (it isn't dependent on a choice of basis or anything like that).

These two are obviously the case, my definition of $\phi_x$ does not depend on any arbitrary choice.

Now, let's come back to your proof attempt. First of, there are a few other mistakes in your proof. The $l_i$ aren't properly defined, you missed a few indices on the basis elements $x_i$ and the $l$ in your defining formula. You also claim the $l_i$ are a basis of $X''$. That is true, but why? Also, the following section doesn't really make sense to me. You are not really clear on what you are trying to show. Try writing the logical flow of your argument out with clear statements of the form "I want to show: insert condition. To do this, i will do the following steps: Insert steps. Then try to write down each step clearly.