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Let $ \mathcal{X} $ be a normed linear space and $ S,T: \mathcal{X} \to \mathcal{X} $ be linear operators such that $ S \circ T- T \circ S=1 $.

  1. Show that $ S \circ T^{n+1}- T^{n+1} \circ S=(n+1)T^n $ for $ n=0,1,2,... $

  2. Deduce that if $ S$ is bounded then $ T$ is unbounded.

For the first part I thought of applying the principle of mathematical induction.Is it alright to get the result like that or is there any other method to get that result?And for the second part, Since $ S \circ T- T \circ S=1 $ is the commutator operator,the result is obvious,but how can I give a proof of this?Please help!! Thanks!!

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+1 for typesetting your question and showing some thoughts! –  Martin Jan 7 '13 at 0:38
    
@Martin,Thanks! –  ccc Jan 7 '13 at 1:30
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This answer to the question Martin already linked to follows the same approach to answering a question essentially equivalent to this one. –  Jonas Meyer Jan 7 '13 at 21:41

1 Answer 1

up vote 5 down vote accepted
  1. Yes, induction is the way to go. The case $n=0$ is the hypothesis $ST - TS = 1$, and for the induction step assume $(n+1)T^{n} = ST^{n+1} - T^{n+1}S$ and use $ST = TS + 1$ to calculate $$\begin{align*}(n +1) T^{n+1} &= (n+1)T^{n}T = (ST^{n+1} - T^{n+1}S)T = ST^{n+2} - T^{n+1}(TS+1) \\&= ST^{n+2} - T^{n+2}S - T^{n+1}\end{align*}$$ which after rearranging is the identity you want to prove.

  2. I do not see how the second part is obvious "[s]ince $S \circ T - T\circ S = 1$ is the commutator", unless you are in positive finite dimension and apply the trace to reach the contradiction $0 = \dim\mathcal{X}$ or you know about the Wielandt-Wintner theorem, which is a consequence of the exercise you're solving.

    Anyway, exclude the case $\mathcal{X} = 0$ and look at the identity from part 1. If $T^{N} = 0$ for some $N$, let $n$ be the smallest $n$ such that $T^{n+1} = 0$. Then $ST^{n+1} - T^{n+1}S = (n+1)T^{n} = 0$ hence $T^{n} = 0$ contradicting the choice of $n$. So, suppose $T^{n} \neq 0$ for all $n$ and assume that $T$ is bounded. Take the norm on both sides. Leave the right hand side as it is and for the left hand side you get the estimate $$\lVert S T^{n+1} - T^{n+1}S\rVert \leq 2 \lVert S\rVert \lVert T^{n} \rVert \lVert T\rVert.$$ Then you can deduce that $2 \lVert S \rVert \lVert T \rVert \geq n+1$ by dividing by $\lVert T^n\Vert$ on both sides. This quickly leads to a contradiction.

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Thanks again!!I understand the method now! –  ccc Jan 7 '13 at 1:30
    
@ Martin,can you please go through this problem and help me to do that? –  ccc Jan 7 '13 at 2:35

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