A question about why a space under a certain norm is complete

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)!

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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
$\| Y\| \geq e^{-K\| P\| _{L^1}}|Y(t)|$ 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
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

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 $\infty$.

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(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