MHB Norm of Integrals: Bounding the Matrix Product

[sarrah1]

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

 

[Euge]

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).