2
$\begingroup$

Let $\Lambda \in \mathbb{R}^{n\times n}$ be a diagonal matrix with positive entries, $B\in\mathbb{R}^{n\times m}$ an arbitrary real matrix, and $u: \mathbb{R} \to \mathbb{R}^{m}$ some continuous, vector-valued, univariate function. Throughout, I mean the spectral norm when using $\| \cdot \|$.

Now I am trying to find an (upper) bound for the following expression, $$ \left\| \int_0^t e^{-(t-\tau)\Lambda} B u(\tau)\, d\tau \right\|$$, in terms of some norm of $u$.

In the scalar-valued case ($n=m=1$), I could use Cauchy-Schwarz (unless I am missing something) to arrive at a bound $$ \left| \int_0^t e^{-(t-\tau)\lambda} b u(\tau)\, d\tau \right| \leq |b| \sqrt{\frac{1-e^{-2\lambda t}}{2\lambda}} \sqrt{\int_0^t |u(\tau)|^2 \, d\tau}$$ which I would be rather happy with (I guess that would even be tight ... so no reason to be unhappy).

Now I am wondering if this can be generalized to the matrix-valued case. More concretely, I am wondering if something along the lines of $$ \left\| \int_0^t e^{-(t-\tau)\Lambda} B u(\tau)\, d\tau \right\| \leq \|B \| \sqrt{\int_0^t \, \| e^{-(t-\tau) \Lambda} \|^2 \, d\tau} \sqrt{\int_0^t \|u(\tau) \|^2 \, d\tau} $$ is valid. That would not be immediately obvious to me for many reasons (most prominently that the left-hand-side above doesn't seem to resemble an inner product on some Hilbert space) but I also can't come up with a counterexample.

Now I would like to ask for help on two questions:

a) Could you sketch an argument why the above bound is valid or not (or point me to references that discusses that or something similar)?

b) If the above bound is not valid, can you think of a bound (or bounding strategy) that would be valid? If so could you share it and sketch an argument why it is valid.

Thanks!

$\endgroup$
3
  • $\begingroup$ What do you mean by the spectral norm? $\endgroup$
    – almosteverywhere
    Commented Aug 26, 2022 at 21:17
  • $\begingroup$ The induced 2-norm. $$\|A\| = \|A\|_2 = \sup_{\|v\|_2 = 1} \frac{\|Av\|_2}{\|v\|_2}$$ $\endgroup$
    – user1089574
    Commented Aug 26, 2022 at 23:08
  • $\begingroup$ Ah I see. Thanks $\endgroup$
    – almosteverywhere
    Commented Aug 27, 2022 at 11:30

1 Answer 1

Reset to default
1
$\begingroup$

Your bound is correct. You can estimate as follows: $$\left\vert\int_0^t e^{-(t-\tau)\Lambda}Bu(\tau)\ \mathrm{d}\tau\right\vert\leq\int_0^t\left\vert e^{-(t-\tau)\Lambda}Bu(\tau)\right\vert\mathrm{d}\tau\leq\int_0^t\left\Vert e^{-(t-\tau)\Lambda}\right\Vert \left\vert Bu(\tau)\right\vert \mathrm{d}\tau\leq\int_0^t \left\Vert e^{-(t-\tau)\Lambda}\right\Vert \left\Vert B\right\Vert\left\vert u(\tau)\right\vert \mathrm{d}\tau\leq\Vert B\Vert\sqrt{\int_0^t \left\Vert e^{-(t-\tau)\Lambda}\right\Vert^2\mathrm{d}\tau}\sqrt{\int_0^t \vert u(\tau)\vert^2\mathrm{d}\tau}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .