# Vector space of all bounded linear operators

• Feb 9th 2009, 11:55 AM
matt33
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.
• Feb 10th 2009, 12:33 PM
Opalg
Quote:

Originally Posted by matt33
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, $\|Tv\|\leqslant\|T\|\|v\|$. Then a/ follows, because $\|(T+T')v\|\leqslant\|Tv\|+\|T'v\|\leqslant(\|T\|+ \|T'\|)\|v\|$.

b/ follows from a/, because $d(T,T'') = \|T-T''\| = \|(T-T')+(T'-T'')\| \leqslant \|T-T'\|+\|T'-T''\|$ $= d(T,T')+d(T',T'')$. 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 $\|T\| = \sup\{\|Tv\|:v\in H, \|v\|\leqslant1\}$.

Given a Cauchy sequence $(T_n)$ in L(H), we must first find a candidate for a limit operator T, and then show that $T_n\to T$ 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 $(T_nv)$ 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) $\|T_n-T\|\to0$.

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