Good definition for $\int_a^\infty X(t)dt$, where $X(t)\in M_n$

Question

Any comments on this? : I'm reading a book, and some exercise says: Provide a good definition for $\int_a^\infty X(t)dt$, where $X(t)\in M_n(\mathbb{R})$ and prove that if $$\int_a^\infty \|X(t)\|dt \ \ \ \ \mathrm{exists\ then\ }\ \int_a^\infty X(t)dt \ \ \mathrm{exists}$$

Idea:

1. We should have $X(t):[a,\infty)\to M_n$ continuous, in order to prove the result, also for the existence of the improper integral (otherwise we should have a lot of extra conditions for the integral to exists).
2. Does it matter which $\|\cdot\|$ to use? There are several $\|\cdot\|_{\sup}$, $\|\cdot\|_1$, $\|\cdot\|_{Frob}$.
3. $\int_a^\infty X(t)dt=\lim_{b\to\infty}\int_a^b X(t)dt$
4. Convergence of $\int_a^\infty X(t)dt$ is proved by showing there is $U\in M_n$ such that $$\left\| \int_a^b X(t)dt-U \right\|\to 0 $$

Answer 1

Using matrix units, you can write $$ X(t)=\sum_{k,j} x_{kj}(t)\,E_{kj}, $$ where $x_{kj}:[a,\infty)\to\mathbb R$ are scalar functions. An integral is a limit of sums of the form "value of the function times size of the region". In particular, you expect an integral to be linear. So you want $$ \int_a^\infty X(t)\,dt=\sum_{kj}\bigg(\int_a^\infty x_{kj}(t)\,dt\bigg)\,E_{kj}. $$ So you need each entry $x_{kj}$ to be integrable (you don't need continuity necessarily, but either Riemann or Lebesgue integrability).

The norm doesn't matter, because all norms in $M_n(\mathbb R)$ are equivalent.

Finally, if $\displaystyle\int_a^\infty\|X(t)\|\,dt$ exists, you can use the norm $$ \|X(t)\|=\sum_{kj}\|x_{kj}(t)\| $$ and again you are integrating coordinate wise.

Answer 2

On way to define the operator-valued integral is as follow:

Define $\sigma:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}$ by $\sigma(x,y)=\int_{a}^{\infty}\langle X(t)x,y\rangle dt.$ (Assuming that $t\mapsto\langle X(t)x,y\rangle$ is continuous if you are working with Riemann integral. However, this assumption can be weaken and only require that $t\mapsto\langle X(t)x,y\rangle$ is Lebesgue measurable if you are working with Lebesgue integral.) Observe that \begin{eqnarray*} \left|\langle X(t)x,y\rangle\right| & \leq & \left\Vert X(t)x\right\Vert \left\Vert y\right\Vert \\ & \leq & \left\Vert X(t)\right\Vert \left\Vert x\right\Vert \left\Vert y\right\Vert \end{eqnarray*} and by assumption, $\int_{a}^{\infty}\left\Vert X(t)\right\Vert dt<\infty$, so the improper integral $\int_{a}^{\infty}\langle X(t)x,y\rangle dt$ converges absolutely.

It can be proved that $\sigma$ is bilinear. From linear algebra, there exists uniquely a linear operator $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ such that $\sigma(x,y)=\langle Tx,y\rangle.$ This operator $T$ is denoted by $\int_{a}^{\infty}X(t)dt.$ This is how we define operator-valued integral.