# Thread: L_2 space for matrix-valued functions

1. ## L_2 space for matrix-valued functions

hi,

here is the standard definition of L_2-space of matrix-valued functions: it's those measurable F that satisfy

(i) $\int_0^1 ||F(x)||^2dx <\infty$

(or equivalently, $\int_0^1 ||F(x)^*F(x)||dx <\infty$, or $\int_0^1 trace(F(x)^*F(x))dx <\infty$)

(here $F^*$ is the adjoint of $F$)

clearly this implies

(ii) $\left\|\int_0^1 F(x)^* F(x)dx\right\| <\infty$

what about the converse -- does (ii) necessarily implies (i), i.e. is (ii) enough for the function to be in L_2?

for the scalar case (i)=(ii), but I suspect (and afraid) that this fails for matrices...

...help...

2. okay, I think I got it, but I've been struggling with it for half of the day, so I might just be fancying -- please let me know if you see a flaw in the following proof.

I claim (ii) implies (i).

$
\left\|\int_0^1 F(x)^* F(x)dx\right\| <\infty
$
, which means in particular that for any vector
$\phi$,

$
\int_0^1 (F(x)^* F(x)\phi,\phi) dx =$
$\int_0^1 ||F(x)\phi||^2 dx <\infty
$

Choose
$\phi=e_{j}$. Then we obtain $\int_0^1 \sum_{k=1}^l |F_{kj}(x)|^2 dx <\infty$. Thus for any $k,j$, $\int_0^1 |F_{kj}(x)|^2 dx <\infty$, which implies $\int_0^1 trace(F(x)^*F(x))dx =\int_0^1 \sum_{j,k=1}^l |F_{kj}(x)|^2dx<\infty$, so $F$ is in $L_2$

3. That proof looks good to me.