So, as a part of a problem, I've been asked to prove that if $H$ is a complex Hilbert space and $T\in L(H)$ is normal, then $\| T^2 \| =\| T \| ^2$ (Operator norm)

Context: This is part (b) in a three part problem obviously designed to build to a final result. The final result being this is good for any integer $n$, and the first part (already proved) was that $$T \text { normal } \iff \forall x\in H,\| T(x)\|=\|T^*(x)\|$$

So, here's my attempt at a proof of part b:
Since the operator norm is submultiplicative, we have $\|T^2\|\le \| T\| ^2$. Then from a theorem in Kreyszig's Introductory Functional Analysis with Applications (Thm 3.94(e), page 198) we have $$\|TT^*\|=\|T\|^2$$, and using submultiplicativity again, combined with them 3.92 (p 196) that states $\|T^*\|=\|T\|$, we have $$\|T\|^2\le \|T\|\cdot \|T^*\|=\|T\|^2$$ hence $\|T^2\|=\|T\|^2$ as desired.

Now, the problem is...none of theorems I have used require $T$ to be normal (Or the space to be complex). Nor did I use part a. So....either my proof is wrong, or this statement actually holds for all bounded linear operators on any Hilbert space...which I doubt, so I think I made a mistake somewhere. Any advice?

  • $\begingroup$ You didn't make any mistakes. The equalities $\|TT^*\| = \|T\|^2$ and $\|T\| = \|T^*\|$ hold for all bounded linear operators $T$ on a hilbert space. $\endgroup$ – Falko Mar 18 '15 at 23:48
  • $\begingroup$ @user49797 So the statement holds for any operator, not just normal ones, in any hilbert space (real or complex)? $\endgroup$ – Alan Mar 18 '15 at 23:50
  • $\begingroup$ Oh my bad, you did make a mistake. Your last inequality doesn't prove $\|T\|^2 \leq \|T^2\|$. $\endgroup$ – Falko Mar 18 '15 at 23:58
  • $\begingroup$ Woops. Yeah, sleep dep. Thanks. $\endgroup$ – Alan Mar 19 '15 at 0:05
  • $\begingroup$ Friend just walked in to the office and showed me the proof, I'm good. $\endgroup$ – Alan Mar 19 '15 at 0:07

So, just to formally complete this thing, the proper proof: Taking the square of the norms, we have

$$\|T^2x\|^2=\langle T^2x,T^2x\rangle=\langle T^*T^2x,Tx\rangle=\langle TT^*Tx,Tx\rangle=\langle T^*Tx,T^*Tx\rangle=\|T^*Tx\|^2$$ since this holds for all $x\in H$, we have $\|T^2\|=\|T^*T\|=\|T\|^2$ as desired

(And yeah, normalcy being used!)

For those playing along with our home game, to prove this extends to any arbitray positive power $n\in \mathbb N$, use forward/backward induction: First it proves for arbitrary powers of 2: $\forall k\in \mathbb N$ it holds for $2^k$ by induction on $k$. Then use backwards induction on $n$: if it holds for $n$, it holds for $n-1$. (this is only the second time I've seen forward/backwards induction used, the other time being in the proof of the arithmetic/geometric mean inequality....does anyone know of other uses?)


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