Under which circumstances it is true that $\|(A+B)^n\|^{1/n}\le \|A^n\|^{1/n}+\|B^n\|^{1/n}$ for elements $A$ and $B$ in a Banach algebra and a natural number $n$?


The case $n=1$ holds always.

For $n\geq2$, i don't have any kind of general answer, but I expect such inequality to hold basically for norms that behave like the one-norm, and little else.

As an example, the inequality fails for any norm in the $2\times2$ matrices. Indeed, if $$ A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ B=\begin{bmatrix}0&0\\1&0\end{bmatrix}, $$ then $$ \|(A+B)^n\|>0, \ \mbox{ while } \ A^n=B^n=0. $$ And of course this idea works for any algebra where you have nilpotent elements with non-nilpotent sum.


Having duly noted Martin Argerami's answer, I thought it worth mentioning that for normal $A,B$ in a $C^*$-algebra, we have $$ \|(A+B)^m\|^{1/m} \leq \|A+B\| \leq \|A\|+\|B\|=\|A^m\|^{1/m}+\|B^m\|^{1/m}. $$ Observe also that for commuting normal operators, we have the following Cauchy-Schwarz type inequality $$ \|AB x\|^2=\langle A^*Ax,B^*Bx\rangle\leq \|A^*Ax\|\|B^*Bx\|=\|A^2x\|\|B^2x\|, $$ which implies $$ \|(A+B)^2x\|\leq \|A(A+B)x\|+\|B(A+B)x\|\leq (\|A^2x\|^{1/2}+\|B^2x\|^{1/2})\|(A+B)x\|^{1/2}, $$ or $$ \|(A+B)^2x\|^{1/2}\leq \|A^2x\|^{1/2}+\|B^2x\|^{1/2}. $$ I'm quite sure more can be done if one employs the functional calculus.

  • $\begingroup$ Did you mean for $A+B$ normal? I assume you want to apply the Gelfand representation. In any case, I was wondering if I may change Cauchy-Schwartz to Cauchy-Schwarz. $\endgroup$ – Rudy the Reindeer Dec 7 '14 at 9:52
  • $\begingroup$ @RudytheReindeer As the answer is stated, I don't think any additional hypothesis are necessary (if $A,B$ are normal and commuting, then Fuglede's Theorem implies that $A+B$ is normal). For the topmost inequality, normality is only invoked to ensure $\|A\|\leq \|A^m\|^{1/m}$. If you invoke the full Gelfand representation (as you may if $A,B$ are normal and commuting), you get the full range of Hölder-type inequalities, from which Minkowski-type inequalities may be deduced. $\endgroup$ – Jonas Dahlbæk Dec 8 '14 at 10:37
  • $\begingroup$ Thank you for your comment, I understand. I have one question though: How do you get $AB^\ast = B^\ast A$ and $A^\ast B = B A^\ast$ from $A,B$ normal and commuting? $\endgroup$ – Rudy the Reindeer Dec 9 '14 at 7:03
  • $\begingroup$ That is the content of Fuglede's Theorem. $\endgroup$ – Jonas Dahlbæk Dec 9 '14 at 7:17
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    $\begingroup$ Thanks. Now what about changing Cauchy-Schwartz to Cauchy-Schwarz? $\endgroup$ – Rudy the Reindeer Dec 9 '14 at 7:24

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