Let $T \in \mathcal L (E)$. Then $\|T^3\|^2 \le \|T^2\|^3 \, \|T\|^2$

Question

Let $E$ be a real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$. Let $\| \cdot\|$ be the operator norm on $\mathcal L(E)$.

I'm trying to solve Brezis' exercise 6.23.2 where the author provides a hint. A special case of the hint can be stated as

Let $T \in \mathcal L (E)$. Then $\|T^3\|^2 \le \|T^2\|^3 \, \|T\|^2$.

Could you provide me some hints (not the full solution) on how to prove above statement?

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Update: Below is a screenshot of the hint.

Answer 1

I hope that the following hints are not too much for your liking.

1. It seems that the "$\frac nm$" within the hint is really supposed to be $q$, i.e. $\lfloor n/m\rfloor$. Plugging $n = 3, m = 2, r = 1$ into this version of the equation from the hint, when cast in terms of the original operator, becomes the following: $$ \log(\|T^3\|) \leq \lfloor3/2\rfloor \log(\|T^2\|) + \log(\|T\|) \iff\\ \log(\|T^3\|) \leq \log(\|T^2\|) + \log(\|T\|) \iff\\ \|T^3\| \leq \|T^2\| \cdot \|T\|, $$ which is straightforward to see.
2. The fact that $a_n \leq \lfloor n/m\rfloor a_m + a_r$ can be deduced as a direct consequence of the fact $a_{i+j} \leq a_i + a_j$