Hahn-Banach From Systems of Linear Equations

[bolbteppa]

In this paper on the history of functional analysis, the author mentions the following example of an infinite system of linear equations in an infinite number of variables $c_i = A_{ij} x_j$:

\begin{align*}
\begin{array}{ccccccccc}
1 & = & x_1 & + & x_2 & + & x_3 & + & \dots \\
1 & = & & & x_2 & + & x_3 & + & \dots \\
1 & = & & & & & x_3 & + & \dots \\
& \vdots & & & & & & & \ddots
\end{array} \to \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \end{bmatrix} = \begin{bmatrix}
1 & 1 & 1 & \dots \\
& 1 & 1 & \dots \\
& & 1 & \dots \\
& & & \ddots
\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \end{bmatrix}
\end{align*}

as an example of a system such that any finite truncation of the system down to an $n \times n$ system has a unique solution $x_1 = \dots = x_{n=1} = 0, x_n = 1$ but for which the full system has no solution.

This book has the following passage on systems such as this one:

The Hahn-Banach theorem arose from attempts to solve infinite systems of linear equations... The key to the solvability is determining "compatibility" of the system of equations. For example, the system $x + y = 2$ and $x + y = 4$ cannot be solved because it requires contradictory things and so are "incompatible". The first attempts to determine compatibility for infinite systems of linear equations extended known determinant and row-reduction techniques. It was a classical analysis - almost solve the problem in a finite situation, then take a limit. A fatal defect of these approaches was the need for the (very rare) convergence of infinite products."​

and then mentions a theorem about these systems that motivates Hahn-Banach:

Theorem 7.10.1 shows that to solve a certain system of linear equations, it is necessary and sufficient that a continuity-type condition be satisfied.

Theorem 7.10.1 The Functional Problem Let $X$ be a normed space over $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$, let $\{x_s \ : \ s \in S \}$ and $\{ c_s \ : \ s \in S \}$ be sets of vectors and scalars, respectively. Then there is a continuous linear functional $f$ on $X$ such
that $f(x_s) = c_s$ for each $s \in S$ iff there exists $K > 0$ such that
\begin{align*}
|\sum_{s \in S} a_s c_s | \leq K || \sum_{s \in S} a_s x_S || \ \ \ \ (1),
\end{align*}
for any choice of scalars $\{a_s \ : \ s \in S \}$ for which $a_s = 0$ for all but finitely many $s \in S$ ("almost all" the $a_s = 0$).

Banach used the Hahn-Banach theorem to prove Theorem 7.10.1 but Theorem 7.10.1 implies the Hahn-Banach theorem: Assuming that Theorem 7.10.1 holds, let $\{ x_s \}$ be the vectors of a subspace $M$, let $f$ be a continuous linear functional on $M$; for each $s \in S$, let $c_s = f(x_s)$. Since $f$ is continuous, $(1)$ is satisfied and $f$ possesses a continuous extension to $X$.​

My question is:

1. If you knew none of the theorems just mentioned, how would one begin from the system $c_i = A_{ij} x_j$ at the beginning of this post and think of setting up the conditions of theorem 7.10.1 as a way to test whether this system has a solution?
2. How does this test show the system has no solution?
3. How do we re-formulate this process as though we were applying the Hahn-Banach theorem?
   
4. Does anybody know of a reference for the classical analysis of systems in terms of infinite products?

 

[Svein]

Does anybody know of a reference for the classical analysis of systems in terms of infinite products?

Since any product with one factor = 0 will have the value 0, the usual way is write the product as $\prod_{n=1}^{\infty}(1+a_{n})$. Then

A necessary and sufficient condition for the absolute convergence of the product $\prod_{n=1}^{\infty}(1+a_{n})$ is the convergence of the series $\sum_{n=1}^{\infty}\lvert a_{n}\rvert$

 

[Erland]

A necessary and sufficient condition for the absolute convergence of the product $\prod_{n=1}^{\infty}(1+a_{n})$ is the convergence of the series $\sum_{n=1}^{\infty}\lvert a_{n}\rvert$

What does it mean that a product converges absolutely?

 

[Svein]

What does it mean that a product converges absolutely?

Again, citing from Ahlfors:

An infinite product is ($\prod_{1}^{\infty}(1+a_{n})$) said to be absolutely convergent if and only if the corresponding series $\sum_{n=1}^{\infty}\log(1+a_{n})$ converges absolutely.

 

[WWGD]

Again, citing from Ahlfors:

An infinite product is ($\prod_{1}^{\infty}(1+a_{n})$) said to be absolutely convergent if and only if the corresponding series $\sum_{n=1}^{\infty}\log(1+a_{n})$ converges absolutely.

I see, and we this result too , towards finding a holomorphic function with a prescribed sequence of zeros, right?

 

[Svein]

I see, and we this result too , towards finding a holomorphic function with a prescribed sequence of zeros, right?

Yes. Example: $\prod_{n=1}^{\infty}(1+\frac{z}{n})$ has all negative integers for zeros.

 

[WWGD]

Yes. Example: $\prod_{n=1}^{\infty}(1+\frac{z}{n})$ has all negative integers for zeros.

And convergence of the product guarantees it is holomorphic, right?

 

[Svein]

And convergence of the product guarantees it is holomorphic, right?

Hm. Since $\frac{1}{n}$ diverges, we need something extra. One way is to introduce a converging factor: $\prod_{n=1}^{\infty}(1+\frac{z}{n})e^{-\frac{z}{n}}$ will converge. Another way is to extend the view a little: $z\prod_{n\neq 0}(1+\frac{z}{n})e^{-\frac{z}{n}}$. Now all integers are zeros...

But the last product can be manipulated a bit. If we combine the elements for n and -n, we end up with the following product $z\prod_{n=1}^{\infty}(1-\frac{z^{2}}{n^{2}})$ which converges (and, incidentally $z\prod_{n=1}^{\infty}(1-\frac{z^{2}}{n^{2}})=\sin(\pi z)$).

 

[WWGD]

Hm. Since $\frac{1}{n}$ diverges, we need something extra. One way is to introduce a converging factor: $\prod_{n=1}^{\infty}(1+\frac{z}{n})e^{-\frac{z}{n}}$ will converge. Another way is to extend the view a little: $z\prod_{n\neq 0}(1+\frac{z}{n})e^{-\frac{z}{n}}$. Now all integers are zeros...

But the last product can be manipulated a bit. If we combine the elements for n and -n, we end up with the following product $z\prod_{n=1}^{\infty}(1-\frac{z^{2}}{n^{2}})$ which converges (and, incidentally $z\prod_{n=1}^{\infty}(1-\frac{z^{2}}{n^{2}})=\sin(\pi z)$.

Ah, yes, I had forgotten about weighing factors. I think that does it. Thanks.

 

[bolbteppa]

Do references even exist that try to study systems like
\begin{align*}
\begin{array}{ccccccccc}
1 & = & x_1 & + & x_2 & + & x_3 & + & \dots \\
1 & = & & & x_2 & + & x_3 & + & \dots \\
1 & = & & & & & x_3 & + & \dots \\
& \vdots & & & & & & & \ddots
\end{array} \to \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \end{bmatrix} = \begin{bmatrix}
1 & 1 & 1 & \dots \\
& 1 & 1 & \dots \\
& & 1 & \dots \\
& & & \ddots
\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \end{bmatrix}
\end{align*}
like you were doing elementary linear algebra and slowly getting more theoretical?