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Thread: Vector space of all bounded linear operators

  1. #1
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    Vector space of all bounded linear operators

    Hi,
    I can figure out how to solve this problem, although it does not seem very hard...

    Let H denote a Hilbert space, and L(H) the vector space of all bounded linear operators on H. Given T in L(H), define the operator norm :
    ||T|| = inf {B : ||Tv|| < B.||v||, for all v in H}.

    a/ Show that ||T+T'|| < ||T|| + ||T'|| whenever T and T' are in L(H).

    b/ Prove that d(T,T') = ||T-T'|| defines a metric on L(H).

    c/ Show that L(H) is complete in the metric d.

    Any ideas ?
    Thanks.
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  2. #2
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    Quote Originally Posted by matt33 View Post
    Hi,
    I can? can't? figure out how to solve this problem, although it does not seem very hard...

    Let H denote a Hilbert space, and L(H) the vector space of all bounded linear operators on H. Given T in L(H), define the operator norm :
    ||T|| = inf {B : ||Tv|| < B.||v||, for all v in H}.

    a/ Show that ||T+T'|| ||T|| + ||T'|| whenever T and T' are in L(H). (Should be a weak inequality.)

    b/ Prove that d(T,T') = ||T-T'|| defines a metric on L(H).

    c/ Show that L(H) is complete in the metric d.
    First show that, for T in L(H) and v in V, $\displaystyle \|Tv\|\leqslant\|T\|\|v\|$. Then a/ follows, because $\displaystyle \|(T+T')v\|\leqslant\|Tv\|+\|T'v\|\leqslant(\|T\|+ \|T'\|)\|v\|$.

    b/ follows from a/, because $\displaystyle d(T,T'') = \|T-T''\| = \|(T-T')+(T'-T'')\| \leqslant \|T-T'\|+\|T'-T''\|$ $\displaystyle = d(T,T')+d(T',T'')$. At any rate, that proves the triangle inequality. The other metric space axioms are straightforward.

    The easiest way to prove the completeness of L(H) is to show first that $\displaystyle \|T\| = \sup\{\|Tv\|:v\in H, \|v\|\leqslant1\}$.

    Given a Cauchy sequence $\displaystyle (T_n)$ in L(H), we must first find a candidate for a limit operator T, and then show that $\displaystyle T_n\to T$ in the operator norm (or more precisely, in the metric d associated with the norm).

    So suppose that we have such a Cauchy sequence. For every v in H, it's not hard to see that $\displaystyle (T_nv)$ is a Cauchy sequence in H. But H is complete, so this sequence converges to a vector that we call Tv. Check that the operator T defined in this way is linear.

    To complete the proof, you have to show that (1) T is bounded, and (2) $\displaystyle \|T_n-T\|\to0$.

    Let $\displaystyle \varepsilon>0$. For all sufficiently large n and m, $\displaystyle \|T_n-T_m\|<\varepsilon$. Therefore, for all v with $\displaystyle \|v\|\leqslant1$, $\displaystyle \|(T_n-T_m)v\|<\varepsilon$. In that inequality, you can let m→∞ to get $\displaystyle \|(T_n-T)v\|\leqslant\varepsilon$. Taking the sup over all v in the unit ball, you get $\displaystyle \|T_n-T\|\leqslant\varepsilon$. That simultaneously establishes (1) and (2).
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