Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A theorem I am reading (about the existence and uniqueness of solutions to Sturm-Liouville intial-value problems) defines a space $B$ consisting of the continuous functions defined on a closed real interval $[a,b]$ and assuming values given by complex matrices $m \times n$.

The author then defines what he calls a Bielecki norm and says that "it is clear" that with this norm $B$ is a Banach space. I am aware of the definition of a Banach space but fail to see why the claim is true. I have tried to find a limit to an arbitrary Cauchy sequence, but without luck so far.

Here it is:

$$ \| Y \| = \sup \left\{|Y(t)| \exp\left(-K \int_a^t |P(s)|ds\right) : K > 1 \text{ and constant, } a \leq t \leq b\right\} $$

To clarify, the Euler constant is raised to the power minus K times the integral from a to t of the norm of a fairly arbitrary complex-valued matrix with variable entries. The matrix arises from the context of the theorem, but is merely defined as being a square matrix of Lebesgue-integrable complex-valued functions. The matrix norm, in case of Y and P on the RHS of the above, is defined as the sum of absolute values of the matrix entries. This is as distinct, of course, from the Bielecki norm of Y expressed on the LHS.

Thanks if you can help (or try)!

share|improve this question
    
What exactly does $\int_a^t (\det P)$ mean? I.e., with respect to what variable does one evaluate the integral? –  JavaMan Apr 19 '11 at 3:27
    
"A theorem I am reading...": Reading where, exactly? –  Jonas Meyer Apr 19 '11 at 3:30
    
The theorem is in "Sturm-Liouville Theory" by Anton Zettl, Theorem 1.2.1. However, in this context he invokes A. Bielecki, Une remarque sur la methode de Banach-Cacciopoli-Tikhonov dans la theorie des equations differentielles ordinaires, Bull. Acad. Polon. Sci. Cl. III 4(1956), 261-264. (Seminal, apparently.) –  Josef K. Apr 19 '11 at 3:38
1  
$\| Y\| \geq e^{-K\| P\| _{L^1}}|Y(t)|$ so $\| Y \| \geq C\| Y \| _{\infty}$, and it's clear that $\| Y\| \leq \| Y \| _{\infty}$ which gives that your norm is equivalent to one that gives a Banach space. ($\| . \| _{\infty}$ is the supremum norm) –  Jose27 Apr 19 '11 at 5:23
1  
Marko Stojovic: $B$ is the space of continous functions on a (I assume) compact interval, so the a.e. is unnecessary. To define it, give the space of matrices any norm $\| . \|$ (they're all equivalent) and define $\| Y\| _{\infty} = \sup_t\{ \| Y(t) \| \}$. In particular, for the calculations I did I used $\| A\| = \sum_{i,j} |a_{ij}|$. –  Jose27 Apr 19 '11 at 17:36
show 3 more comments

1 Answer

up vote 3 down vote accepted

The definition of the norm can be simplified.

$$\| Y \| = \sup \left\{|Y(t)| \exp\left(- \int_a^t |P(s)|ds\right) : a \leq t \leq b\right\}$$

Proof: Suppose $r < \| Y \|$. Then $r < |Y(t)|\exp\left(- K\int_a^t |P(s)|ds\right)$ for some $K > 1$ and $a \leq t \leq b$. But then $r < |Y(t)|\exp\left( - \int_a^t |P(s)|ds\right)$ and so $r$ is less than the right hand side of the equation above.

The converse can be shown easily.

So we can rewrite the norm:

$$\| Y \| = \sup \left\{|Y(t)|f(t) : a \leq t \leq b\right\}$$

where $f : [a,b] \to [r,1]$ is a continuous decreasing surjection with $r = \exp\left(-\int_a^b |P(s)|ds\right)$. Now suppose $\langle Y_n : n\in \mathbb{N}\rangle$ is a Cauchy sequence with respect to $\|\cdot\|$. Then I claim this sequence has a limit, and that this limit is the entry-wise limit of the $Y_n$.

First let's see that the entry-wise limit exists: Since $C[a,b]$ is complete, it suffices to show that the entry-wise sequences are Cauchy. Suppose not, that is suppose for some $i,j$, we have that $(Y_n)_{i,j}$ is not Cauchy. Then $$\lim _{N\to\infty}\sup\{\sup\{|(Y_m)_{i,j}(x) - (Y_n)_{i,j}(x)| : x \in [a,b]\} : m,n > N\} = L$$

for some $L > 0$. Now fix $N > 0$. Pick $m,n > N$ such that $\sup _{x\in [a,b]}|(Y_m)_{i,j}(x) - (Y_n)_{i,j}(x)| > L$. So we can pick $x \in [a,b]$ such that $|(Y_m)_{i,j}(x) - (Y_n)_{i,j}(x)| > L$. But then $|(Y_m-Y_n)(x)|>L$. Then $|(Y_m-Y_n)(x)|f(x) > Lr$, and so $\|Y_m - Y_n\| > Lr$. This contradicts the assumption that the $Y_n$ formed a Cauchy sequence.

Now that we know the entry-wise limit $Y$ exists, we have to show that the $Y_n$ converge to $Y$ with respect to $\|\cdot\|$. Well $$\|Y - Y_n\| \leq \sup _{x \in [a,b]}|Y(x) - Y_n(x)| \leq \sum_{i,j} \sup_x |Y_{i,j}(x) - (Y_n)_{i,j}(x)|$$ Since for each $i,j$ we have that $\lim _n |Y_{i,j}(x) - (Y_n)_{i,j}(x)| = 0$, the right hand side of the above inequality goes to $0$ as $n$ goes to $\infty$.

share|improve this answer
    
(I left the following comment earlier but then deleted it in error.) Many thanks for your clear and detailed proof. I will credit you in my essay! –  Josef K. Apr 20 '11 at 3:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.