Norm of Integrals: Bounding the Matrix Product

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SUMMARY

The discussion focuses on bounding the norm of the integral of the product of two continuous matrices, A(t) and B(t), defined on the interval [0,1]. It establishes that if the matrices are of sizes m x n and n x k respectively, and the Frobenius norm is used, then the inequality ||integral (AB)|| ≤ ||B(t)|| * integral (A) holds under specific conditions. The proof utilizes Minkowski's inequality and the submultiplicative property of the Frobenius norm to derive the conclusion definitively.

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sarrah1
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Hi
I have an integral over [0,1] of product of two matrices say A(t). B(t) and I wish to bound its norm. Can you say that
||integral (AB)||<||B(t)||.||integral (A)|.
is there some conditions on that to occur
thanks sarrah
 
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Hi Sarrah,

Suppose $A(t)$ and $B(t)$ are continuous on $[0,1]$, of sizes $m \times n$ and $n \times k$, respectively. If the matrix norm is Frobenius, then

\[
\left\|\int_0^1 A(t) B(t)\, dt\right\| \le (\max_{t\in [0,1]} \|B(t)\|) \int_0^1 \|A(t)\|\, dt.
\]

To see this, note that by Minkowski's inequality,

\[
\left\|\int_0^1 A(t)B(t)\, dt\right\| \le \int_0^1 \|A(t)B(t)\|\, dt \qquad (1)
\]

Since the Frobenius norm is submultiplicative,

\[
\|A(t)B(t)\| \le \|A(t)\| \|B(t)\| \le (\max_{t\in [0,1]} \|B(t)\|) \|A(t)\|
\]

for all $t$ in $[0,1]$. Hence

\[
\int_0^1 \|A(t)\| \|B(t)\|\, dt \le (\max_{t\in [0,1]} \|B(t)\|) \int_0^1 \|A(t)\|\, dt. \qquad (2)
\]

The result is obtained by combining (1) and (2).
 

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