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Math Help - L_2 space for matrix-valued functions

  1. #1
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    Unhappy 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...
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  2. #2
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    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).

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

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

    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
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  3. #3
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    That proof looks good to me.
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